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!