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Is there a known geometric proof for this famous problem? $$\zeta(2)=\sum_{n=1}^\infty n^{-2}=\frac16\pi^2$$

Moreover we can consider possibilities of geometric proofs of the following identity for positive even inputs of the Zeta function: $$ \zeta(2n)=(-1)^{n+1} \frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$ and negative inputs: $$ \zeta(-n)=-\frac{B_{n+1}}{n+1}$$

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There are many nice proofs at… . Is one of them geometric enough for you? (David Speyer's answer might fit the bill.) –  Qiaochu Yuan Aug 27 '12 at 1:04
Here's a proof by Prof. Greene at UCLA: –  Isaac Solomon Aug 27 '12 at 2:20
thank you Qiaochu one of them is ;) –  progressiveforest Aug 27 '12 at 2:44
I wanted to ask this question some time ago. I'm glad to see it posted. (I thought of an elementary geometrical proof) –  Chris's sis the artist Aug 27 '12 at 5:25
do you mean that you have an elementary geometric proof? Would you mind sharing it? –  progressiveforest Aug 27 '12 at 5:42

3 Answers 3

up vote 1 down vote accepted

Yes, there is a geometric proof for $\zeta(2)=\sum_{n=1}^\infty n^{-2}=\frac16\pi^2$, and, on top of that, it is a very unusual and, in my opinion, beautiful one.

In his gorgeous paper "How to compute $\sum \frac{1}{n^2}$ by solving triangles", Mikael Passare offers this idea for proving $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$:

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Proof of equality of square and curved areas is based on another picture:

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Recapitulation of Passare's proof using formulas is as follows: (for each step there is a geometric justification)

$$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx}}{n}\; dx\; = -\int_0^\infty log(1-e^{-x})\; dx\; = \frac{\pi^2}{6}$$

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I'm not sure what you mean by a geometric proof, but the following should fit the bill, as here the identity is deduced from a comparison of two areas: the first area is

$\displaystyle\int_{[0,1]^2} \frac{1}{1 - xy} \frac{dx dy}{\sqrt{xy}}$

and the second is

$4 \displaystyle \int_{\substack{\xi, \eta>0 \\ \xi \eta \leq 1}} \frac{ d \xi \, d \eta}{(1+\xi^2)(1+\eta^2)}$;

They are equal by a change of variables. For the first quantity, expand $(1-xy)^{-1}$ as a geometric series and integrate term-wise to get $3 \cdot \zeta(2)$. The second can be computed to $\pi^2/2$. This can be found in detail in Kontsevich and Zagier's Periods (bottom of page 8), where they attribute the idea to Calabi. (You should easily be able to hunt down an identity of $\zeta (n)$ being equal to an integral over the unit square in $\mathbb{R}^n$, like the first above. It might be a fun exercise to see if this idea is can be adapted for even $n$.)

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Euler’s Solution of the Basel Problem – The Longer Story

You could look at that paper.

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the link is dead... –  Grigory M Aug 9 at 8:04
I think I found the paper at another location. i'll fix the link. –  VividD Sep 12 at 16:31
I don't think it's dead... –  ciceksiz kakarot Sep 15 at 18:51

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