Summing a trigonometric series, $\sum\limits_{n＝1}^{\infty} {\frac{\sin nx}{n}}$

Question

$$\sum_{n＝1}^{\infty} {\frac{\sin nx}{n}}$$ I tried $(\frac{\sin nx}n)'＝\cos nx$，then consider $\;\sum\limits_{n＝1}^\infty\cos nx\;$ , but I didn't succeed. Any help will be appreciated.

Possibly related: http://math.stackexchange.com/questions/71706/using-fourier-series-to-calculate-an-infinite-sum or http://math.stackexchange.com/questions/1514850/compute-infinite-sum-using-fourier-series — Gerry Myerson, Jan 01 '17 at 15:25

lab bhattacharjee · Accepted Answer · 2017-01-01T15:50:46.750

3

$\displaystyle\sum_{n＝1}^{\infty} {\frac{\sin nx}n}=$ imaginary part of $\displaystyle\sum_{n＝1}^{\infty}\dfrac{(e^{ix})^n}n$

$\displaystyle-\sum_{n＝1}^{\infty}\dfrac{(e^{ix})^n}n=\ln(1-e^{ix})$

$\displaystyle=\ln(e^{ix/2})+\ln(-1)+\ln(e^{ix/2}-e^{-ix/2})$

$\displaystyle=ix/2+\ln(-i)+\ln(2\sin x/2)$

$\displaystyle=ix/2-i\pi/2+\ln(2\sin x/2)$

as $e^{(4m-1)i\pi/2}=-i$ for some integer $m$

the principal value of $\ln(-i)=-\dfrac{i\pi}2$

edited Jan 01 '17 at 15:50

answered Jan 01 '17 at 15:29

lab bhattacharjee

274,582

2

If "$m$ is any integer" then aren't you saying the sum has infinitely many different values? – Gerry Myerson Jan 01 '17 at 15:38
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@GerryMyerson, https://en.wikipedia.org/wiki/Principal_value#Complex_argument – lab bhattacharjee Jan 01 '17 at 15:41
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I know about principal values, but you have $\sum n^{-1}\sin nx=x/2+(4m-1)\pi/2$, where $m$ is any integer. Can that be right? – Gerry Myerson Jan 01 '17 at 15:46
You should justify why the principal branch of log is the right choice. (e.g. using something like Abel's theorem for power series) – achille hui Jan 01 '17 at 15:47

score 2 · Answer 2 · answered Jan 01 '17 at 15:26

2

Hint

$$\log(1-z) =- \sum_{n=1}^\infty \frac{z^n}{n}$$

where $z = e^{ix}$.

answered Jan 01 '17 at 15:26

Zaid Alyafeai

14,343

score 1 · Answer 3 · answered Jan 01 '17 at 15:34

1

Note the following: $$\sum_{n=1}^{\infty} \frac{\sin nx}{n}=\textrm{Im} \left(\sum_{n=1}^\infty \frac{z^n}{n} \right)=\textrm{Im} \left( - \log(1-z) \right).$$

$\textrm{Im}(z)$ stands for the Imaginary part of the complex number $z$ as usual.

answered Jan 01 '17 at 15:34

And how does $x$ relate to $z$? – Gerry Myerson Jan 01 '17 at 15:39
$z^n=e^{nix}=\cos nx + i \sin nx.$ – Jan 01 '17 at 15:41
In other words, $z=e^{ix}$, right? – Gerry Myerson Jan 01 '17 at 15:47
https://math.stackexchange.com/questions/879726/evaluation-of-sum-of-sum-n-1-infty-frac-sin-nn/3319218#3319218 – Ahmed Hejazi Aug 22 '19 at 15:15

score 0 · Answer 4 · answered Oct 21 '20 at 19:35

my attemp:

$$\ I=\sum_{n=1}^{\infty }\frac{sin(n)}{n} \ \ \ \ \ \ \ \ \ \ \ (*)\\ \\ we\ know\ that: \ \ \ \ \ \ \ \ \ \ \ \sum_{n=1}^{\infty }x^n=\frac{x}{1-x}\ \ \ ,|x|<1 \ \ \\ \\ \\ let\ x=re^{i\theta }\ , \ \ \ \ \ \ \ \ \ \ \ for \ \ \ all\ \ \ \ \ |r|<1\\ \\ \therefore \sum_{n=1}^{\infty }r^{n}(cosn\theta +isin(n\theta ))=\frac{rcos\theta +irsin\theta }{(1-rcos\theta )-irsin\theta }\\ \\ \therefore \sum_{n=1}^{\infty }r^{n-1}(cosn\theta +isinn\theta )=\frac{(cos\theta +isin\theta )(1-rcos\theta +irsin\theta )}{(1-rcos\theta )^2+r^2sin^2\theta }\\ \\ \therefore \sum_{n=1}^{\infty }r^{n-1}sin(n\theta )=\frac{sin\theta }{r^2-2rcos\theta +1}\ \ \ \ \therefore \sum_{n=1}^{\infty }sin(n\theta )\int_{0}^{1}r^{n-1}dr=\int_{0}^{1}\frac{sin\theta }{sin \theta ^ 2+(r-cos\theta )^2}dr\\ \\ =\int_{0}^{1}\frac{1}{sin\theta +sin\theta (\frac{r-cos\theta }{sin\theta })^2}dr=arctan(\frac{r-cos\theta }{sin\theta })\\ \\ \\ \sum_{n=1}^{\infty }\frac{sin(n\theta )}{n}=arctan(\frac{1-cos\theta }{sin\theta })+arctan(\frac{1}{tan\theta })=arctantan\frac{\theta }{2}+\frac{\pi }{2}-\theta \ \ \ \ \ \ \ \\ now \ put\ \theta =1\\ \\ \\ \therefore \sum_{n=1}^{\infty }\frac{sinn}{n}=\frac{\pi -1}{2}\\ \\ \\ $$

Summing a trigonometric series, $\sum\limits_{n＝1}^{\infty} {\frac{\sin nx}{n}}$

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