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

May 6, 2017

The series:

${\sum}_{n = 0}^{\infty} \frac{1}{\sqrt{{n}^{3} + 4}}$

is convergent.

#### Explanation:

The series:

${\sum}_{n = 0}^{\infty} \frac{1}{\sqrt{{n}^{3} + 4}}$

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

$0 < \frac{1}{\sqrt{{n}^{3} + 4}} < \frac{1}{\sqrt{{n}^{3}}}$

We know however that the series:

${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{{n}^{3}}} = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{3}{2}\right)$

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