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Question Number 41051 by turbo msup by abdo last updated on 01/Aug/18

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

Commented by math khazana by abdo last updated on 01/Aug/18

$$letS=\sum _{n=1}^{\mathrm{\infty }}\frac{2n+3}{n^2{(n+1)}^2}and{S}_n=\sum _{k=1}^n\frac{2k+3}{k^2{(k+1)}^2}$$$$wehave{lim}_{n\to +\mathrm{\infty }}{S}_n=Sletdecompose$$$$F\left(x\right)=\frac{2x+3}{x^2{(x+1)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x+1}+\frac{d}{{(x+1)}^2}$$$$b={lim}_{x\to 0}{x}^{2}F\left(x\right)=3$$$$d={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=1\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{3}{x^2}+\frac{c}{x+1}+\frac{1}{{(x+1)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{3}{x^2}-\frac{a}{x+1}+\frac{1}{{(x+1)}^2}$$$$F\left(1\right)=\frac{5}{4}=a+3-\frac{a}{2}+\frac{1}{4}=\frac{a}{2}+\frac{13}{4}$$$$\Rightarrow 5=2a+13\Rightarrow 2a=-8\Rightarrow a=-4\Rightarrow$$$$F\left(x\right)=-\frac{4}{x}+\frac{3}{x^2}+\frac{4}{x+1}+\frac{1}{{(x+1)}^2}\Rightarrow$$$$S_n=\sum _{k=1}^{n}F\left(k\right)=-4\sum _{k=1}^n\frac{1}{k}+3\sum _{k=1}^n\frac{1}{k^2}$$$$+4\sum _{k=1}^n\frac{1}{k+1}+\sum _{k=1}^n\frac{1}{{(k+1)}^2}$$$$=-4H_n+3{\xi }_n\left(2\right)+4(H_{n+1}-1)+{\xi }_{n+1}\left(2\right)-1$$$$=4(H_{n+1}-H_n)+3{\xi }_n\left(2\right)+{\xi }_{n+1}\left(2\right)-5$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=4\times 0+3\frac{{\pi }^2}{6}+\frac{{\pi }^2}{6}-5$$$$=\frac{2{\pi }^2}{3}-5\Rightarrow S=\frac{2{\pi }^2}{3}-5.$$

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