Value of $\sum^{\infty}_{n=1}\frac{1}{n} e^{\frac{2\pi i n}{3}}$ Hello I am trying to find a closed form for $S=\sum^{\infty}_{n=1}\frac{1}{n}\cdot e^{\frac{2\pi i n}{3}}\, $ My thought was to use $$\frac{1} {1 - x} =\sum_{k\ge 0} x^k$$ And by integrating to get $$-\log(1-x)=\sum_{k\ge 1} \frac{x^k} {k} $$ Now here is the problem, this converges for $|x|<1$ but clearly in the original series the absolute value is $1$ So I cant just plug the value for x. Am I on the right track? Can I somehow show that I can plug into $x$ even though it's absolute value it's not in the radius of convergence ? Many thanks in advance. Edit: How can I finish it? 
 A: A less formal approach, assume it converges and reorganize the sum,
\begin{equation}
\sum^\infty_{n=1} \left\{ \left[-\frac{1}{2} \frac{1}{3n-2} - \frac{1}{2} \frac{1}{3n-1} + \frac{1}{3n} \right] + i \frac{\sqrt{3}}{2}\left[\frac{1}{3n-2}-\frac{1}{3n-1} \right] \right\} \, = \, - \frac{1}{2}\log(3) + i \frac{\pi}{6} \, .
\end{equation}
A: Yes, you're on the right track. For every $z\in\mathbb C$ with $|z|\leqslant1$, except for $z=1$,$$-\log(1-z)=\sum_{n=1}^\infty\frac{z^n}n,$$where $\log\colon\mathbb{C}\setminus(-\infty,0]\longrightarrow\mathbb C$ is the main branch of the logarithm.
A: If you multiply the $n$th term in the series by $r^n$ with $r\in (0,\,1)$, the $r\to 1^-$ limit is the finite quantity $-\ln (1-e^{2\pi i/3})$ (although its definition requires care). So you can in fact use the logarithm in this case.
A: Yes, you can use the Abel's summation method in order to show that the series convergences uniformly for $x_r = r \exp(2 \pi i/3)$ with $r \in (0,1]$. Thus, the limes is continuous in $r \in (0,1]$. Now your argument can be used to show this identity for $r \in (0,1)$. By continuity this formula holds also for $r=1$.
