How do you test the series #Sigma n/sqrt(n^3+1)# from n is #[0,oo)# for convergence?

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Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

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Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

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Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

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Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

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Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

NowTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \limTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now,To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, weTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compareTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare theTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the givenTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given seriesTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sumTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} )To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) withTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with (To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sumTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} =To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \limTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \toTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ).To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). ByTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the LimitTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit ComparisonTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison TestTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test,To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, ifTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdotTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if theTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limitTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit (To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \toTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \inTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

NowTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now,To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplifyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify andTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalizeTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ =To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \limTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \toTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdotTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} )To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) existsTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists andTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and isTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finiteTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and nonTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zeroTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, thenTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdotTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both seriesTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series convergeTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge orTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or divergeTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

NowTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now,To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, letTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let'sTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplifyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limitTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \limTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \toTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ =To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \inTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \toTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdotTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 +To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} =To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \limTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ =To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \toTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \limTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \inTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \toTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \limTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \toTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \limTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \toTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrtTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} =To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \limTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \toTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \inTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ =To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \limTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 +To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

SinceTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since theTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limitTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is aTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finiteTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite nonTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zeroTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number,To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1),To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), andTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \toTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and theTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \inTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series (To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sumTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} )To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is aTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a pTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series withTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \inftyTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p =To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

SinceTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since theTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \fracTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limitTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit isTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infiniteTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite andTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zeroTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), whichTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero,To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and theTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges,To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series (To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by theTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sumTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the LimitTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit ComparisonTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \fracTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison TestTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test,To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the givenTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series (To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sumTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}}To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} )To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) convergesTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, byTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\inTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by theTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\inftyTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the LimitTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit ComparisonTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison TestTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{nTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test,To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the seriesTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series (To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}}To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} \To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrtTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} )To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrt{nTo test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) alsoTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrt{n^To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) also converTo test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrt{n^3To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) also converges.To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrt{n^3 +To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) also converges.To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrt{n^3 + To test the series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) for convergence, we can use the Limit Comparison Test.

Let's compare it with a simpler series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ).

Taking the limit as ( n ) approaches infinity of the ratio of the terms of the two series:

[ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3+1}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} ]

Now, simplify and rationalize:

[ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{\sqrt{n^3+1}} \cdot \frac{\sqrt{n^3+1}}{\sqrt{n^3+1}} ] [ = \lim_{n \to \infty} \frac{n \cdot \sqrt{n^3+1}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6+n^2}}{n^3 + 1} ] [ = \lim_{n \to \infty} \frac{\sqrt{n^6}}{n^3 + 1} ]

[ = \lim_{n \to \infty} \frac{n^3}{n^3 + 1} ]

Since the limit is a finite non-zero number, namely (1), and the series ( \sum_{n=0}^{\infty} \frac{1}{\sqrt{n}} ) is a p-series with (p = \frac{1}{2}), which converges, by the Limit Comparison Test, the given series ( \sum_{n=0}^{\infty} \frac{n}{\sqrt{n^3+1}} ) also converges.To test the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) for convergence, we can use the Limit Comparison Test.

Let's consider the series ( \sum \frac{1}{\sqrt{n^3}} ), which is easier to evaluate. This series is a p-series with ( p = \frac{3}{2} ), and since ( \frac{3}{2} > 1 ), it converges.

Now, we compare the given series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) with ( \sum \frac{1}{\sqrt{n^3}} ). By the Limit Comparison Test, if the limit ( \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} ) exists and is finite and non-zero, then both series converge or diverge.

Now, let's simplify this limit: [ \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^3 + 1}}}{\frac{1}{\sqrt{n^3}}} = \lim_{n \to \infty} \frac{n \sqrt{n^3}}{\sqrt{n^3 + 1}} ] [ = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^4}{n^3} \cdot \frac{1}{1 + \frac{1}{n^3}}} ] [ = \lim_{n \to \infty} \sqrt{n} \cdot \sqrt{\frac{1}{1 + \frac{1}{n^3}}} = \infty ]

Since the limit is infinite and non-zero, and the series ( \sum \frac{1}{\sqrt{n^3}} ) converges, by the Limit Comparison Test, the series ( \sum \frac{n}{\sqrt{n^3 + 1}} ) also converges.

I come up with this by looking at dominant terms in the numerator and denominator of the nth term of the given series: