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Question Number 115732 by gab last updated on 28/Sep/20

Commented by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{2\mathrm{n}}^2}{\mathrm{e}}^{\pi \mathrm{in}}$$$$=\frac{1}{2}(\frac{{\mathrm{e}}^{\mathrm{i}\pi }}{1^2}-\frac{{\mathrm{e}}^{2\mathrm{i}\pi }}{2^2}+\frac{{\mathrm{e}}^{3\mathrm{i}\pi }}{3^2}-.....)$$$$=\frac{1}{2}(-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-....)$$$$=-\frac{{\pi }^2}{12}$$

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