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

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

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

$$=\sum _{n=2p}(...)+\sum _{n=2p+1}(...)$$$$=-\frac{1}{4}\sum _{p=1}^{\propto }\frac{1}{p^2}+\sum _{p=0}^{\propto }\frac{1}{{(2p+1)}^2}$$$$=-\frac{1}{4}\frac{{\pi }^2}{6}+\frac{{\pi }^2}{8}$$$$=\frac{{\pi }^2}{8}-\frac{{\pi }^2}{24}=\frac{{\pi }^2}{12}.$$

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