Sum of $\frac{1}{n^{2}}$ for $n = 1 ,2 ,3, ...$? [duplicate]

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Possible Duplicate:
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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1 Answer

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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:

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