# Determine whether the series is conditionally convergent, absolutely divergent, or divergent.

## Question:

Determine whether the series is conditionally convergent, absolutely convergent, or divergent.

{eq}\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}\arctan (n)}{n^{2}} {/eq}

## Answer and Explanation:

We want to study the convergence of

{eq}\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}\arctan (n)}{n^{2}} \, . {/eq}

We'll start by checking whether it is absolutely convergent. This means determining if the series

{eq}\displaystyle \sum_{n=1}^{\infty}\left|\frac{(-1)^{n}\arctan (n)}{n^{2}} \right| =\sum_{n=1}^{\infty}\frac{\arctan (n)}{n^{2}} {/eq}

converges.

For any {eq}n {/eq}, we have {eq}\arctan(n) < \frac{\pi}{2} {/eq} (since in fact this is true for all real {eq}x {/eq}). So

{eq}\displaystyle \frac{\arctan (n)}{n^{2}} < \frac{\pi}{2}\frac{1}{n^2} {/eq}.

But the series

{eq}\displaystyle \sum_{n=1}^\infty \frac{\pi}{2}\frac{1}{n^2}=\frac{\pi}{2}\sum_{n=1}^\infty \frac{1}{n^2} {/eq}

converges by the {eq}p {/eq}-test. So {eq}\sum_{n=1}^{\infty}\frac{\arctan (n)}{n^{2}} {/eq} converges by the comparison test.

Since {eq}\sum_{n=1}^{\infty}\frac{\arctan (n)}{n^{2}} {/eq} converges, the original series

{eq}\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n \arctan (n)}{n^{2}} {/eq}

converges absolutely.