# How do you use the limit comparison test on the series #sum_(n=1)^oo1/sqrt(n^3+1)# ?

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To use the limit comparison test on the series (\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}), we compare it with a known series whose convergence or divergence is known.

Let's denote the given series as (a_n = \frac{1}{\sqrt{n^3 + 1}}).

Choose a series (b_n) such that:

- (b_n > 0) for all (n),
- (\lim_{n \to \infty} \frac{a_n}{b_n}) exists and is positive and finite.

One common choice for (b_n) that satisfies these conditions is (b_n = \frac{1}{n^{3/2}}).

Now, we find the limit:

[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n^3 + 1}}}{\frac{1}{n^{3/2}}} ]

[ = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3 + 1}} ]

Now, we can rationalize the denominator:

[ = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3 + 1}} \cdot \frac{\sqrt{n^3 + 1}}{\sqrt{n^3 + 1}} ]

[ = \lim_{n \to \infty} \frac{n^{3/2}\sqrt{n^3 + 1}}{n^3 + 1} ]

Since (n^{3/2}) grows slower than (\sqrt{n^3 + 1}) as (n \to \infty), we can simplify the limit to:

[ = \lim_{n \to \infty} \frac{n^{3/2}\sqrt{n^3 + 1}}{n^3 + 1} = 0 ]

Since the limit is a finite positive number, the series (\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}) converges or diverges in the same way as the series (\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}).

The series (\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}) is a (p)-series with (p = \frac{3}{2}), and since (p > 1), it converges.

Therefore, by the limit comparison test, the original series (\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}) also converges.

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