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

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.

## marked as duplicate by Jonas Meyer, user649, Ross Millikan, Timothy Wagner, AryabhataDec 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

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