Convergence of alternating series

**hjortur** · November 28th 2009, 11:32 AM

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.

**Opalg** · November 28th 2009, 12:12 PM

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

**hjortur** · November 28th 2009, 12:18 PM

Thank you Opalg.

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

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