Use Fourier series for computing $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$ I need to compute Fourier series for the following function: $f(x)=\frac{-\pi}{4}
$ for $-\pi \leq x <0$, and $\frac{\pi}{4} $ for $ 0 \leq x \leq \pi$, and then to use it and compute $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
I tried to use Parseval equality:
$$\widehat{f(n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}=\frac{1}{4in}-\frac{(-1)^n}{4in}, \sum_{-\infty}^{\infty}|\widehat{f(n)}|^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2.$$
$$\sum_{-\infty}^{\infty}|\frac{1}{4in}-\frac{(-1)^n}{4in}|=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)=\frac{\pi^2}{16}.$$
Does someone see how can I compute form that the requsted sum?
Thanks!

Or you might want to think that
$$\eqalign{
  & \omega  = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} +  \cdots  = \frac{{{\pi ^2}}}{6}  \cr 
  & \frac{\omega }{4} = \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} +  \cdots  = \frac{{{\pi ^2}}}{{24}}  \cr 
  & \omega  - \frac{\omega }{4} = 1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} +  \cdots  = \frac{{{\pi ^2}}}{6} - \frac{{{\pi ^2}}}{{24}} = \frac{{{\pi ^2}}}{8} \cr} $$
