prove that :

$\displaystyle \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}$

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- Nov 29th 2009, 10:43 PMdhiabprove that series
prove that :

$\displaystyle \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}$ - Nov 29th 2009, 11:40 PMsimplependulum
- Nov 30th 2009, 01:03 AMchisigma
- Nov 30th 2009, 01:13 AMsimplependulum
Oh , yes

If i say comparing with the series , it should be $\displaystyle x^3 $ , but not $\displaystyle x^2 $ .

Or say compare the coefficient of $\displaystyle x^2$ after dividing the factor $\displaystyle x $ - Nov 30th 2009, 01:54 AMtonio

Take a peek at this very nice paper: http://www.math.titech.ac.jp/~inoue/...lder/zeta2.pdf

You have there no less than 14 different proofs of the result $\displaystyle \zeta(2)=\sum\limits_{n+1}^{\infty}\frac{1}{n^2}=\ frac{\pi^2}{6}$, from rather basic ones to those that require more knowledge....enjoy!

Tonio - Nov 30th 2009, 01:56 AMKrizalid
- Nov 30th 2009, 01:59 AMJameson
I believe the OP mostly posts challenge problems but places them in the regular forums. This thread has been answered many times over and I am going to PM the OP about using the Challenge Problems forum to save everyone time.