# Thread: Convergence of alternating series

1. ## Convergence of alternating series

Let $a_n\in\mathbb{R}$ and $\lim_{n\to\infty}a_n=0$.
Prove or provide a counter example that $\sum_{n=1}^{\infty}(-1)^{n+1}a_n^2$ converges.

I was wondering wether this is a valid counter example:

Let $a_n=\begin{cases}\frac{1}{\sqrt[4]{n}}& n \text{ odd }\\\frac{1}{\sqrt{n}}& n\text{ even }\end{cases}$.

Then

$a_n^2=\begin{cases}\frac{1}{\sqrt{n}}& n \text{ odd }\\ \frac{1}{n}& n\text{ even }\end{cases}$

And $\sum_{n=1}^{\infty}(-1)^{n+1}a_n^2=\sum_{n=1}^{\infty}\left(\frac{1}{\s qrt{2n+1}}-\frac{1}{2n}\right)$.

Now using the integral test we find that this series converges if and only if

$\int_1^{\infty}\left(\frac{1}{\sqrt{2x+1}}-\frac{1}{2x}\right)dx$ converges.

Now:

$\int_1^{\infty}\left(\frac{1}{\sqrt{2x+1}}-\frac{1}{2x}\right)dx=\left[\sqrt{2x+1}-\frac{1}{2}ln(2x)\right]_1^{\infty}=\left[\sqrt{2x+1}-ln(\sqrt{2x})\right]_1^{\infty}=\infty$

So that the series diverges.

Is this correct, and if so is there a more straight forward way to do this?

Thank you.

2. Good example. You could have made it slightly simpler by taking $a_n=\begin{cases}0& n \text{ odd,}\\ \frac{1}{\sqrt n}& n\text{ even.}\end{cases}$

3. Thank you Opalg.

Now I am surprised I didn't figure out the example you gave.