What is the value of $\sum_{n \geq 1} \epsilon^n/n$, where $\epsilon$ is a $k^\text{th}$ root of $1$

Question

I wonder what is the value of $$\sum_{n \geq 1} \frac{\epsilon^n}{n}$$ where $\epsilon$ is a $k^\text{th}$ root of $1$, say $\epsilon=e^{(2\pi i l)/k}$ for a fixed $l=1,\ldots,k-1$.

My idea is that the sum is one of the values of $$-\log(1-\epsilon)=-\log(re^{i \phi})=-\log r -i \phi-2\pi i \mathbb Z,$$ but which one?

Answer 1

You can find in the literature of Complex Analysis, that

$\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{e^{i2\pi x n}}{n}=i\pi(\frac{1}{2}-x)-\ln(2\sin(\pi x))\enspace$ for $\enspace 0<x<1$ .

To answer your question set $\displaystyle x:=\frac{l}{k}$ with $k\geq 2$.

Note:

$i2\pi \mathbb{Z}$ disappears because the main branch of the logarithm has to be used. E.g set $x=\frac{1}{2}$ then you know why.