Lizeth Jackson

2022-01-26

Evaluating $\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(nx\right)}{n}$ without integrating $\sum _{n=1}^{\mathrm{\infty }}{e}^{nx}$
My solution proceeds by integrating
$\sum _{n=1}^{\mathrm{\infty }}{e}^{nx}=\frac{{e}^{ix}}{1-{e}^{ix}}$

Emilie Booker

Note that
$\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(nx\right)}{n}=\text{Im}\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{e}^{\in x}}{n}\right)$
$=\text{Im}\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{\left({e}^{ix}\right)}^{n}}{n}\right)$
If , then and therefore
$\sum _{n=1}^{\mathrm{\infty }}\frac{{\left({e}^{ix}\right)}^{n}}{n}=\mathrm{log}\left(\frac{1}{1-{e}^{ix}}\right)$
(where $\mathrm{log}$ is the principal branch of the logarithm), since
$\mathrm{log}\left(1+x\right)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{x}+\cdots ⇒\mathrm{log}\left(1-x\right)=-\left(x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}+\cdots \right)$
$⇒\mathrm{log}\left(\frac{1}{1-x}\right)=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}+\cdots$
Therefore
$\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(nx\right)}{n}=-\mathrm{arctan}\left(\frac{-\mathrm{sin}x}{1-\mathrm{cos}\left(x\right)}\right)$
$=\mathrm{arctan}\left(\mathrm{cot}\left(\frac{x}{2}\right)\right)$
Of course, if x=0, then this equality also holds.

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