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Question Number 31979 by abdo imad last updated on 17/Mar/18

$$calculate\sum _{n=2}^{\mathrm{\infty }}\frac{1}{{(n^2-1)}^2}.$$

Commented by abdo imad last updated on 22/Mar/18

$$letput{S}_n=\sum _{k=2}^n\frac{1}{{(k^2-1)}^2}$$$$S_n=\sum _{k=2}^n\frac{1}{{(k-1)}^2{(k+1)}^2}letdecompose$$$$F\left(x\right)=\frac{1}{{(x-1)}^2{(x+1)}^2}=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{c}{(x+1)}+\frac{d}{{(x+1)}^2}$$$$b={lim}_{x\to 1}{(x-1)}^{2}F\left(x\right)=\frac{1}{4}$$$$d={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=\frac{1}{4}wehaveF(-x)=F\left(x\right)\Rightarrow$$$$\frac{-a}{x+1}-\frac{c}{x-1}+\frac{b}{{(x+1)}^2}+\frac{d}{{(x-1)}^2}=F\left(x\right)\Rightarrow c=-a\Rightarrow F\left(0\right)$$$$F\left(x\right)=\frac{a}{x-1}-\frac{a}{x+1}+\frac{1}{4{(x-1)}^2}+\frac{1}{4{(x+1)}^2}$$$$F\left(0\right)=1=-2a+\frac{1}{2}\Rightarrow 2a=-\frac{1}{2}\Rightarrow a=-\frac{1}{4}and$$$$F\left(x\right)=\frac{-1}{4(x-1)}+\frac{1}{4(x+1)}+\frac{1}{4{(x-1)}^2}+\frac{1}{4{(x+1)}^2}$$$$4S_n=4\sum _{k=2}^{n}F\left(k\right)=-\sum _{k=2}^n\frac{1}{k-1}+\sum _{k=2}^n\frac{1}{k+1}+\sum _{k=2}^n\frac{1}{{(k-1)}^2}$$$$+\sum _{k=2}^n\frac{1}{{(k+1)}^2}$$$$=-\sum _{k=1}^{n-1}\frac{1}{k}+\sum _{k=3}^{n+1}\frac{1}{k}+\sum _{k=1}^{n-1}\frac{1}{k^2}+\sum _{k=3}^{n+1}\frac{1}{k^2}$$$$=-H_{n-1}+H_{n+1}-\frac{3}{2}+{\xi }_{n-1}\left(2\right)+{\xi }_{n+1}\left(2\right)-\frac{5}{4}$$$$=H_{n+1}-H_{n-1}-\frac{11}{4}+{\xi }_{n-1}\left(2\right)+{\xi }_{n+1}\left(2\right)but$$$$H_{n+1}-H_{n-1}{}_{n\to \mathrm{\infty }}\to 0{\xi }_{n-1}\left(2\right)\to \frac{{\pi }^2}{6}$$$${\xi }_{n+1}\left(2\right)\to \frac{{\pi }^2}{6}\Rightarrow 4S_n\to \frac{{\pi }^2}{3}-\frac{11}{4}\Rightarrow$$$${lim}_{n\to \mathrm{\infty }}{S}_n=\frac{{\pi }^2}{12}-\frac{11}{16}.letrememberthat{H}_n=\sum _{k=1}^n\frac{1}{k}$$$$and{\xi }_n\left(x\right)=\sum _{k=1}^n\frac{1}{k^x}.$$$$$$

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