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

Question Number 27388 by macanudo last updated on 06/Jan/18

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n^{2}}

Commented by abdo imad last updated on 06/Jan/18

= \sum_{n = 2 p} (\dots) + \sum_{n = 2 p + 1} (\dots)

= - \frac{1}{4} \sum_{p = 1}^{\propto} \frac{1}{p^{2}} + \sum_{p = 0}^{\propto} \frac{1}{{(2 p + 1)}^{2}}

= - \frac{1}{4} \frac{π^{2}}{6} + \frac{π^{2}}{8}

= \frac{π^{2}}{8} - \frac{π^{2}}{24} = \frac{π^{2}}{12} .