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