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Question Number 86958 by abdomathmax last updated on 01/Apr/20

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

Commented by mathmax by abdo last updated on 04/Apr/20

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

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