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Question Number 42187 by maxmathsup by imad last updated on 19/Aug/18

$$studytheconvergenceof\sum _{k=0}^{\mathrm{\infty }}{e}^{-i\frac{k\pi }{x}}andfinditssum$$

Commented by maxmathsup by imad last updated on 22/Aug/18

$$letS=\sum _{k=0}^{\mathrm{\infty }}{e}^{-i\frac{k\pi }{x}}(withx\ne 0)$$$$S=\sum _{k=0}^{\mathrm{\infty }}{\left(e^{-i\frac{\pi }{x}}\right)}^k=\frac{1}{1-e^{-\frac{i\pi }{x}}}=\frac{1}{1-cos\left(\frac{\pi }{x}\right)+isin\left(\frac{\pi }{x}\right)}$$$$=\frac{1-cos\left(\frac{\pi }{x}\right)-isin\left(\frac{\pi }{x}\right)}{{(1-cos\left(\frac{\pi }{x}\right))}^2+{sin}^2\left(\frac{\pi }{x}\right)}=\frac{1-cos\left(\frac{\pi }{x}\right)-isin\left(\frac{\pi }{x}\right)}{2-2cos\left(\frac{\pi }{x}\right)}$$$$=\frac{1}{2}-\frac{i}{2}\frac{sin\left(\frac{\pi }{x}\right)}{2{sin}^2\left(\frac{\pi }{2x}\right)}=\frac{1}{2}-\frac{i}{2}\frac{2sin\left(\frac{\pi }{2x}\right)cos\left(\frac{\pi }{2x}\right)}{2{sin}^2\left(\frac{\pi }{2x}\right)}=\frac{1}{2}-\frac{i}{2}cotan\left(\frac{\pi }{2x}\right)$$$$finallyS=\frac{1}{2}-\frac{i}{2}cotan\left(\frac{\pi }{2x}\right).$$

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