Evaluation of Sum of $ \sum_{n=1}^{\infty}\frac{\sin (n)}{n}$. If $\displaystyle S = \sum_{n=1}^{\infty}\frac{\sin (n)}{n}.$ Then value of $2S+1 = $
Using Fourier Series Transformation I am Getting $2S+1=\pi$
But I want to solve it Using Euler  Method and Then Use Logarithmic Series.
$\bf{My\; Try::}$ Using $\displaystyle \sin (n) = \left(\frac{e^{in}-e^{-in}}{2i}\right)$. So $\displaystyle S = \sum_{n=1}^{n}\frac{\sin (n)}{n} = \frac{1}{2i}\sum_{n=1}^{\infty}\frac{e^{in}}{n}-\frac{1}{2i}\sum_{n=1}^{\infty}\frac{e^{-in}}{n}$
Now Using $\displaystyle \ln(1-x) = -x-\frac{x^2}{2}-\frac{x^3}{3}...............\infty$
So Let $\displaystyle S = -\frac{1}{2i}\ln(1-e^{i})+\frac{1}{2i}\ln(1-e^{-i})$
Now How can I solve after that
Help me
Thanks
 A: Rewrite what you have as:
\begin{align} 2S =& i \ln(1-e^i) -i \ln(1 - e^{-i} )  \\
=& i \ln \left (\frac{1-e^i}{1-e^{-i}} \right ) \\
=&i \ln \left ( -e^{i} \frac{1- e^{-i}}{1-e^{-i}} \right ) \\
=& i \big ( \ln ( -e^{i}) \big)  \\
=& i \big ( \ln ( e^{-i \pi} e ^{i} \big ) \\
=& i \big ( \ln ( e^{ i(1 -\pi)} \big ) \\ 
=& i  \big( i ( 1- \pi) \big)\\
=& \pi -1 \\
\end{align}
by choosing the right branch. 
A: my attempt :
$$\  \ S=\sum_{n=1}^{\infty } \frac{sin(n)}{n}=\sum_{n=1}^{\infty }
= \int_{0}^{\infty } e^{-nw}\sin(n)dw\\ \\ \\$$
$$\therefore S=Im\int_{0}^{\infty }\sum_{n=1}^{\infty
 }e^{-(w-i)n}dw=Im\ \int_{0}^{\infty
 }\frac{1}{e^{w-i}}dw=Im\int_{0}^{\infty
 }\frac{dw}{cos(1)e^{w}-isin(1)e^{w}-1}\\ \\ \\$$
$$\therefore S=Im\int_{0}^{\infty
 }\frac{e^{w}cos(1)-1+isin(1)e^{w}}{(cos(1)e^{w}-1)^{2}+(sin(1)e^{w})^{2}}dw=\int_{0}^{\infty
 }\frac{sin(1)\ e^{-w}}{sin^{2}(1)+(cos(1)-e^{-w})^{2}}dw\\ \\ \\$$
$$\therefore S=\lim_{w\rightarrow \infty }\ tan^{-1}\left (
 \frac{cos(1)-e^{-w}}{sin(1)} \right )-tan^{-1}\left (
 \frac{cos(1)-1}{sin(1)} \right )\\ \\ \\$$
$$\therefore S=tan^{-1}\left ( cot(1) \right )-tan^{-1}\left (  cot(1)-csc(1) \right )=\frac{\pi }{2}-1+\frac{1}{2}=\frac{1}{2}\left ( \pi -1 \right )$$
