# The value of $\sum_{n=1}^{\infty} \frac{\sin(n)}{n}$

What we know is that $$x = \sum_{n=1}^{\infty}\frac{\sin(nx)}{n}(-2\cos(n\pi)), x \in (-\pi, \pi)$$ it was calculated using Fourier's series.
The task is to find the limit of
$$\sum_{n=1}^{\infty} \frac{\sin(n)}{n}$$ using the sum above.
I noticed that for $x = 1$ we have something similar in the first sum but I don't know how to finish the problem.

• try $x=-1$ and think about $\cos(n\pi)$ – Marcel May 30 '17 at 22:04

Set $\pi-x$ as $x$ in your formula to get \begin{align} \pi-x&=\sum_{n=1}^\infty\frac{\sin(n\pi-nx)}{n}(-2\cos n\pi)= \sum_{n=1}^\infty\frac{\sin n\pi\cos nx-\cos n\pi\sin nx}{n}(-2\cos n\pi)=\\ &=\sum_{n=1}^\infty\frac{\sin nx}{n}(2\cos^2n\pi)=2\sum_{n=1}^\infty\frac{\sin nx}{n}. \end{align}

• Very nice solution! :) Thank you – Hendrra May 31 '17 at 7:27

For any $z\in\mathbb{C}$ such that $|z|<1$ we have $$\sum_{n\geq 1}\frac{z^n}{n} = -\log\left(1-z\right) \tag{1}$$ and if we set $z=\rho e^i$ for some $\rho=1-\varepsilon<1$ we get: $$\sum_{n\geq 1}\frac{e^{in}}{n}\rho^n = -\log\left|1-\rho e^{i}\right|+i\varphi_\rho \tag{2}$$ where $\varphi_\rho$ is the angle between the line joining $\rho e^i$ with $1$ and the line joining $-1$ to $1$.
By switching to the imaginary parts we get: $$\sum_{n\geq 1}\frac{\sin n}{n}\rho^n = \varphi_\rho \tag{3}$$ and by summation by parts / Abel's summation we are allowed to replace $\rho$ with $1$ in both sides: $$\sum_{n\geq 1}\frac{\sin n}{n}=\color{red}{\frac{\pi-1}{2}}.\tag{4}$$


This becomes a straightforward application of the Abel-Plana Formula:

\begin{align} \left.\sum_{n = 1}^{\infty}{\sin\pars{n\theta} \over n}\right\vert_{\ \theta\ \not=\ 0} & = \theta\sum_{n = 1}^{\infty}\mrm{sinc}\pars{n\verts{\theta}} = -\theta + \theta\sum_{n = 0}^{\infty}\mrm{sinc}\pars{n\verts{\theta}} \\[5mm] & = -\theta + \,\mrm{sgn}\pars{\theta}\int_{0}^{\infty} {\sin\pars{n\verts{\theta}} \over n\verts{\theta}}\,\verts{\theta}\,\dd n + \theta\bracks{{1 \over 2}\,\mrm{sinc}\pars{0}} \\[5mm] & = -\theta + {1 \over 2}\,\pi\,\mrm{sgn}\pars{\theta} + {1 \over 2}\,\theta = \bbx{\pi\,\mrm{sgn}\pars{\theta} - 1 \over 2} \end{align}

The result is a valid one whenever

\begin{align} 0 & = \lim_{y \to \pm\infty}\bracks{% {\sin\pars{\theta\bracks{x + \ic y}} \over x + \ic y}\,\expo{-2\pi\verts{y}}} _{\ x\ \geq\ 0} \\[5mm] & = \lim_{y \to \pm\infty}\bracks{% {\sin\pars{\theta x}\cosh\pars{\theta y} + \ic\cos\pars{\theta x}\sinh\pars{\theta y}\over x + \ic y}\,\expo{-2\pi\verts{y}}} \end{align} It becomes clear that the above expression is true whenever $\ds{2\verts{\theta} - 2\pi < 0 \implies \bbx{\verts{\theta} < \pi}}$