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In literature, there are many proofs of the well-known result $$\zeta(2) = \frac{\pi^2}{6}.$$

However, as far as I know, they do not offer an intuitive explanation of why this result should be true.

So my question is:

What is the key intuition -- that is, the picture -- behind the result? Are there any visual of anyway intuitive proofs of the statement?

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    $\begingroup$ Euler's first proof is very intuitive to me: find a "polynomial" whose roots are the non-zero integers (so your "polynomial" is sin/x) then use the relation between the sum of the inverse of the square of the roots and the first coefficients. Voilà. $\endgroup$
    – Olivier
    Oct 21, 2014 at 23:18
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    $\begingroup$ @Olivier I see what you mean, but honestly, from my limited experience, that is not what I call intuitive. $\endgroup$
    – user60665
    Oct 21, 2014 at 23:57
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    $\begingroup$ Not everything in maths can be justified by a reason that feels "intuitive", and "intuition" is highly subjective $\endgroup$
    – Yemon Choi
    Oct 22, 2014 at 0:08
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    $\begingroup$ @Dal, you don't need intuition, you need Euler. $\endgroup$
    – Włodzimierz Holsztyński
    Oct 22, 2014 at 0:24
  • $\begingroup$ John von Neumann said something like this: young man, you don't understand (things) in mathematics, you get used to (them). $\endgroup$
    – Włodzimierz Holsztyński
    Oct 22, 2014 at 1:45

2 Answers 2

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The word intuitive by definition means that something "feels correct", and is very myopic. The fact that there are so many different proofs is in itself a gift because you can pick your favorite and intuit all you want.

I like Euler's proof because it shows how it's related to the taylor series of $\sin(x)$ and so the mystery of $\pi$ and 6 becomes less mysterious.

On the other hand someone with lots of experience in Fourier analysis will see it as the Fourier transform of something and again the appearance of $\pi$, particular $\pi^2$ becomes apparent from Parseval's identity.

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Here is a short note by Robert E. Greene (UCLA):

How Geometry implies $\sum \frac{1}{k^2} = \pi^2/6$

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    $\begingroup$ I feel that this proof only uses geometry in name, kind of like the "topological" proof of Euclids theorem of infinitely many primes. It's a neat proof nevertheless. $\endgroup$
    – Alex R.
    Oct 22, 2014 at 2:25