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:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$

Hint: Prove that $\bigg(1+\dfrac13-\dfrac15-\dfrac17+\dfrac19+\dfrac1{11}\pm\cdots\bigg)^2=1+\dfrac1{3^2}+\dfrac1{5^2}+\dfrac1{7^2}+\dfrac1{9^2}+\cdots$
