find-the-value-of-n-1-1-n-n-2-n-1-n-2-n-3-

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}