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Question Number 65674 by mathmax by abdo last updated on 01/Aug/19

$$findthevalueof\sum _{n=2}^{\mathrm{\infty }}\frac{n}{{(n+1)}^2{(n-1)}^3}$$

Commented by mathmax by abdo last updated on 04/Aug/19

$$letS=\sum _{n=2}^{\mathrm{\infty }}\frac{n}{{(n+1)}^2{(n-1)}^3}\Rightarrow S=\sum _{n=1}^{\mathrm{\infty }}\frac{n+1}{{(n+2)}^2{n}^3}$$$$letdecomposeF\left(x\right)=\frac{x+1}{x^3{(x+2)}^2}\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3}+\frac{d}{x+2}+\frac{e}{{(x+2)}^2}$$$$c={lim}_{x\to 0}{x}^{3}F\left(x\right)=\frac{1}{4}$$$$e={lim}_{x\to -2}{(x+2)}^{2}F\left(x\right)=\frac{-1}{-8}=\frac{1}{8}\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{1}{4x^3}+\frac{d}{x+2}+\frac{1}{8{(x+2)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+d\Rightarrow d=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{1}{4x^3}-\frac{a}{x+2}+\frac{1}{8{(x+2)}^2}$$$$F\left(1\right)=\frac{2}{9}=a+b+\frac{1}{4}-\frac{a}{3}+\frac{1}{72}=\frac{2}{3}a+b+\frac{1}{72}\Rightarrow$$$$2=6a+9b+\frac{1}{8}\Rightarrow 6a+9b=2-\frac{1}{8}=\frac{15}{8}$$$$F(-1)=0=-a+b-\frac{1}{4}-a+\frac{1}{8}=-2a+b-\frac{1}{8}\Rightarrow$$$$-2a+b=\frac{1}{8}\Rightarrow b=2a+\frac{1}{8}\Rightarrow 6a+9(2a+\frac{1}{8})=\frac{15}{8}\Rightarrow$$$$24a+\frac{9}{8}=\frac{15}{8}\Rightarrow 24a=\frac{6}{8}=\frac{3}{4}\Rightarrow a=\frac{3}{4.8.3}=\frac{1}{32}$$$$b=\frac{1}{16}+\frac{1}{8}=\frac{3}{16}\Rightarrow F\left(x\right)=\frac{1}{32x}+\frac{3}{16x^2}+\frac{1}{4x^3}-\frac{1}{32(x+2)}+\frac{1}{8{(x+2)}^2}$$$$S_n=\sum _{k=1}^{n}F\left(k\right)=\frac{1}{32}\sum _{k=1}^n\frac{1}{k}+\frac{3}{16}\sum _{k=1}^n\frac{1}{k^2}+\frac{1}{4}\sum _{k=1}^n\frac{1}{k^3}$$$$-\frac{1}{32}\sum _{k=1}^n\frac{1}{k+2}+\frac{1}{8}\sum _{k=1}^n\frac{1}{{(x+2)}^2}$$$$\sum _{k=1}^n\frac{1}{k+2}=\sum _{k=3}^{n+2}\frac{1}{k}=\sum _{k=1}^n\frac{1}{k}-\frac{3}{2}+\frac{1}{n+1}+\frac{1}{n+2}\Rightarrow$$$$\frac{1}{32}\sum _{k=1}^n\frac{1}{k}-\frac{1}{32}\sum _{k=1}^n\frac{1}{k+2}=-\frac{1}{32}(-\frac{3}{2}+\frac{1}{n+1}+\frac{1}{n+2})\to \frac{3}{64}\Rightarrow$$$$\sum _{k=1}^n\frac{1}{{(x+2)}^2}=\sum _{k=3}^{n+2}\frac{1}{k^2}={\xi }_n\left(2\right)+\frac{1}{{(n+1)}^2}+\frac{1}{{(n+2)}^2}-1-\frac{1}{4}$$$$\to \xi \left(2\right)-\frac{5}{4}\Rightarrow S={lim}_{n\to +\mathrm{\infty }}{S}_n$$$$=\frac{3}{64}+\frac{3}{16}\xi \left(2\right)+\frac{1}{4}\xi \left(3\right)+\frac{1}{8}\{\xi \left(2\right)-\frac{5}{4}\}$$$$=\frac{3}{64}+\frac{5}{16}\frac{{\pi }^2}{6}+\frac{1}{4}\xi \left(3\right)-\frac{5}{32}=-\frac{7}{64}+\frac{5{\pi }^2}{96}+\frac{1}{4}\xi \left(3\right)$$

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