Assume the formula $$\sum_{n=-\infty}^\infty\frac{1}{(n+u)^2}=\frac{\pi^2}{(\sin \pi u)^2},$$ where $u\notin\Bbb Z$. I have been trying to prove that $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}.$$Setting $u=1/2$, I was able to derive $$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8},$$but I haven't made any progress toward proving the value of $\zeta(2)$. Any suggestions?
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In sum of squares, divide it into sum of even terms and sum of odd terms. Then take the 1/4 out from the sum of even terms. – Mar 04 '13 at 07:56
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the assumption formula seems interesting ... however I can't find it on google? can you give me links? (+1) – S L Mar 04 '13 at 08:26
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@experimentX: It comes from Stein and Shakarchi's complex analysis textbook. I was able to prove the formula by integrating $$f(z)=\frac{\pi\cot(\pi z)}{(u+z)^2}$$ over the contour made up of a circle of radius $N+1/2$ with $N\geq|u|$. With the correct bounds, you can show the integral goes to $0$, then use the residue formula. The summation question came as additional exercise from my professor. – Clayton Mar 04 '13 at 12:37
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Splitting the sum into odd and even parts:
$$\sum_{n=1}^{\infty}\frac 1{n^2}=\sum_{n=1}^{\infty}\frac 1{(2n)^2}+\sum_{n=0}^{\infty}\frac 1{(2n+1)^2}$$
$$\sum_{n=1}^{\infty}\frac 1{n^2}=\frac 14\sum_{n=1}^{\infty}\frac 1{n^2}+\sum_{n=0}^{\infty}\frac 1{(2n+1)^2}$$
$$\frac 34 \sum_{n=1}^{\infty}\frac 1{n^2}=\sum_{n=0}^{\infty}\frac 1{(2n+1)^2}$$
and it's trivial from there.
Robert Mastragostino
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A more direct way of seeing this is to do the following. Break up the sum as $$\sum_{n=-\infty}^{-1}\frac{1}{(n+u)^2}+\sum_{n=1}^\infty\frac{1}{(n+u)^2}=\frac{\pi^2}{\sin^2(\pi u)}-\frac{1}{u^2}$$ and take $u\to 0$. This is easiest to do by Taylor expanding $$\frac{\pi^2}{\sin^2(\pi u)}=\frac{1}{u^2}+\frac{\pi^2}{3}+\frac{\pi^4 u^2}{15}+\cdots$$ so we see that the singularity cancels exactly and we have $$2\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{3}.$$
Jonathan
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