# 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

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• You can use Fourier_transformation – Jlamprong May 8 '14 at 12:10
• @Jlamprong Fourier-transformation without complexes??? – peterh May 8 '14 at 12:11
• Yes. see here – Jlamprong May 8 '14 at 12:14
• @Jlamprong Is it me, or complex numbers are clearly mentioned on pages 3, 5, 6, 7 and 8 of the slides you link to? – Did May 8 '14 at 12:41

## 2 Answers

Evaluating ζ(2) by Robin Chapman contains several proofs (~14 altogether). You can have a look through and find a nice one.

• They are really good proofs. Thanks! – peterh May 8 '14 at 12:42

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.