Answer to Question #258288 in Complex Analysis for yuvasri

Question

1. Comparing coefficients in the Laurent developments of cot (pi*z) and of its expression as a sum of partial fractions, find the values of sum 1 to infinity(1/n^ 2) ,

sum 1 to infinity(1/n^ 4) , sum 1 to infinity (1/n^ 6) .

Answer 1

$\pi cot(\pi z)=\frac{1}{z}+2z\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2-n^2}$

$z\pi cot(\pi z)=1-2\displaystyle{\sum^{\infin}_{n=1}}\frac{z^2}{n^2-z^2}$

expand $\frac{z^2}{n^2-z^2}$ in Taylor series:

$\frac{z^2}{n^2-z^2}=\displaystyle{\sum^{\infin}_{i=1}}\frac{z^{2i}}{n^{2i}}$

So:

$z\pi cot(\pi z)=1-2\displaystyle{\sum^{\infin}_{i=1}}(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{2i}})z^{2i}$

$(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{2}})z^{2}+(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{4}})z^{4}+(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}})z^{6}+...=\displaystyle{\sum^{\infin}_{n=1}}\frac{z^2}{n^2-z^2}$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{2}}+(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{4}})z^{2}+(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}})z^{4}+...=\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^2-z^2}$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{2}}=\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^2-z^2}-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{4}})z^{2}-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}})z^{4}-...=$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{4}}=\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2(n^2-z^2)}-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2n^{2}})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}})z^{2}-...$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}}=\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^4(n^2-z^2)}-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^4n^{2}})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2n^{4}})-...$

$\frac{1}{n^2-z^2}=\frac{1}{2z(n-z)}-\frac{1}{2z(n+z)}$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{2}}=\displaystyle{\sum^{\infin}_{n=1}}(\frac{1}{2z(n-z)}-\frac{1}{2z(n+z)})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{4}})z^{2}-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}})z^{4}-...=$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{4}}=\displaystyle{\sum^{\infin}_{n=1}}(\frac{1}{2z(n-z)}-\frac{1}{2z(n+z)})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2n^{2}})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}})z^{2}-...$

$\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{6}}=\displaystyle{\sum^{\infin}_{n=1}}(\frac{1}{2z(n-z)}-\frac{1}{2z(n+z)})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^4n^{2}})-(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2n^{4}})-...$