Compute the residue of $ 1/(z^2 \sin z) $ at $z=0$ I'm trying to compute the following:

$$ \operatorname*{Res}_{z=0} \frac{1}{z^2 \sin z} $$

I know I can either find an analytic function and use the equation 
$$ \operatorname*{Res}_{z=z_0} \frac{p(z)}{q(z)}=\frac{p(z_0)}{q(z_0)} $$
so long as $p(z_0)\ne0, \, q(z_0)=0, \, q'(z_0)\ne0 $, or figure out some new expansion.
My first attempt at this was to choose $\csc(z)$ as my analytic function, $p(z)$, and $z^2$ as $q(z)$. But $q'(0)=0$, which fails to satisfy the requirements of the equation. 
I could really just use a hint to get started, I'm not sure how else to approach this without doing an expansion. 
Any help would be appreciated, thank you.
 A: Elaborating on the comment by Daniel Fischer:
$$\begin{align}
 f(z) &= \frac{1}{z^2\sin z} = \frac{1}{z^3(1-\frac{z^2}{6}+O(z^4))}=\frac{1}{z^3}(1-\frac{z^2}{6}+O(z^4))^{-1}\\
&=\frac{1}{z^3}(1+\frac{z^2}{6}+O(z^4)) = \frac{1}{z^3}+\frac{1}{6z}+O(z).
\end{align}$$
Where I have used the expansion $(1+z)^\alpha = 1 +\alpha z +\frac{\alpha(\alpha-1)}{2}z^2 +\dots$ for $\alpha=-1$.
The coefficient of the $1/z$ term is the residue, which in this case is $1/6$.
Because $z^3f(z)$ is for this function analytic at $z=0$, the function has a pole of order 3. You could also use the general formula for poles of order $m$, with $m=3$ to get the residue at $z=0$.
$$ \mathrm{Residue} = \lim_{z\to0}\frac{1}{2}\frac{d^2}{dz^2}z^3f(z),$$
which after a longer calculation will also give you 1/6 (easier and shorter with the expansion suggested by Daniel Fisher).
A: It is not hard to calculate the residue without using the series expansion of the sine.  
You only need that $\lim_{z \to 0}\frac z {\sin z} =1$ (l'Hospital).
It then follows that $z^2 \sin z$ has a zero of order 3 at $0$. Thus 
$$\operatorname{Res}_{z=0}(\frac 1 {z^2 \sin(z)})=\lim_{z \to 0} \frac 1 {(3-1)!} \frac {z^3}{z^2 \sin z}=\frac 1 6 \lim_{z \to 0} \frac z {\sin z}=\frac 1 6 $$
