find the value of Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)(n+2)(n+3)))


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Question Number 143254 by Mathspace last updated on 12/Jun/21

$f i n d t h e v a l u e o f \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} (n + 1) (n + 2) (n + 3)}$
	Answered by Olaf_Thorendsen last updated on 12/Jun/21

	$R (n) = \frac{1}{n^{2} (n + 1) (n + 2) (n + 3)}$ $R (n) = \frac{1}{6 n^{2}} - \frac{11}{36 n} + \frac{1}{2 (n + 1)} - \frac{1}{4 (n + 2)} + \frac{1}{18 (n + 3)}$ $Let = A_{n} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} = \ln 2$ $Let = B_{n} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2}} = - \frac{π^{2}}{12}$ $Let = S_{n} = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} R (n)$ $S_{n} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{6 n^{2}} - \underset{n = 1}{\sum^{\infty}} \frac{11 {(- 1)}^{n}}{36 n} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 (n + 1)}$ $- \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{4 (n + 2)} + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{18 (n + 3)}$ $S_{n} = \frac{1}{6} B_{n} - \frac{11}{36} [- 1 - A_{n}] + \frac{1}{2} A_{n}$ $- \frac{1}{4} [- \frac{1}{2} - A_{n}] + \frac{1}{18} [\frac{1}{2} - \frac{1}{3} + A_{n}]$ $S_{n} = \frac{11}{36} + \frac{1}{8} + \frac{1}{36} - \frac{1}{54} + \frac{1}{6} B_{n}$ $+ (\frac{11}{36} + \frac{1}{2} + \frac{1}{4} + \frac{1}{18}) A_{n}$ $S_{n} = \frac{95}{216} + \frac{10}{9} A_{n} + \frac{1}{6} B_{n}$ $S_{n} = \frac{95}{216} + \frac{10}{9} \ln 2 + \frac{1}{6} (- \frac{π^{2}}{12})$ $S_{n} = \frac{95}{216} + \frac{10}{9} \ln 2 - \frac{π^{2}}{72}$
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