# How do I determine if the alternating series sum_(n=1)^oo(-1)^n/sqrt(3n+1) is convergent?

Oct 18, 2014

Alternating Series Test

An alternating series ${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {b}_{n}$, ${b}_{n} \ge 0$ converges if both of the following conditions hold.

$\left\{\begin{matrix}{b}_{n} \ge {b}_{n + 1} \text{ for all } n \ge N \\ {\lim}_{n \to \infty} {b}_{n} = 0\end{matrix}\right.$

Let us look at the posted alternating series.

In this series, ${b}_{n} = \frac{1}{\sqrt{3 n + 1}}$.

${b}_{n} = \frac{1}{\sqrt{3 n + 1}} \ge \frac{1}{\sqrt{3 \left(n + 1\right) + 1}} = {b}_{n + 1}$ for all $n \ge 1$.

and

${\lim}_{n \to \infty} {b}_{n} = {\lim}_{n \to \infty} \frac{1}{\sqrt{3 n + 1}} = \frac{1}{\infty} = 0$

Hence, we conclude that the series converges by Alternating Series Test.

I hope that this was helpful.