Sum of $1/(1 - x)$ over roots of unity Define $r_\ell=e^{2 \sqrt{-1} \pi \ell /N}$.
I believe the relation
$$ \sum_{\ell=1}^N \frac 1 {1 - \gamma r_\ell} = \frac N {1- \gamma^N}$$
holds for every positive integer $N$ and complex number $\gamma$ such that $\gamma^N \neq 1$.
Is there a simple proof of it, if it is true?
 A: Notice $z^N - 1 = \prod\limits_{\ell=1}^N (z - r_\ell)$. Taking logarithm and differentiate, one get
$$\frac{Nz^{N-1}}{z^N-1} = \sum_{\ell=1}^N \frac{1}{z - r_\ell}$$
Multiply both sides by $z$ and then take $\displaystyle\;z = \frac{1}{\gamma}\;$, this becomes
$$\frac{Nz^N}{z^N-1} = \sum_{\ell=1}^N \frac{z}{z-r_\ell}
\quad\iff\quad
\frac{N}{1-\gamma^N} = \sum_{\ell=1}^N\frac{1}{1-\gamma r_\ell}$$
A: Consider the rational function
$$ f(z) = \frac{N}{1 - z^N} $$
It has simple poles at the $N$-th roots of unity, which I will write as $\zeta$. To compute the residue at $\zeta$, the formula for the residues says it is
$$ \mathrm{Res}(f, \zeta) = \frac{N}{-N \zeta^{N-1}} = - \zeta^{N-1} $$
Another function with the same residue is
$$ \mathrm{Res}\left( \frac{1}{1 - \zeta z}, \zeta \right) = \frac{1}{-\zeta} = -\zeta^{N-1} $$
We can then conclude that the partial fraction decomposition of $f(z)$ is given by
$$ \frac{N}{1 - z^N} = \sum_\zeta \frac{1}{1 - \zeta z} $$

Instead of invoking partial fractions, we can argue as follows. We can use your sum to subtract off the poles:
$$ r(z) = f(z) - \sum_\zeta \frac{1}{1 - \zeta z} $$
Since this function has only removable singularities, it must be entire. Since it is a rational function with no poles, it must be a polynomial.
Since every term goes to zero and there are only finitely many, we can compute the limit
$$ \lim_{z \to \infty} r(z) = 0 $$
A polynomial with a limit at $\infty$ is a constant function; therefore $r(z) = 0$, and we infer an equation of rational functions
$$ \sum_\zeta \frac{1}{1 - \zeta z} = \frac{N}{1 - z^N} $$
