0
$\begingroup$

$$\sum_{n＝1}^{\infty} {\frac{\sin nx}{n}}$$ 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.

  • 3
    What is the problem...? – 2017-01-01
  • 0
    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 – 2017-01-01

3 Answers 3

5

$\displaystyle\sum_{n＝1}^{\infty} {\frac{\sin nx}n}=$ 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 $\ln(-i)=-\dfrac{i\pi}2$

  • 2
    If "$m$ is any integer" then aren't you saying the sum has infinitely many different values? – 2017-01-01
  • 1
    @GerryMyerson, https://en.wikipedia.org/wiki/Principal_value#Complex_argument – 2017-01-01
  • 1
    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? – 2017-01-01
  • 0
    You should justify why the principal branch of log is the right choice. (e.g. using something like [Abel's theorem for power series](https://en.wikipedia.org/wiki/Abel's_theorem)) – 2017-01-01
3

Hint

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

where $z = e^{ix}$.

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.

  • 0
    And how does $x$ relate to $z$? – 2017-01-01
  • 0
    $z^n=e^{nix}=\cos nx + i \sin nx.$ – 2017-01-01
  • 0
    In other words, $z=e^{ix}$, right? – 2017-01-01