Evaluate $ sum_{n=1}^{infty} frac{sin n}{ n } $ using the fourier series

I am a beginner with Fourier series and I have to evaluate the sum

$$\sum_{n =1}^{\infty}{\sin\left(n\right) \over n}$$

I don't know which function I have to take to evaluate the fourier series ...
Someone can give me a hint ?

Thanks in advance!

Answer

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$\ds{\sum_{n = 1}^{\infty}{\sin\pars{n} \over n} = \half\pars{\,\sum_{n = -\infty}^{\infty}{\sin\pars{n} \over n} - 1}.\quad}$ See $\large\tt details$
over here .

\begin{align}
\sum_{n = -\infty}^{\infty}{\sin\pars{n} \over n}&=
\int_{-\infty}^{\infty}{\sin{x} \over x}\sum_{n = -\infty}^{\infty}\expo{2n\pi x\ic}
\,\dd x
=
\int_{-\infty}^{\infty}\half\int_{-1}^{1}\expo{\ic kx}\,\dd k

\sum_{n = -\infty}^{\infty}\expo{-2n\pi x\ic}\,\dd x
\\[3mm]&=
\pi\sum_{n = -\infty}^{\infty}\int_{-1}^{1}\dd k
\int_{-\infty}^{\infty}\expo{\ic\pars{k - 2n\pi}x}\,{\dd x \over 2\pi}
=
\pi\sum_{n = -\infty}^{\infty}\int_{-1}^{1}\delta\pars{k - 2n\pi}\,\dd k
\\[3mm]&=
\pi\sum_{n = -\infty}^{\infty}\Theta\pars{{1 \over 2\pi} - \verts{n}}
= \pi\,\Theta\pars{1 \over 2\pi} = \pi
\end{align}

Then,
$$\color{#0000ff}{\large%
\sum_{n = 1}^{\infty}{\sin\pars{n} \over n} = \half\pars{\pi - 1}}
$$