How to convert an alternating sum into a positive sum?

[Eclair_de_XII]

Homework Statement

"Given that $FS f(x)=\frac{L^2}{3}+(\frac{2L}{\pi})^2⋅\sum_{n=1}^\infty \frac{(-1)^n}{n^2}cos(\frac{n\pi}{L}x)$, prove that $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}=\frac{\pi^2}{12}$ and $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$.

Homework Equations

$FS f(x) = f(x) = x^2$

The Attempt at a Solution

Setting $x=0$, we have here $f(0)=0$ and $FS f(x)=\frac{L^2}{3}+(\frac{2L}{\pi})^2⋅\sum_{n=1}^\infty \frac{(-1)^n}{n^2}=0$. Now we have $-\frac{L^2}{3}=\frac{4L^2}{\pi^2}⋅\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$, so $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$, or $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$.

I basically got the first half. I took Calculus II over a year ago, and cannot exactly remember how to do this. Either that, or I would prove that the value of $\frac{1}{4} \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{24}$ and add two of that to the first part of the problem. Can anyone point me in the right direction as to what I should do?

 

[vela]

Choose values of $x$ and $L$ so that the cosine is equal to $(-1)^n$.

 

[Eclair_de_XII]

Oh, so set $x=L$. Thanks!