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Question Number 116555 by Bird last updated on 04/Oct/20

$$find\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2{(n+1)}^3{(n+2)}^4}$$

Answered by Olaf last updated on 05/Oct/20

$$\mathrm{R}\left(n\right)=\frac{1}{n^2{(n+1)}^3{(n+2)}^4}$$$$\mathrm{R}\left(n\right)=-\frac{5}{16}.\frac{1}{n}+\frac{1}{16}.\frac{1}{n^2}+\frac{5}{n+1}-\frac{2}{{(n+1)}^2}$$$$+\frac{1}{{(n+1)}^3}-\frac{75}{16}.\frac{1}{n+2}-\frac{39}{16}.\frac{1}{{(n+2)}^2}$$$$-\frac{1}{{(n+2)}^3}-\frac{1}{4}.\frac{1}{{(n+2)}^4}$$$$$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}=-\ln 2$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2}=-\frac{{\pi }^2}{12}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^3}=-\frac{3}{12}\xi \left(3\right)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^4}=-\frac{7{\pi }^4}{720}$$$$$$$$\mathrm{S}=-\frac{5}{16}(-\ln 2)+\frac{1}{16}(-\frac{{\pi }^2}{12})+5(-\ln 2+1)$$$$-2(-\frac{{\pi }^2}{12}+1)+(-\frac{3}{12}\xi \left(3\right)+1)-\frac{75}{16}(-\ln 2+1-\frac{1}{2})$$$$-\frac{39}{16}(-\frac{{\pi }^2}{12}+1-\frac{1}{4})-(-\frac{3}{12}\xi \left(3\right)+1-\frac{1}{8})$$$$-\frac{1}{4}(-\frac{7{\pi }^4}{720}+1-\frac{1}{16})$$$$\mathrm{S}=\ln 2(\frac{5}{16}-5+\frac{75}{16})$$$$-\frac{{\pi }^2}{12}(\frac{1}{16}-2-\frac{39}{16})$$$$-\frac{3}{12}\xi \left(3\right)(1-1)+\frac{7{\pi }^4}{2880}$$$$+5-2+1-\frac{75}{32}-\frac{117}{64}-\frac{7}{8}-\frac{15}{64}$$$$\mathrm{S}=\frac{35{\pi }^2}{96}+\frac{7{\pi }^4}{2880}-\frac{41}{32}$$$$$$$$\mathrm{Please}\mathrm{mister}\mathrm{verify}\mathrm{the}\mathrm{calculous}.$$

Commented by mathmax by abdo last updated on 05/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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