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Question Number 69338 by Rasheed.Sindhi last updated on 22/Sep/19

Commented by Prithwish sen last updated on 22/Sep/19

it is the series of 1−(1/3)+(1/5)−(1/7)+...... tan^(−1) x=x−(x^3 /3)+(x^5 /5)−(x^7 /7)+...... Now put x=1

$$\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{series}\:\mathrm{of} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+...... \\ $$$$\mathrm{tan}^{−\mathrm{1}} \mathrm{x}=\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{7}}+...... \\ $$$$\mathrm{Now}\:\mathrm{put}\:\mathrm{x}=\mathrm{1} \\ $$

Answered by mind is power last updated on 22/Sep/19

Σ_(n=1) ^(+∞) ((sin(na))/n)=ImΣ_(n≥1) (e^(ian) /n)=Im(Σ_(n≥1) (((e^(ia) )^n )/n))=Im(−ln(1−e^(ia) )) =Im(−ln(e^(i(a/2)) (e^(−((ia)/2)) −e^((ia)/2) ))=Im{(−ln(e^(i(a/2)) )−ln(−2isin((a/2)))} =im(−i(a/2)−ln(−2i)−ln(sin((a/2))) =im(−i(a/2)−ln(2)+i(π/2)−ln((a/2))−ln(sin((a/2))) =((π−a)/2) for a=(π/2) we get =(π/4)

$$\sum_{{n}=\mathrm{1}} ^{+\infty} \frac{{sin}\left({na}\right)}{{n}}={Im}\sum_{{n}\geqslant\mathrm{1}} \frac{{e}^{{ian}} }{{n}}={Im}\left(\sum_{{n}\geqslant\mathrm{1}} \frac{\left({e}^{{ia}} \right)^{{n}} }{{n}}\right)={Im}\left(−{ln}\left(\mathrm{1}−{e}^{{ia}} \right)\right) \\ $$$$={Im}\left(−{ln}\left({e}^{{i}\frac{{a}}{\mathrm{2}}} \left({e}^{−\frac{{ia}}{\mathrm{2}}} −{e}^{\frac{{ia}}{\mathrm{2}}} \right)\right)={Im}\left\{\left(−{ln}\left({e}^{{i}\frac{{a}}{\mathrm{2}}} \right)−{ln}\left(−\mathrm{2}{isin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\right\}\right.\right. \\ $$$$={im}\left(−{i}\frac{{a}}{\mathrm{2}}−{ln}\left(−\mathrm{2}{i}\right)−{ln}\left({sin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\right. \\ $$$$={im}\left(−{i}\frac{{a}}{\mathrm{2}}−{ln}\left(\mathrm{2}\right)+{i}\frac{\pi}{\mathrm{2}}−{ln}\left(\frac{{a}}{\mathrm{2}}\right)−{ln}\left({sin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\right. \\ $$$$=\frac{\pi−{a}}{\mathrm{2}} \\ $$$${for}\:{a}=\frac{\pi}{\mathrm{2}}\:{we}\:{get}\:=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$

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