TITLE: How to prove that $ \sum_{k=1}^{n-1} \frac{1}{1-e^{2 \pi i k/n}} = \frac{n-1}{2}$?
QUESTION [9 upvotes]: I came across the fact that
$$ \sum_{k=1}^{n-1} \frac{1}{1-e^{2 \pi i k/n}} = \frac{n-1}{2}.$$
How can we prove this identity?

REPLY [6 votes]: One has for $f(X)=\frac{X^n-1}{X-1}=\prod\limits_{k=1}^{n-1}(X-e^{\frac{2ik\pi}{n}})$
$$\frac{f'(X)}{f(X)}=\sum\limits_{k=1}^{n-1}\frac{1}{X-e^{\frac{2ik\pi}{n}}}$$
So we have $\sum\limits_{k=1}^{n-1}\frac{1}{1-e^{\frac{2ik\pi}{n}}}=\frac{f'(1)}{f(1)}$
Now let us compute $\frac{f'(X)}{f(X)}=\frac{(n-1)X^{n-2}+(n-2)X^{n-3}+\cdots+2X+1}{X^{n-1}+\cdots+1}$ so we have
$$\frac{f'(1)}{f(1)}=\frac{(n-1)+(n-2)+\cdots+2+1}{n}=\frac{n^2-\frac{n(n+1)}{2}}{n}=n-\frac{n+1}{2}=\frac{n-1}{2}$$