Residue of $f(z) = \frac{1}{z-\sin z}$ at $z=0$ My attempt:
$$ f(z) = \frac{1}{z-\sin z}$$
$$\frac{1}{z-(z-\frac{z^3}{6}+\frac{z^5}{120}-...)}$$
$$\frac{1}{z(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$Res(f(z),0) = \lim_{z \to 0} z \cdot \frac{1}{z(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$ = \lim_{z \to 0} \frac{1}{(1-(1-\frac{z^2}{6}+\frac{z^4}{120}-...))}$$
$$=\frac{1}{0}$$
This is where I'm stuck. Using L'Hopital's rule doesn't help. The answer should be $\frac{3}{10}$.
 A: You're looking for the $z^{-1}$ term in a Laurent series of $f$. The formula 
\begin{align*}
\operatorname{Res}(f, 0) = \lim_{z\to 0} zf(z)
\end{align*}
you give on the fourth line only applies to a function of the form $a_{-1} z^{-1} + a_0 + a_1 z + \cdots$ near $z = 0$; if $f$ has higher $z^{-n}$ terms, then the limit is going to diverge.
A: Note that $$(z-\sin(z))'=1-\cos(z),$$
$$(z-\sin(z))''=\sin(z),$$
$$(z-\sin(z))'''=\cos(z).$$ Hence $0$ is a root of multiplicity $3$ of the denominator and thus a triple pole of your function. The residue is then given by $$\textrm{Res}(f,0) = \lim_{z \to 0} \frac{1}{2!}\left(\frac{z^3}{z-\sin(z)}\right)''.$$
A: Two small hints in addition to the nice answer of @C.Dubussy

  
*
  
*If $f$ has a pole of order $n$ in $z_0$, the $Res$ of $f$ at $z_0$ is
  \begin{align*}
Res(f,z_0)=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\left[(z-z_0)^nf(z)\right]
\end{align*}
  
*From your  series expansion 
  \begin{align*}
f(z)&=\frac{1}{z-\sin z}\\
&=\frac{1}{z-(z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots)}\\
&=\frac{1}{z^3\left(\frac{1}{6}-\frac{z^2}{120}-\cdots\right)}
\end{align*}
  you can immediately deduce that $z_0=0$ is a pole of order $3$.

