Lizeth Jackson

Answered

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

$\sum _{n=1}^{\mathrm{\infty}}{e}^{nx}=\frac{{e}^{ix}}{1-{e}^{ix}}$

My solution proceeds by integrating

Answer & Explanation

Emilie Booker

Expert

2022-01-27Added 14 answers

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$x\in \mathbb{R}\text{}\left\{0\right\}$ , then $e}^{ix}\in {S}^{1}\text{}\left\{1\right\$ 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

$\mathrm{log}(1+x)=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{x}+\cdots \Rightarrow \mathrm{log}(1-x)=-(x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}+\cdots )$

$\Rightarrow \mathrm{log}\left(\frac{1}{1-x}\right)=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\frac{{x}^{4}}{4}+\cdots$

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.

If

(where

Therefore

Of course, if x=0, then this equality also holds.

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