# What is the sum of the series $\sum\limits _{k=1}^\infty \frac{1}{k^2}$? [duplicate]

Possible Duplicate:
Different methods to compute $\sum_{n=1}^\infty \frac{1}{n^2}$.

What is $\lim \limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k^2}$ as an exact value?

## marked as duplicate by Asaf Karagila♦, Aryabhata, Hans Lundmark, J. M. is a poor mathematician, Mike SpiveyApr 20 '11 at 16:56

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## 1 Answer

This is a very well-known series. Astoundingly enough, the exact value is $\dfrac{\pi^2}{6}$.

There are proofs here: http://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

A reasonable video presentation here: http://mathnotations.blogspot.com/.../pi-squared-over-6-algebraic-genius-of.html

And a duplication of Euler's original proof here: http://www.cs.nthu.edu.tw/~wkhon/random/tutorial/pi-squared-over-six.pdf

I prefer the last one, as it's closely related to Viete's infinite product for $\pi$.