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Question Number 104891 by mathmax by abdo last updated on 24/Jul/20

$$1)\mathrm{decompose}\mathrm{the}\mathrm{fraction}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^3{(\mathrm{x}+1)}^3}$$$$2)\mathrm{find}\mathrm{the}\mathrm{sum}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^3{(\mathrm{n}+1)}^3}$$

Answered by mathmax by abdo last updated on 26/Jul/20

$$1)\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^3{(\mathrm{x}+1)}^3}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)={\left(\frac{1}{\mathrm{x}(3+1)}\right)}^3={(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1})}^3$$$$=\frac{1}{{\mathrm{x}}^3}-\frac{3}{{\mathrm{x}}^2(\mathrm{x}+1)}+\frac{3}{\mathrm{x}{(\mathrm{x}+1)}^2}-\frac{1}{{(\mathrm{x}+1)}^3}$$$$=\frac{1}{{\mathrm{x}}^3}-\frac{3}{\mathrm{x}}(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1})+\frac{3}{\mathrm{x}+1}(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1})-\frac{1}{{(\mathrm{x}+1)}^3}$$$$=\frac{1}{{\mathrm{x}}^3}-\frac{3}{{\mathrm{x}}^2}+\frac{3}{\mathrm{x}(\mathrm{x}+1)}+\frac{3}{\mathrm{x}(\mathrm{x}+1)}-\frac{3}{{(\mathrm{x}+1)}^2}-\frac{1}{{(\mathrm{x}+1)}^3}$$$$=\frac{1}{{\mathrm{x}}^3}-\frac{3}{{\mathrm{x}}^2}+6(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1})-\frac{3}{{(\mathrm{x}+1)}^2}-\frac{1}{{(\mathrm{x}+1)}^3}$$$$\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{6}{\mathrm{x}}-\frac{3}{{\mathrm{x}}^2}+\frac{1}{{\mathrm{x}}^3}-\frac{6}{\mathrm{x}+1}-\frac{3}{{(\mathrm{x}+1)}^2}-\frac{1}{{(\mathrm{x}+1)}^3}$$

Commented by mathmax by abdo last updated on 26/Jul/20

$$2)\mathrm{let}\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^3{(\mathrm{n}+1)}^3}\Rightarrow \mathrm{S}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3{(\mathrm{k}+1)}^3}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3{(\mathrm{k}+1)}^3}=6\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}-3\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}+\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3}-6\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+1}$$$$-3\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+1)}^2}-\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+1)}^3}\mathrm{but}\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\to -\ln 2(\mathrm{n}\to +\mathrm{\infty })$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}\to \sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^2}=\delta \left(2\right)=(2^{1-2}-1)\xi \left(2\right)=-\frac{1}{2}\times \frac{{\pi }^2}{6}=-\frac{{\pi }^2}{12}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3}\to \sum _{\mathrm{k}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{{\mathrm{k}}^3}=\delta \left(3\right)=(2^{1-3}-1)\xi \left(3\right)=-\frac{3}{4}\xi \left(3\right)$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+1}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}}-1=-\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}-1$$$$\to \ln \left(2\right)-1$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+1)}^2}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{{\mathrm{k}}^2}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{{\mathrm{k}}^2}-1\to -\delta \left(2\right)-1=\frac{{\pi }^2}{12}-1$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{{(\mathrm{k}+1)}^3}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{{\mathrm{k}}^3}=\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{{\mathrm{k}}^3}-1\to -\delta \left(3\right)-1$$$$=\frac{3}{4}\xi \left(3\right)-1\Rightarrow$$$$\mathrm{S}=-6\ln \left(2\right)-3(-\frac{{\pi }^2}{12})-\frac{3}{4}\xi \left(3\right)-6(\ln 2-1)-3(\frac{{\pi }^2}{12}-1)+\xi \left(3\right)+1$$$$=-12\ln \left(2\right)+\frac{{\pi }^2}{4}+\frac{1}{4}\xi \left(3\right)+6-\frac{{\pi }^2}{4}+4$$$$\mathrm{S}=-12\ln \left(2\right)+10+\frac{1}{4}\xi \left(3\right)$$

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