# Proof for $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ without complexes? [duplicate]

This is what I needed. Practically, a link were also okay.

$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

## marked as duplicate by user91500, Jean-Sébastien, Did, M Turgeon, Claude LeiboviciMay 8 '14 at 12:37

I think you have to compute the Fourier series of either $\sin$ or $x$ on $(0,2\pi)$ extended to $\mathbb R$ periodically to get the left hand side and then use Parseval's theorem to prove equivalence to the right hand side.