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Question Number 74887 by abdomathmax last updated on 03/Dec/19

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

Commented by mathmax by abdo last updated on 21/Dec/19

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

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