How do you test the series #Sigma 1/sqrt(n^3+4)# from n is #[0,oo)# for convergence?

To test the convergence of the series (\sum_{n=0}^{\infty} \frac{1}{\sqrt{n^3 + 4}}) from (n = 0) to infinity, we can use the Limit Comparison Test or the Comparison Test.

1. Limit Comparison Test: Compare the given series to a known series whose convergence behavior is known.

   Let's consider the series (\sum_{n=0}^{\infty} \frac{1}{n^{3/2}}). This is a p-series with (p = \frac{3}{2}), which converges because (p > 1).

   Now, take the limit as (n) approaches infinity of the ratio of the terms of the given series to the terms of the comparison series: [\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n^3 + 4}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3 + 4}}]

   If this limit is a finite positive number, then both series either converge or diverge. If the limit is zero, the series converges; if the limit is infinity or does not exist, the series diverges.

2. Comparison Test: If the Limit Comparison Test seems inconclusive or difficult to apply, we can use the Comparison Test directly.

   Here, observe that for all (n \geq 0): [\frac{1}{\sqrt{n^3 + 4}} < \frac{1}{\sqrt{n^3}} = \frac{1}{n^{3/2}}]

   The series (\sum_{n=0}^{\infty} \frac{1}{n^{3/2}}) converges (as mentioned before), so by the Comparison Test, the given series (\sum_{n=0}^{\infty} \frac{1}{\sqrt{n^3 + 4}}) also converges.

Therefore, the series (\sum_{n=0}^{\infty} \frac{1}{\sqrt{n^3 + 4}}) from (n = 0) to infinity converges.

The series:

has positive terms. Now consider that if we lower the denominator the quotient increases, so:

We know however that the series:

is convergent based on the p-series test, so also our series is convergent based on direct comparison.