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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

Σ_(n=1) ^∞ (((−1)^(n+1) )/(2n^2 ))e^(πin)   =(1/2)((e^(iπ) /1^2 )−(e^(2iπ) /2^2 )+(e^(3iπ) /3^2 )−.....)  =(1/2)(−(1/1^2 )−(1/2^2 )−(1/3^2 )−....)  =−(π^2 /(12))

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

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