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!
 A: $\newcommand{\+}{^{\dagger}}%
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$\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,
$$\color{#0000ff}{\large%
\sum_{n = 1}^{\infty}{\sin\pars{n} \over n} = \half\pars{\pi - 1}}
$$
A: What is the function that has its fourier coefficients $A_n=0$ and $B_n = \frac{1}{n}$, i.e
$$
f(x) = \sum_{n=1}^{\infty} \frac{1}{n} \sin(n x)$$
Once you have figured out $f(x)$, find $f(1)$.
By the way $f(x)$ is a "standard" function in engineering analysis.
A: By the way, note that $\frac{\sin n}{n}$ is exactly the n-th Fourier coefficient of the function 
$
 \sqrt{\frac{\pi}{2}}\chi_{[-1,1]}(x).
$
Since $\chi_{[-1,1]}(x)$ has a compact support, one can use the Poisson formula: 
$$
    \sum_{n\in\mathbb{Z}} \widehat{\chi_{[-1,1]}}(n) = \sqrt{2\pi} \sum_{n \in \mathbb{Z}} \chi_{[-1,1]}(2\pi n) = \sqrt{2\pi},
$$
and get
$$
   \sum_{n\in\mathbb{Z}} \frac{\sin n}{n} = \pi.
$$
A: Hint: $$\sum_{n=1}^\infty \frac{\sin (n)}{n}= \text{Im }\sum_{n=1}^\infty \frac{e^{in}}{n}=\text{Im } \sum_{n=1}^\infty \int_0^1 x^{n-1} \mathrm{d} x \bigg|_{x=e^i}=\text{Im } \int_0^1 \frac{\mathrm{d} x}{1-x} \bigg|_{x=e^i}=\text{Im Log }(1-e^i) $$
