How to show that $\sum\limits_{n=1}^\infty \exp(i\,nz)$ converges? How do I show that $$ f(z) = \exp(i\, z) + \exp(i\, 2z) + \ldots + \exp(i\, nz) + \ldots $$ converges? Problem is taken from a Yahoo! Answers question: "Find the infinite sum of sin(n)/n?".
 A: For the series to converge, you have the following condition,
$$ |{\rm e}^{iz}| < 1 \,.$$
Assuming $z=x+iy$, we have,
$$ \left|{\rm e}^{i(x+iy)} \right| =   \left|{\rm e}^{ix-y)} \right| =  {\rm e}^{-y} < 1 \Rightarrow  -y < 0 \Rightarrow y > 0 \,.$$ 
A: It looks like a geometric series to me. Recall that a geometric series is of the form
$S = a + ar + ar^2 + ar^3 + \cdots \, .$
In this case, the first term $a = e^{iz}$ while the common ratio $r = e^{iz}.$
$S = \frac{a}{1-r} = \frac{e^{iz}}{1-e^{iz}} \, . $
This function defines the analytic continuation of $f$ provided $e^{iz} \neq 1$.
A: The radius of convergence of this series is $1$, due to the pole at $1$ of the geometric series. Thus, for any $z$ such that: 
$$ |\exp(iz)| < 1$$
the series converges. This is true for every $y = \Im(z) > 0$, since for these values, you have a factor $e^{i(iy)}=e^{-y} < 1$.
A: The general proof for the formula for the geometric series given here on Wikipedia works for complex numbers as well. If you look at real numbers: $1 + r + r^2 + ...$, then we need $\lvert r \lvert < 1$. If $r$ is a complex number, then you still need $\lvert r \lvert < 1$, where now the $\lvert \cdot\lvert$ is the complex norm.
