How to find this infinite summation: $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$? 
It is given that $f(x)=|x|$ in $-\pi\le x\le \pi$ then the value of 
  $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$
  is equal to 
a) $\Large \frac{\pi^2}{2}\qquad$ b) $\Large\frac{\pi^2}{4}\qquad$ c) $\Large\frac{\pi^2}{6}\qquad$ d) $\Large\frac{\pi^2}{8}$

I expanded the series $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$
$$=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\ldots$$ Now, I don't have any clue how to proceed.
 A: If we compute the Fourier cosine series of $f(x)$ over $(-\pi,\pi)$ through
$$\forall n\geq 1,\qquad \int_{-\pi}^{\pi}f(x)\cos(nx) = 2\int_{0}^{\pi}x\cos(nx)\,dx = 2\,\frac{-1+\cos(\pi n)}{n^2}\tag{1}$$
we get:
$$ |x|=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos((2n-1)x)}{(2n-1)^2} \tag{2} $$
and by evaluating both sides at $x=0$ (pointwise convergence - as a by-product of uniform convergence - is ensured by the fact that the Fourier coefficients give a summable sequence)
$$ \sum_{n\geq 1}\frac{1}{(2n-1)^2}=\color{red}{\frac{\pi^2}{8}}\tag{3} $$
follows. On the other hand, it is trivial that
$$ \sum_{n\geq 1}\frac{1}{(2n-1)^2}\leq 1+\int_{1}^{+\infty}\frac{dx}{(2x-1)^2}=\frac{3}{2} \tag{4}$$
hence $(d)$ is the only reasonable option.
A: Hint: $$ \sum_{n=1}^\infty \frac1{n^2} = \sum_{n=1}^\infty \frac1{(2n)^2}+\sum_{n=1}^\infty \frac1{(2n-1)^2}.$$
A: Use the Fourier Series of the function $f$ in the interval $[-\pi, \pi]$
And substitute $x = 0$.
The series converges at this point to the value of $f$.
This gives you the desired result.
