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Question Number 36820 by maxmathsup by imad last updated on 06/Jun/18

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

Commented by math khazana by abdo last updated on 24/Jun/18

$$let{S}_n=\sum _{k=2S}^n\frac{1}{{(n-1)}^2{(n+1)}^2}wehaveS={lim}_{n\to +\mathrm{\infty }}{S}_n$$$$letdecomposeF\left(x\right)=\frac{1}{{(x-1)}^2{(x+1)}^2}$$$$F\left(x\right)=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{c}{x+1}+\frac{d}{{(x+1)}^2}$$$$b={lim}_{x\to 1}{(x-1)}^{2}F\left(x\right)=\frac{1}{4}$$$$d={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=\frac{1}{4}\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}+\frac{1}{4{(x-1)}^2}+\frac{c}{x+1}+\frac{1}{4{(x+1)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x-1}-\frac{a}{x+1}+\frac{1}{4{(x-1)}^2}+\frac{1}{4{(x+1)}^2}$$$$F\left(0\right)=1=-2a+\frac{1}{2}\Rightarrow -2a=\frac{1}{2}\Rightarrow a=-\frac{1}{4}\Rightarrow$$$$F\left(x\right)=\frac{1}{4(x+1)}-\frac{1}{4(x-1)}+\frac{1}{4{(x-1)}^2}+\frac{1}{4{(x+1)}^2}$$$$4S_n=\sum _{k=2}^n\frac{1}{k+1}-\sum _{k=2}^n\frac{1}{k-1}+\sum _{k=2}^n\frac{1}{{(k-1)}^2}$$$$+\sum _{k=2}^n\frac{1}{{(k+1)}^2}but$$$$\sum _{k=2}^n\frac{1}{k+1}=\sum _{k=3}^{n+1}\frac{1}{k}=H_{n+1}-\frac{3}{2}$$$$\sum _{k=2}^n\frac{1}{k-1}=\sum _{k=1}^{n-1}\frac{1}{k}=H_{n-1}$$$$\sum _{k=2}^n\frac{1}{{(k-1)}^2}=\sum _{k=1}^{n-1}\frac{1}{k^2}={\xi }_{n-1}\left(2\right)$$$$\sum _{k=2}^n\frac{1}{{(k+1)}^2}=\sum _{k=3}^{n+1}\frac{1}{k^2}={\xi }_{n+1}\left(2\right)-\frac{5}{4}\Rightarrow$$$$4S_n=H_{n+1}-\frac{3}{2}-H_{n-1}+{\xi }_{n-1}\left(2\right)+{\xi }_{n+1}\left(2\right)-\frac{5}{4}$$$$4S_n=\gamma +ln(n+1)+o\left(\frac{1}{n}\right)-\gamma -ln(n-1)-o\left(\frac{1}{n}\right)$$$$+{\xi }_{n-1}\left(2\right)+{\xi }_{n+1}\left(2\right)-\frac{11}{4}\Rightarrow$$$$4S_n\to \frac{2{\pi }^2}{6}-\frac{11}{4}=\frac{{\pi }^2}{3}-\frac{11}{4}\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{{\pi }^2}{12}-\frac{11}{16}=S.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18

$$T_n=\frac{1}{{(n+1)}^2{(n-1)}^2}=\frac{1}{2}\frac{(n+1)-(n-1)}{{(n+1)}^2{(n-1)}^2}$$$$=\frac{1}{2}\frac{1}{(n+1){(n-1)}^2}-\frac{1}{2}\frac{1}{{(n+1)}^2(n-1)}$$$$=\frac{1}{4}\times \frac{(n+1)-(n-1)}{(n+1){(n-1)}^2}-\frac{1}{4}\times \frac{(n+1)-(n-1)}{{(n+1)}^2(n-1)}$$$$$$$$=\frac{1}{4}\times \frac{1}{{(n-1)}^2}-\frac{1}{4}\times \frac{1}{(n+1)(n-1)}-$$$$\frac{1}{4}\times \frac{1}{(n+1)(n-1)}+\frac{1}{4}\times \frac{1}{{(n+1)}^2}$$$$=\frac{1}{4}\times \frac{1}{{(n-1)}^2}-\frac{1}{2}\times \frac{1}{(n+1)(n-1)}+\frac{1}{4}\times \frac{1}{{(n+1)}^2}$$$$so$$$$S_n=S_1-S_2+S_3(n>1)$$$$S_1=\frac{1}{4}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....)=\frac{1}{4}\times \frac{\prod ^2}{6}$$$$S_2=\frac{1}{4}\left\{\frac{(n+1)-(n-1)}{(n+1)(n-1)}\right\}$$$$=\frac{1}{4}\{\frac{1}{n-1}-\frac{1}{n+1}\}$$$$=\{\frac{1}{4}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...)-\frac{1}{4}(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}...)\}$$$$=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$$$$S_3=\frac{1}{4}\times \left\{\frac{1}{{(n+1)}^2}\right\}$$$$=\frac{1}{4}\times (\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....)$$$$=\frac{1}{4}\times (\frac{\prod ^2}{6}-\frac{1}{1^2}-\frac{1}{2^2})=\frac{1}{4}(\frac{\prod ^2}{6}-\frac{5}{4})$$$$S_n=S_1-S_2+S_3$$$$=\frac{1}{4}\times \frac{\prod ^2}{6}-\frac{1}{4}\times \frac{3}{2}+\frac{1}{4}(\frac{\prod ^2}{6}-\frac{5}{4})$$$$=\frac{1}{4}\times \{\frac{\prod ^2}{6}-\frac{3}{2}+\frac{\prod ^2}{6}-\frac{5}{4}\}$$$$=\frac{1}{4}\times \{\frac{2\prod ^2}{6}-\frac{11}{4}\}$$

Commented by math khazana by abdo last updated on 24/Jun/18

$$correctanswersirTanmay...$$

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