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Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$.

I just got the "New and Revised" edition of "Mathematics: The New Golden Age", by Keith Devlin. On p. 64 it says the sum is $\pi^2/6$, but that's way off. $\pi^2/6 \approx 1.64493406685$ whereas the sum in question is $\approx 1.29128599706$. I'm expecting the sum to be something interesting, but I've forgotten how to do these things.

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NotSuper
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    Devlin is right, and 1.29128599706 is incorrect. Since many proofs of this are given at http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2, I think this should be closed as duplicate. – Jonas Meyer Dec 16 '10 at 05:22
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    There are plenty of ingenious proofs of this you should read the posts of http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-n-1-infty-frac1n2 and check the links. – AD - Stop Putin - Dec 16 '10 at 05:35
  • Could someone please fix the LaTeX of the question/title? – AD - Stop Putin - Dec 16 '10 at 05:36
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    Oh, I see, your incorrect approximation is the approximate value of $$\sum_{n=1}^\infty \frac{1}{n^n}.$$ http://oeis.org/A073009 – Jonas Meyer Dec 16 '10 at 06:32
  • I can not add a answer to question but I just know have it. The property that I used there was a Fourier sum for a function that just have value equal to abs(x) in [-1,1] and alternating it in R. It first find the above sum for odd number and then try to extend that to all number by just converting the above sum to sum of even and odd and then show that sum of even is 1/4 of above sum. So we have all thing to calculate some for even, odd and all n. – Mehdi Mowlavi Aug 06 '23 at 16:50

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The answer is indeed pretty interesting!

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

This can be proven using complex analysis or calculus, or probably in many hundreds of other ways. One example of how to prove this is given here:

http://www.math.uu.se/~bjorklund/euler.pdf

Eric Haengel
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