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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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marked as duplicate by Jonas Meyer, Chandru1, Ross Millikan, Timothy Wagner, Aryabhata Dec 16 '10 at 5:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

1 Answer 1

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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