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.
 A: 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}
A: 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:
$$ \sum_{n\geq 1}\frac{\sin n}{n}=\color{red}{\frac{\pi-1}{2}}.\tag{4}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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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
$\ds{2\verts{\theta} - 2\pi < 0 \implies \bbx{\verts{\theta} < \pi}}$
