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

Jan 4, 2018

The series diverges.

#### Explanation:

I would limit compare to $\sum \frac{1}{\sqrt{n}}$.

I come up with this by looking at dominant terms in the numerator and denominator of the nth term of the given series:

numerator: dominant term is $n$
denominator: dominant term is $\sqrt{{n}^{3}} = {n}^{\frac{3}{2}}$

So the given series can be compared to $\sum \left(\frac{n}{n} ^ \left(\frac{3}{2}\right)\right) = \sum \left(\frac{1}{n} ^ \left(\frac{1}{2}\right)\right) = \sum \left(\frac{1}{\sqrt{n}}\right)$.

We know that $\sum \frac{1}{\sqrt{n}}$ is a divergent $p$-series with $p = \frac{1}{2}$ which is less than 1.

So we calculate ${\lim}_{n \to \infty} \left(\frac{\frac{n}{\sqrt{{n}^{3} + 1}}}{\frac{1}{\sqrt{n}}}\right)$

$= {\lim}_{n \to \infty} \left(\frac{n}{\sqrt{{n}^{3} + 1}} \cdot \sqrt{n}\right) = {\lim}_{n \to \infty} \left({n}^{\frac{3}{2}} / \sqrt{{n}^{3} + 1}\right)$

$= {\lim}_{n \to \infty} \left(\sqrt{{n}^{3} / \left({n}^{3} + 1\right)}\right) = 1$

Since the limit is positive and finite and $\sum \frac{1}{\sqrt{n}}$ diverges, so does $\sum \left(\frac{n}{\sqrt{{n}^{3} + 1}}\right)$ by the limit comparison test.