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Question Number 115736 by gab last updated on 28/Sep/20

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

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{2\mathrm{n}}^2\left(2\mathrm{i}\right)}.({\mathrm{e}}^{\mathrm{i}\pi \mathrm{n}}-{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{n}})(\sin \left(\pi \mathrm{n}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{n}}-{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{n}}}{2\mathrm{i}})$$$$\frac{1}{4\mathrm{i}}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{\mathrm{n}}^2}{\mathrm{e}}^{\mathrm{i}\pi \mathrm{n}}-\frac{1}{4\mathrm{i}}\sum ^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}+1}}{{\mathrm{n}}^2}.{\mathrm{e}}^{-\mathrm{i}\pi \mathrm{n}}$$$$\frac{1}{4\mathrm{i}}(-\frac{1}{1^2}-\frac{1}{2^2}-...)-\frac{1}{4\mathrm{i}}(-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-...)$$$$=0$$$$$$$$\mathrm{Or}$$$$\mathrm{Another}\mathrm{method}$$$$\sin \left(\mathrm{k}\pi \right)=0(\mathrm{k}\in \mathbb{Z})$$$$\mathrm{So}\mathrm{every}\mathrm{term}\mathrm{is}0$$$$\mathrm{Sum}\mathrm{is}\mathrm{zero}$$

Commented by gab last updated on 28/Sep/20

$$$$$$thankyousir.$$

Answered by mathmax by abdo last updated on 28/Sep/20

$$\sin \left(\mathrm{n}\pi \right)=0\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}+1}\sin \left(\mathrm{n}\pi \right)}{{2\mathrm{n}}^2}=0$$

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