Find singularities and their type of $\frac{1}{z(e^z-1)}$ I'm a little bit confused about finding singularities, poles etc. So let's take an example of $\frac{1}{z(e^z-1)}$ at $z=0$.
To do this problem, I've expanded $e^z-1$:
$$e^z-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+...$$
And so
$$\frac{1}{z(e^z-1)}=\frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}$$
I know that if
$$\lim_{z\rightarrow z_0} (z-z_0)^M f(z)=A_{-M} \ne 0$$
$$\lim_{z\rightarrow z_0} (z-z_0)^{M+k} f(z)=0$$
then we have a pole of order $M$ at $z_0$.
So I take $m=1$:
$$\lim_{z\rightarrow 0} \ z^1 \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}=\infty $$
Take $m=2$:
$$\lim_{z\rightarrow 0} \ z^2 \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}=1=A_{-2}=A_{-M} $$
$m=3$:
$$\lim_{z\rightarrow 0} \ z^3 \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+...)}=0 $$
So it seems that it is an isolated singularity and the function is meromorphic.
I don't understand what this $\infty$ means for $m=1$. Shouldn't this be equal to $A_{-1}$? Or it's only true for a pole, so $m=M$?
 A: 
So it seems that it is an isolated singularity and the function is meromorphic.

Yes, the function has an isolated singularity at $z=0$, which is a pole of order 2, because  $\lim_{z\to 0}=z^2f(z)=1$, as $z$ approaches $0$, where $f(z)=\dfrac{1}{z(e^z-1)}$, i.e.
\begin{eqnarray*}
\lim_{z\to 0}z^{2}f(z) &=&\lim_{z\to 0}\frac{z}{e^{z}-1}=\lim_{z\to 0}\frac{1}{e^{z}}=1=A_{-2} \end{eqnarray*}
and for $k=1,2,\ldots$ we have
\begin{eqnarray*}
\lim_{z\to 0}z^{2+k}f(z) &=&\lim_{z\to 0}z^{k}\times\lim_{z\to 0}z^{2}f(z)=0\times 1=0=A_{-2-k}.
\end{eqnarray*}

I don't understand what this $\infty$ means for $m=1$.

It just means that $zf(z)$ behaves like $1/z$ as $z$ tends to $0$.

Shouldn't this be equal to $A_{-1}$?

No, $A_{-1}=-1/2$ rather than $\infty$, because it is the coefficient of $z^{-1}$ in the expansion as a Laurent series of
$f(z)$, which is
\begin{eqnarray*}
f(z) &=&z^{-2}-\frac{1}{2}z^{-1}+\frac{1}{12}-\frac{
1}{720}z^{2}+\ldots,\tag{1}\\ \\
&:=&A_{-2}z^{-2}+A_{-1}z^{-1}+A_{0}+A_{2}z^{2}+\ldots \qquad z\neq 0 \tag{2}. 
\end{eqnarray*}
Moreover, by definition, the coefficient of $z^{n}$ in $(1)$ is  $A_{n}$. So, as computed by you, $A_{-2}=1$. The same result is obtained if we use $(2)$ to find the limits $\lim_{z\to 0}z^2f(z)$ and $\lim_{z\to 0}z^3f(z)$. We get $\lim_{z\to 0}z^2f(z)=1$ and $\lim_{z\to 0}z^3f(z)=0$. Please note that $A_{-1}$ is the coefficient of $z^{-1}$ in $(1)$ and not in $zf(z)$:
\begin{eqnarray*}
zf(z)&=&z^{-1}-\frac{1}{2}+\frac{1}{12}z-\frac{
1}{720}z^{3}+\ldots \tag{3} .
\end{eqnarray*}
