Question Number 69338 by Rasheed.Sindhi last updated on 22/Sep/19
Commented by Prithwish sen last updated on 22/Sep/19
$$\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{series}\:\mathrm{of} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+…… \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}=\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{7}}+…… \\ $$$$\mathrm{Now}\:\mathrm{put}\:\mathrm{x}=\mathrm{1} \\ $$
Answered by mind is power last updated on 22/Sep/19
$$\sum_{{n}=\mathrm{1}} ^{+\infty} \frac{{sin}\left({na}\right)}{{n}}={Im}\sum_{{n}\geqslant\mathrm{1}} \frac{{e}^{{ian}} }{{n}}={Im}\left(\sum_{{n}\geqslant\mathrm{1}} \frac{\left({e}^{{ia}} \right)^{{n}} }{{n}}\right)={Im}\left(−{ln}\left(\mathrm{1}−{e}^{{ia}} \right)\right) \\ $$$$={Im}\left(−{ln}\left({e}^{{i}\frac{{a}}{\mathrm{2}}} \left({e}^{−\frac{{ia}}{\mathrm{2}}} −{e}^{\frac{{ia}}{\mathrm{2}}} \right)\right)={Im}\left\{\left(−{ln}\left({e}^{{i}\frac{{a}}{\mathrm{2}}} \right)−{ln}\left(−\mathrm{2}{isin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\right\}\right.\right. \\ $$$$={im}\left(−{i}\frac{{a}}{\mathrm{2}}−{ln}\left(−\mathrm{2}{i}\right)−{ln}\left({sin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\right. \\ $$$$={im}\left(−{i}\frac{{a}}{\mathrm{2}}−{ln}\left(\mathrm{2}\right)+{i}\frac{\pi}{\mathrm{2}}−{ln}\left(\frac{{a}}{\mathrm{2}}\right)−{ln}\left({sin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\right. \\ $$$$=\frac{\pi−{a}}{\mathrm{2}} \\ $$$${for}\:{a}=\frac{\pi}{\mathrm{2}}\:{we}\:{get}\:=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$