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Question Number 70595 by mathmax by abdo last updated on 06/Oct/19

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

Commented by mathmax by abdo last updated on 08/Oct/19

$$letS=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2{(n+1)}^3}letdecomposeF\left(x\right)=\frac{1}{x^2{(x+1)}^3}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x+1}+\frac{d}{{(x+1)}^2}+\frac{e}{{(x+1)}^3}$$$$b=1ande=1\Rightarrow F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}+\frac{c}{x+1}+\frac{\mathrm{d}}{{(\mathrm{x}+1)}^2}+\frac{1}{{(\mathrm{x}+1)}^3}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}-\frac{a}{x+1}+\frac{d}{{(x+1)}^2}+\frac{1}{{(x+1)}^3}$$$$F\left(1\right)=\frac{1}{8}=a+1-\frac{a}{2}+\frac{d}{4}+\frac{1}{8}\Rightarrow \frac{a}{2}+\frac{d}{4}=-1\Rightarrow 2a+d=-4$$$$F(-2)=-\frac{1}{4}=-\frac{a}{2}+\frac{1}{4}+a+d-1\Rightarrow$$$$-1=-2a+1+4a+4d-4\Rightarrow -1=2a+4d-3\Rightarrow 2a+4d=2\Rightarrow$$$$a+2d=1\Rightarrow a=1-2d\Rightarrow 2(1-2d)+d=-4\Rightarrow 2-3d=-4\Rightarrow$$$$-3d=-6\Rightarrow d=2\Rightarrow a=1-4=-3\Rightarrow$$$$F\left(x\right)=-\frac{3}{x}+\frac{1}{x^2}+\frac{3}{x+1}+\frac{2}{{(x+1)}^2}+\frac{1}{{(x+1)}^3}\Rightarrow$$$$S=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n}F\left(n\right)=-3\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}$$$$+3\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^3}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}=-ln\left(2\right)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=(2^{1-2}-1)\xi \left(2\right)=-\frac{1}{2}\frac{{\pi }^2}{6}=-\frac{{\pi }^2}{12}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}-1=ln\left(2\right)-1$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^3}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^3}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^3}-1$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^3}-1=-(2^{1-3}-1)\xi \left(3\right)-1$$$$=-(\frac{1}{4}-1)\xi \left(3\right)-1=\frac{3}{4}\xi \left(3\right)-1\Rightarrow$$$$S=3ln\left(2\right)-\frac{{\pi }^2}{12}+3ln\left(2\right)-3+\frac{3}{4}\xi \left(3\right)-1$$$$S=6ln\left(2\right)+\frac{3}{4}\xi \left(3\right)-4-\frac{{\pi }^2}{12}.$$

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