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

$$calculate\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^2}$$

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

$$letS=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(n+1)}^2}\Rightarrow S=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=\sum _{p=1}^{\mathrm{\infty }}\frac{-1}{{\left(2p\right)}^2}+\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}$$$$=-\frac{1}{4}\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^2}+\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}but\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^2}=\frac{{\pi }^2}{6}$$$$\sum _{p=1}^{\mathrm{\infty }}\frac{1}{p^2}=\sum _{p=1}^{\mathrm{\infty }}\frac{1}{{\left(2p\right)}^2}+\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}\Rightarrow \sum _{p=0}^{\mathrm{\infty }}\frac{1}{{(2p+1)}^2}=\frac{3}{4}\frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}\Rightarrow$$$$S=-\frac{{\pi }^2}{24}+\frac{{\pi }^2}{8}=\frac{3{\pi }^2-{\pi }^2}{24}=\frac{{\pi }^2}{12}$$$$S=\frac{{\pi }^2}{12}.$$

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