1
$\begingroup$

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.

$\endgroup$

marked as duplicate by Jonas Meyer, user649, 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.

  • 5
    $\begingroup$ 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
  • 5
    $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.