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

Question

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

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

@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 08 '14 at 12:41
I think my answer, while not the shortest, seems to require the fewest prerequisites. Definitely, no complex numbers. — robjohn, May 29 '23 at 21:08

score 3 · Accepted Answer · edited May 23 '15 at 23:30

3

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

edited May 23 '15 at 23:30

VividD

15,966

answered May 08 '14 at 12:11

Paulistic

276

2

They are really good proofs. Thanks! – peterh May 08 '14 at 12:42

score 3 · Answer 2 · answered May 08 '14 at 12:16

3

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.

answered May 08 '14 at 12:16

dcs24

533

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

2 Answers2