Find the value of $2S+1$

We know that : $\sum_{k=1}^{\infty} \frac{x^k }{k} = -\log(1-x)$ For any $x$ such that $|x|\leq 1$ and $x\neq 1$ , then $S= \text{Im} \sum_{n=1}^{\infty} \frac{e^{i n}}{n} = -\text{Im}(\log(1-e^{i} ))$ Now we use some trigonometry to get : $1-e^i = 1-\cos 1 -i \sin 1 = 2\sin \frac{1}{2}\left(\sin \frac{1}{2} - i \cos \frac{1}{2} \right) = 2\exp\left(\frac{(1-\pi)i}{2}\right) \sin \frac{1}{2}$ So $S= \text{Im}(\log(1-e^{i} )) = \text{Im}\left(\frac{(\pi-1)i}{2} - \log \left(2\sin \frac{1}{2}\right) \right) = \frac{\pi-1}{2}$ Then the result is $\boxed{\pi}$ .

Generalization : for any $x\in (0,2\pi)$ $f(x)=\sum_{n=1}^{\infty} \frac{\sin (nx)}{n} = \frac{\pi-x}{2}$ Note that $f$ is $2\pi$ -periodic.

Another method : Think about the $2\pi$ -Fourier series of $x\mapsto \frac{\pi-x}{2}$ on $[0,2\pi)$ with canceled continuity at $0$ .