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Question Number 95217 by mathmax by abdo last updated on 24/May/20

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

Answered by mathmax by abdo last updated on 25/May/20

$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2{(\mathrm{x}+2)}^3}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\sum _{\mathrm{i}=1}^2\frac{{\mathrm{a}}_{\mathrm{i}}}{{\mathrm{x}}^{\mathrm{i}}}+\sum _{\mathrm{i}=1}^3\frac{{\mathrm{b}}_{\mathrm{i}}}{{(\mathrm{x}+2)}^{\mathrm{i}}}$$$${\mathrm{a}}_{\mathrm{i}}?\mathrm{we}\mathrm{find}{\mathrm{D}}_1\left(0\right)\mathrm{for}\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+2)}^{-3}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-3{(\mathrm{x}+2)}^{-4}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{o}\right)+{\mathrm{xf}}^{{}^{\prime }}\left(0\right)+\frac{{\mathrm{x}}^2}{2!}\xi \left(\mathrm{x}\right)=2^{-3}-3.2^{-4}\mathrm{x}+\frac{{\mathrm{x}}^2}{2}\xi \left(\mathrm{x}\right)\Rightarrow$$$$\frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}=\frac{2^{-3}}{{\mathrm{x}}^2}-\frac{3.2^{-4}}{\mathrm{x}}+\frac{\mathrm{x}}{2}\xi \left(\mathrm{x}\right)\Rightarrow {\mathrm{a}}_1=-\frac{3}{2^4}\mathrm{and}{\mathrm{a}}_2=\frac{1}{2^3}$$$${\mathrm{b}}_{\mathrm{i}}\mathrm{we}\mathrm{find}{\mathrm{D}}_2(-2)\mathrm{for}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{-2}\Rightarrow {\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=-{2\mathrm{x}}^{-3}\Rightarrow {\mathrm{g}}^{\left(2\right)}\left(\mathrm{x}\right)={6\mathrm{x}}^{-4}$$$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{g}(-2)+\frac{\mathrm{x}+2}{!}{\mathrm{g}}^{{}^{\prime }}(-2)+\frac{{(\mathrm{x}+2)}^2}{2!}{\mathrm{g}}^{\left(2\right)}(-2)+\frac{{(\mathrm{x}+2)}^3}{3!}\xi \left(\mathrm{x}\right)$$$$={(-2)}^{-2}+(\mathrm{x}+2)(-2){(-2)}^{-3}+\frac{{(\mathrm{x}+2)}^2}{2}\left(6\right){(-2)}^{-4}+\frac{{(\mathrm{x}+2)}^3}{6}\xi \left(\mathrm{x}\right)$$$$\frac{\mathrm{g}\left(\mathrm{x}\right)}{{(\mathrm{x}+2)}^3}=\frac{{(-2)}^{-2}}{{(\mathrm{x}+2)}^3}+\frac{{(-2)}^{-2}}{{(\mathrm{x}+2)}^2}+\frac{3{(-2)}^{-4}}{\mathrm{x}+2}+\frac{1}{6}\xi \left(\mathrm{x}\right)\Rightarrow$$$${\mathrm{b}}_1=\frac{3}{16},{\mathrm{b}}_2=\frac{1}{4},{\mathrm{b}}_3=\frac{1}{4}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=-\frac{3}{16\mathrm{x}}+\frac{1}{{8\mathrm{x}}^2}+\frac{3}{16(\mathrm{x}+2)}+\frac{1}{4{(\mathrm{x}+2)}^2}+\frac{1}{4{(\mathrm{x}+2)}^3}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2{(\mathrm{k}+1)}^3}=-\frac{3}{16}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}+\frac{1}{8}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}+\frac{3}{16}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+2}$$$$+\frac{1}{4}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+2)}^2}+\frac{1}{4}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+2)}^3}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\to -\ln \left(2\right)(\mathrm{n}\to +\mathrm{\infty })$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}\to \delta \left(2\right)=(2^{1-2}-1)\xi \left(2\right)=-\frac{1}{2}\times \frac{{\pi }^2}{6}=-\frac{{\pi }^2}{12}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\to \sum _{\mathrm{k}=3}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}$$$$=-\ln 2-(-1+\frac{1}{2})=-\ln 2+\frac{1}{2}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+2)}^2}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}\to \sum _{\mathrm{k}=3}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}=-\frac{{\pi }^2}{12}-(-1+\frac{1}{4})$$$$=\frac{3}{4}-\frac{{\pi }^2}{12}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+2)}^3}=\sum _{\mathrm{k}=3}^{\mathrm{n}+2}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3}\to \sum _{\mathrm{k}=3}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3}-(-1+\frac{1}{8})$$$$=(2^{1-3}-1)\xi \left(3\right)+\frac{7}{8}=(\frac{1}{4}-1)\xi \left(3\right)+\frac{7}{8}=\frac{7}{8}-\frac{3}{4}\xi \left(3\right)$$$$\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{sum}\mathrm{is}\mathrm{known}...$$

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