<s>[INST] 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:
$$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$
 [/INST] A possible short proof comes from Parseval's identity. 
It is well known that the Fourier sine series of the $\text{sign}$ function over $(-\pi,\pi)$ is given by:
$$ \text{sign}(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} \tag{1}$$
hence by Parseval's identity:
$$ 2\pi = \int_{-\pi}^{\pi} 1\,dx = \frac{16}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\tag{2}$$
and the claim readily follows.
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