Question #258288

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) .


1
Expert's answer
2021-11-02T13:29:42-0400

πcot(πz)=1z+2zn=11z2n2\pi cot(\pi z)=\frac{1}{z}+2z\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{z^2-n^2}


zπcot(πz)=12n=1z2n2z2z\pi cot(\pi z)=1-2\displaystyle{\sum^{\infin}_{n=1}}\frac{z^2}{n^2-z^2}


expand z2n2z2\frac{z^2}{n^2-z^2} in Taylor series:


z2n2z2=i=1z2in2i\frac{z^2}{n^2-z^2}=\displaystyle{\sum^{\infin}_{i=1}}\frac{z^{2i}}{n^{2i}}


So:

zπcot(πz)=12i=1(n=11n2i)z2iz\pi cot(\pi z)=1-2\displaystyle{\sum^{\infin}_{i=1}}(\displaystyle{\sum^{\infin}_{n=1}}\frac{1}{n^{2i}})z^{2i}


(n=11n2)z2+(n=11n4)z4+(n=11n6)z6+...=n=1z2n2z2(\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}


n=11n2+(n=11n4)z2+(n=11n6)z4+...=n=11n2z2\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}


n=11n2=n=11n2z2(n=11n4)z2(n=11n6)z4...=\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}-...=


n=11n4=n=11z2(n2z2)(n=11z2n2)(n=11n6)z2...\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}-...


n=11n6=n=11z4(n2z2)(n=11z4n2)(n=11z2n4)...\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}})-...


1n2z2=12z(nz)12z(n+z)\frac{1}{n^2-z^2}=\frac{1}{2z(n-z)}-\frac{1}{2z(n+z)}


n=11n2=n=1(12z(nz)12z(n+z))(n=11n4)z2(n=11n6)z4...=\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}-...=


n=11n4=n=1(12z(nz)12z(n+z))(n=11z2n2)(n=11n6)z2...\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}-...


n=11n6=n=1(12z(nz)12z(n+z))(n=11z4n2)(n=11z2n4)...\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}})-...



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