# Use divergent test to determine convergence of the following series?

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

Mar 25, 2018

Diverges.

#### Explanation:

The Divergence Test tells us that to determine the divergence of a series $\sum {a}_{n} ,$ take ${\lim}_{n \to \infty} {a}_{n}$.

If ${\lim}_{a \to \infty} {a}_{n} \ne 0 ,$ the series $\sum {a}_{n}$ diverges.

The converse is NOT true; if the limit is zero, the series may or may not converge.

Here, ${a}_{n} = \frac{n}{\sqrt{2 {n}^{2} + 1}}$. Take the limit:

${\lim}_{n \to \infty} \frac{n}{\sqrt{2 {n}^{2} + 1}} = {\lim}_{n \to \infty} \frac{\frac{n}{n}}{\sqrt{\frac{2 {n}^{2} + 1}{n} ^ 2}} = {\lim}_{n \to \infty} \frac{1}{\sqrt{2 + \frac{1}{n} ^ 2}} = \frac{1}{\sqrt{2}} \ne 0$

Thus, the series diverges.