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Question Number 57529 by maxmathsup by imad last updated on 06/Apr/19

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

Commented by maxmathsup by imad last updated on 09/Apr/19

$$letdecomposeF\left(x\right)=\frac{1}{(x+1)x^3}\Rightarrow F\left(x\right)=\frac{a}{x+1}+\frac{b}{x}+\frac{c}{x^2}+\frac{d}{x^3}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=-1$$$$d={lim}_{x\to 0}{x}^{3}F\left(x\right)=1\Rightarrow F\left(x\right)=\frac{-1}{x+1}+\frac{b}{x}+\frac{c}{x^2}+\frac{1}{x^3}$$$$F\left(1\right)=\frac{1}{2}=-\frac{1}{2}+b+c+1\Rightarrow b+c=0\Rightarrow c=-b$$$$F(-2)=\frac{1}{8}=1-\frac{b}{2}+\frac{c}{4}-\frac{1}{8}\Rightarrow 1=8-4b+2c-1\Rightarrow -6=-4b+2c\Rightarrow -3=-2b+c$$$$b=1andc=-1\Rightarrow F\left(x\right)=-\frac{1}{x+1}+\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(n+1)n^3}=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^3}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}=\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}-1=ln\left(2\right)-1$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}=-ln\left(2\right)$$$$let\delta \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^x}letdetermine\delta \left(x\right)intermsof\xi \left(x\right)(x>1)$$$$wehave\delta \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^x{n}^x}-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}and$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^x{n}^x}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^x}=\xi \left(x\right)-2^{-x}\xi \left(x\right)$$$$=(1-2^{-x})\xi \left(x\right)\Rightarrow \delta \left(x\right)=2^{-x}\xi \left(x\right)-(1-2^{-x})\xi \left(x\right)=(2^{-x}-1-2^{-x})\xi \left(x\right)$$$$=(2^{1-x}-1)\xi \left(x\right)\Rightarrow \delta \left(x\right)=(2^{1-x}-1)\xi \left(x\right)\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=\delta \left(2\right)=(2^{1-2}-1)\xi \left(2\right)=(\frac{1}{2}-1)\frac{{\pi }^2}{6}=-\frac{{\pi }^2}{12}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^3}=\delta \left(3\right)=(2^{1-3}-1)\xi \left(3\right)=(\frac{1}{4}-1)\xi \left(3\right)=-\frac{3}{4}\xi \left(3\right)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(n+1)n^3}=1-ln\left(2\right)-ln\left(2\right)-(-\frac{{\pi }^2}{12})-\frac{3}{4}\xi \left(3\right)$$$$=1-2ln\left(2\right)+\frac{{\pi }^2}{12}-\frac{3}{4}\xi \left(3\right)with\xi \left(3\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}.$$

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