fecundavai3c

2022-02-27

Prove that:
$\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(n\theta \right)}{n}=\frac{\pi }{2}-\frac{\theta }{2}$

besplodnexkj

Let,
${S}_{1}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{cos}n\theta }{n}$
${S}_{2}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}n\theta }{n}$
Then
${S}_{1}+i{S}_{2}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)}{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{{e}^{\in \theta }}{n}$
Now, from the Taylor expansion, $\mathrm{ln}\left(1+x\right)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}\dots$
$⇒-\mathrm{ln}\left(1-x\right)=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}\dots =\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n}$
${S}_{1}+i{S}_{2}=-\mathrm{ln}\left(1-{e}^{i\theta }\right)$
$=-\mathrm{ln}\left(1-\mathrm{cos}\theta -i\mathrm{sin}\theta \right)$
$=-\mathrm{ln}\left(2\frac{{\mathrm{sin}}^{2}\theta }{2}-2i\mathrm{sin}\left(\frac{\theta }{2}\right)\mathrm{cos}\left(\frac{\theta }{2}\right)\right)$
$=-\mathrm{ln}\left(2\frac{\mathrm{sin}\theta }{2}\right)-\mathrm{ln}\left(\frac{\mathrm{sin}\theta }{2}-i\frac{\mathrm{cos}\theta }{2}\right)$
$=-\mathrm{ln}\left(2\frac{\mathrm{sin}\theta }{2}\right)+\mathrm{ln}\left({e}^{i\left(\frac{\pi }{2}-\frac{\theta }{2}\right)}\right)$
$=-\mathrm{ln}\left(2\frac{\mathrm{sin}\theta }{2}\right)+i\left(\frac{\pi }{2}-\frac{\theta }{2}\right)$
Taking the imaginary part of both sides,
${S}_{2}=\frac{\pi }{2}-\frac{\theta }{2}$

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