how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously : 
$$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$
Now how can we prove that:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$

This is a well known double integral proof by Beukers, Kolk, and Calabi.  First consider the double integral:
$$\int_{0}^{1}\int_{0}^{1} \frac{1}{1-x^2y^2} dydx.$$
Since $0<x,y<1$, rewrite the integrand as a geometric series:
$$\frac{1}{1-x^2y^2}=\sum_{n=0}^{\infty}(xy)^{2n}.$$
Now notice:
$$\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(xy)^{2n}dydx$$
is the same as:
$$\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}(xy)^{2n}dydx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}.$$
Make the change of variables:
$$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(u)}.$$
The Jacobian Determinant is:
$$\det\frac{\partial (x,y)}{\partial(u,v)}=\begin {vmatrix} \frac{\cos(u)}{\cos(v)}&&\frac{\sin(u)\sin(v)}{(\cos(v))^2} \\ \frac{\sin(v)\sin(u)}{(\cos(u))^2}&&\frac{\cos(v)}{\cos(u)}\\\end{vmatrix}=1-x^2y^2,$$
which cancels with the integrand, and the region of integration is the open iscosceles triangle formed by the inequalities: $$0<u+v<\frac{\pi}{2},0<u,v<\frac{\pi}{2}.$$
Using either geometry or evaluating the double integral:
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}-v}1dudv,$$
the area of the isosceles triangle is $\frac{\pi^2}{8}$. So the result is that:
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$
