Type of singularities of $\frac{z}{e^z-1}$ I don't really understand how one can find the type of singularities for a given function. Say if 
$$f(z) = \frac{z}{e^z-1}$$
then I know that the singularities are at $z = 2n\pi i$
However, how do I find the type? If I try to write out Laurent series for this by using
$$e^z = 1 + z + z^2/2! + z^3/3! + \cdots $$, then I got a summation in the denominator, which I don't know how to rearrange as a Laurent series.
Can anyone please give me a hint?
 A: Some of key points to decide the type of the singularity 
I am not claiming that these points are complete. But certainly these point may help you to decide about the types of singularity of a function at a point.
1:  A singular point $z_0 $is called an isolated singular point of an analytic function $f(z)$ if there exists a deleted $\epsilon$-spherical neighborhood of $z_0$ that contains no singularity. If no such neighborhood can be found, $z_0 $is called a non-isolated singular point. Thus an isolated singular point is a singular point that stands completely by itself, embedded in regular points.
2. An isolated singular point $z_0$ such that $f$ can be defined, or redefined, at $z_0$ in such a way as to be make it analytic at $z_0$ is called removable singularity. A singular point $z_0$ is removable if  $\lim _{z\to z_{0}} f(z)$ exists. For example The singular point $z = 0$ is a removable singularity of $f(z) = \frac{\sin z}{z}$
 since $\lim _{z\to z_{0}} \frac{\sin z}{z} = 1$. Dominic's answer says about when a singular point $z_0$ is a pole of order $k$.
3. A singular point that is not a pole or removable singularity is called an essential singular point. 
4 Isolated essential singularity
Limit points of the sequence of a zeros of the nonzero analytic function $f(z)$ is an isolated singularity. Since it is not a pole it is an isolated essential singularity. For example consider the function $\sin \frac{1}{z}$ it has zeros at $z = \frac{1}{k\pi}$ . Limit points of the zeros is the zeros is the point $z =0$ hence, $z = 0$ is an isolated essential singularity of $f(z)$.
5 . Non -isolated essential singularity
Let $z_1, z_2,\ldots, z_n, \ldots$ be the set of points having limit point $z_0$ in a region $D$. Let $f$ be analytic in $D$ except at the poles  $z_1, z_2,\ldots, z_n, \ldots$, then $z_0$  is also singularity of $f(z)$. The reason is that $f$ is unbounded in the neighborhood of $z_0$. But  $z_0$ is not isolated because it is limit points of the poles. Hence $f(z)$ has a non-isolated singularity at $z_0$. For example consider the function $\frac{1}{\cos(1/z)}$ this function has poles at $k = \frac{2}{(2k+1)\pi}$ the point $z =0$
is the limit point of these poles hence given function has non -isolated essential singularity at $z = 0$.
A: On $z=0$ you can remove the singularity.
For the other use that a singularity at $z_0$ is a pole of order $k$ when $f(z)\cdot (z-z_0)^k$ does have a removable singularity at $z_0$. 
