So I have been solving various problems on Fourier series and this particular one got me struggling a bit.
Given a function $f(x) = |x|$ find a Fourier series on $[-\pi,\pi]$ and find the sum of following series: $\sum_{0}^\infty{\frac{1}{(2n-1)^2}}$.
So I successfully found a Fourier series of given function which is: $f(x) = \frac{\pi}{2} - \sum_{0}^{\infty}{\frac{1}{(2n+1)^2}}\pi\cos{nx}$. I have no problem of finding sum of $\frac{1}{(2n+1)^2}$ but how could I find the sum of $\sum_{0}^\infty{\frac{1}{(2n-1)^2}}$ ? I googled before posting this and all I could find is Parseval identity but that's not something I am familiar with. Can someone help me finding the sum or at least explain how could I do it with/without Parseval identity?
$$f(x)=|x| \sim \frac{\pi}{2} - \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}\cos (2n-1) x.$$
Evaluating this at $x=0$:
$$\frac{\pi}{2} = \sum_{n=1}^\infty \frac{4}{\pi (2n-1)^2}$$
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$
$$\sum_{n=0}^\infty \frac{1}{(2n-1)^2}=1+\frac{\pi^2}{8}$$