# prove that series

• Nov 29th 2009, 10:43 PM
dhiab
prove that series
prove that :
$\displaystyle \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}$
• Nov 29th 2009, 11:40 PM
simplependulum
Quote:

Originally Posted by dhiab
prove that :
$\displaystyle \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}$

Consider $\displaystyle \sin(x) = x \prod_{n=1}^{\infty} \left ( 1 - \frac{x^2}{(n\pi)^2} \right )$

Then compare the coefficient of $\displaystyle x^2$ with its Taylor's series .
• Nov 30th 2009, 01:03 AM
chisigma
Quote:

Originally Posted by simplependulum
Consider $\displaystyle \sin(x) = x \prod_{n=1}^{\infty} \left ( 1 - \frac{x^2}{(n\pi)^2} \right )$

Then compare the coefficient of $\displaystyle x^2$ with its Taylor's series .

... $\displaystyle x^2$ or $\displaystyle x^3$? (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 30th 2009, 01:13 AM
simplependulum
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 AM
tonio
Quote:

Originally Posted by dhiab
prove that :
$\displaystyle \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}$

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 AM
Krizalid
• Nov 30th 2009, 01:59 AM
Jameson
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.