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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.


marked as duplicate by Jonas Meyer, user649, Ross Millikan, Timothy Wagner, Aryabhata Dec 16 '10 at 5:32

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    $\begingroup$ Devlin is right, and 1.29128599706 is incorrect. Since many proofs of this are given at math.stackexchange.com/questions/8337/…, I think this should be closed as duplicate. $\endgroup$ – Jonas Meyer Dec 16 '10 at 5:22
  • $\begingroup$ There are plenty of ingenious proofs of this you should read the posts of math.stackexchange.com/questions/8337/… and check the links. $\endgroup$ – AD. Dec 16 '10 at 5:35
  • $\begingroup$ Could someone please fix the LaTeX of the question/title? $\endgroup$ – AD. Dec 16 '10 at 5:36
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    $\begingroup$ Oh, I see, your incorrect approximation is the approximate value of $$\sum_{n=1}^\infty \frac{1}{n^n}.$$ oeis.org/A073009 $\endgroup$ – Jonas Meyer Dec 16 '10 at 6:32

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:



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