calculate-n-2-1-n-n-2-1-

Question Number 43259 by maxmathsup by imad last updated on 08/Sep/18

c a l c u l a t e \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - 1}

Commented by maxmathsup by imad last updated on 09/Sep/18

l e t S = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n^{2} - 1} \Rightarrow S = \frac{1}{2} \sum_{n = 2}^{\infty} {(- 1)}^{n} {\frac{1}{n - 1} - \frac{1}{n + 1}}

\Rightarrow 2 S = \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n - 1} - \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n + 1} b u t

\sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n - 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2)

(\sum_{n = 1}^{\infty} \frac{x^{n}}{n} = - l n (1 - x) w i t h - 1 < x < 1) a l s o

\sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} - {1 - \frac{1}{2}}

= l n (2) - \frac{1}{2} \Rightarrow 2 S = l n (2) - l n (2) + \frac{1}{2} \Rightarrow S = \frac{1}{4} .