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Question Number 140636 by Mathspace last updated on 10/May/21

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

Answered by Dwaipayan Shikari last updated on 10/May/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2(n+1)(2n+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2(2n+1)}-\frac{{(-1)}^n}{n(n+1)(2n+1)}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}(\frac{1}{n}-\frac{2}{2n+1})-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n(2n+1)}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(n+1)(2n+1)}$$$$=-\frac{{\pi }^2}{12}-3\mathrm{\Sigma }\frac{{(-1)}^n}{n(2n+1)}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2{(-1)}^n}{2n+1}-\frac{{(-1)}^n}{n+1}$$$$=-\frac{{\pi }^2}{12}-3\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}-\frac{{(-1)}^{n}2}{2n+1}-2(\frac{1}{3}-\frac{1}{5}+..)+(\frac{1}{2}-\frac{1}{3}+...)$$$$=-\frac{{\pi }^2}{12}+3log\left(2\right)+8(\frac{\pi }{4}-1)+(1-log\left(2\right))$$$$=\frac{-{\pi }^2}{12}+2log\left(2\right)+2\pi -7$$$$$$

Answered by mathmax by abdo last updated on 11/May/21

$$\mathrm{let}\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2(\mathrm{n}+1){(2\mathrm{n}+1)}^2}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2(\mathrm{x}+1)(2\mathrm{x}+1)}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{{\mathrm{x}}^2}+\frac{\mathrm{c}}{\mathrm{x}+1}+\frac{\mathrm{d}}{2\mathrm{x}+1}$$$$\mathrm{b}=1,\mathrm{c}=-1,\mathrm{d}=\frac{1}{\left(\frac{1}{4}\right)\left(\frac{1}{2}\right)}=8\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}+1}+\frac{8}{2\mathrm{x}+1}$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\mathrm{a}-1+4\Rightarrow \mathrm{a}=-3\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=-\frac{3}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}+1}+\frac{8}{2\mathrm{x}+1}\Rightarrow$$$$\mathrm{S}=-3\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}+8\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}=-\log 2$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}=(2^{1-2}-1)\xi \left(2\right)=-\frac{1}{2}.\frac{{\pi }^2}{6}=-\frac{{\pi }^2}{12}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}=-\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}$$$$=-(-\log 2+1)=-1+\log 2$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}=\frac{\pi }{4}-1\Rightarrow$$$$\mathrm{S}=-3(-\log 2)-\frac{{\pi }^2}{12}+1-\log 2+2\pi -8$$$$=2\log 2-\frac{{\pi }^2}{12}+2\pi -7$$

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