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

study the convergence of Σ_(k=0) ^∞   e^(−i((kπ)/x))     and find its sum

$${study}\:{the}\:{convergence}\:{of}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{e}^{−{i}\frac{{k}\pi}{{x}}} \:\:\:\:{and}\:{find}\:{its}\:{sum}\: \\ $$

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

let S =Σ_(k=0) ^∞   e^(−i((kπ)/x))       (with x≠0)  S = Σ_(k=0) ^∞  (e^(−i(π/x)) )^k   = (1/(1−e^(−((iπ)/x)) )) = (1/(1−cos((π/x))+isin((π/x))))  =((1−cos((π/x)) −isin((π/x)))/((1−cos((π/x)))^2  +sin^2 ((π/x)))) =((1−cos((π/x)) −isin((π/x)))/(2−2cos((π/x))))  =(1/2) −(i/2)    ((sin((π/x)))/(2sin^2 ((π/(2x))))) =(1/2) −(i/2) ((2sin((π/(2x)))cos((π/(2x))))/(2sin^2 ((π/(2x))))) =(1/2) −(i/2) cotan((π/(2x)))  finally  S =(1/2) −(i/2)cotan((π/(2x))) .

$${let}\:{S}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{e}^{−{i}\frac{{k}\pi}{{x}}} \:\:\:\:\:\:\left({with}\:{x}\neq\mathrm{0}\right) \\ $$$${S}\:=\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\left({e}^{−{i}\frac{\pi}{{x}}} \right)^{{k}} \:\:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\frac{{i}\pi}{{x}}} }\:=\:\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)+{isin}\left(\frac{\pi}{{x}}\right)} \\ $$$$=\frac{\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)\:−{isin}\left(\frac{\pi}{{x}}\right)}{\left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \left(\frac{\pi}{{x}}\right)}\:=\frac{\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)\:−{isin}\left(\frac{\pi}{{x}}\right)}{\mathrm{2}−\mathrm{2}{cos}\left(\frac{\pi}{{x}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}\:\:\:\:\frac{{sin}\left(\frac{\pi}{{x}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{x}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}\:\frac{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}{x}}\right){cos}\left(\frac{\pi}{\mathrm{2}{x}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{x}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}\:{cotan}\left(\frac{\pi}{\mathrm{2}{x}}\right) \\ $$$${finally}\:\:{S}\:=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}{cotan}\left(\frac{\pi}{\mathrm{2}{x}}\right)\:. \\ $$

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