calculate-n-0-1-n-n-1-2-

Question Number 42401 by abdo.msup.com last updated on 24/Aug/18

c a l c u l a t e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}}

Commented by maxmathsup by imad last updated on 25/Aug/18

l e t S = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} \Rightarrow S = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = \sum_{p = 1}^{\infty} \frac{- 1}{{(2 p)}^{2}} + \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}}

= - \frac{1}{4} \sum_{p = 1}^{\infty} \frac{1}{p^{2}} + \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} b u t \sum_{p = 1}^{\infty} \frac{1}{p^{2}} = \frac{π^{2}}{6}

\sum_{p = 1}^{\infty} \frac{1}{p^{2}} = \sum_{p = 1}^{\infty} \frac{1}{{(2 p)}^{2}} + \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} \Rightarrow \sum_{p = 0}^{\infty} \frac{1}{{(2 p + 1)}^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8} \Rightarrow

S = - \frac{π^{2}}{24} + \frac{π^{2}}{8} = \frac{3 π^{2} - π^{2}}{24} = \frac{π^{2}}{12}

S = \frac{π^{2}}{12} .