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 November 11th, 2010, 03:16 AM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Series computation with residue theory! [color=#000000]Compute with the help of complex analysis the series $\bf \fbox{\sum_{k=1}^{+\infty}\frac{1}{k^{2}}}$.[/color]
 November 11th, 2010, 01:22 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Series computation with residue theory! [color=#000000]A hint: Take the function $\bf f(z)=\frac{\pi \cot(\pi z)}{z^2}$ and then the contour integral over the circle with radius $\bf n+\frac{1}{2}$ with $\bf n\in \mathbb{N}$ and center $\bf(0,0)$.[/color]
 November 13th, 2010, 08:38 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Series computation with residue theory! [color=#000000]The function $\bf f$ has a simple pole in every single $\bf k \in \mathbb{Z}$ with $\bf k\in[-n,n]$ inside the circle mentioned above which we name $\bf C_{n+\frac{1}{2}}$. We will use the formula $\fbox{\bf \frac{1}{2\pi i } \oint_{C_{n+\frac{1}{2}}}f(z)\;dz=\sum_{k=-n}^{n}Res\left(f(z);z=k\right)}$. For the left side of the boxed equation we have, $\bf \sum_{k=-n}^{n}Res\left(f(z);z=k\right)=\sum_{k=-n}^{-1}Res\left(f(z);z=k\right)+Res(f(z);z=0)+\sum_{k=1 }^{n}Res\left(f(z);z=k\right)$ $\bf=\sum_{k=-n}^{-1}\frac{1}{k^2}-\frac{\pi^2}{3}+\sum_{k=1}^{n}\frac{1}{k^2}=2\sum_ {k=1}^{n}\frac{1}k^{2}-\frac{\pi^2}{3}$. Letting $\bf n\rightarrow +\infty$ we take the result, $\bf \lim_{n\rightarrow +\infty}\oint_{n+\frac{1}{2}}f(z)\;dz=2\sum_{k=1}^ {+\infty}\frac{1}{k^{2}}-\frac{\pi^2}{3}\Rightarrow 0=2\sum_{k=1}^{+\infty}\frac{1}{k^{2}}-\frac{\pi^2}{3}\Rightarrow \sum_{k=1}^{+\infty}\frac{1}{k^{2}}=\frac{\pi^2}{6 }$. The line integral goes to zero as n goes to infinity because $\bf f$is bounded , it can be proved easily.[/color]

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