Residue of $f(z) = \frac{1}{z(1-\cos(z))}$ I tried to find the residue of the function $$f(z) = \frac{1}{z(1-\cos(z))}$$ at the $z=0$. So I did:
\begin{align}
\hbox{Res}(0)&=\lim_{x \to 0}(z)(f(z))\\
\hbox{Res}(0)&= \lim_{x \to 0} \frac{1}{1-\cos(z)}
\end{align}
and I got that the residue to be $\infty$, and it seems to be wrong.

$$\cos z=1-\dfrac{1}{2!}z^2+\dfrac{1}{4!}z^4-\dfrac{1}{6!}z^6+\cdots$$
then
\begin{align}
\dfrac{1}{z(1-\cos z)}
&= \dfrac{1}{z^3\left(\dfrac{1}{2!}-\dfrac{1}{4!}z^2+\dfrac{1}{6!}z^4+\cdots\right)} \\
&= \dfrac{1}{z^3}\left(2+\dfrac{1}{6}z^2+\dfrac{1}{120}z^4+\cdots\right) \\
&= \dfrac{2}{z^3}+\dfrac{1}{6z}+\dfrac{z}{120}+\cdots
\end{align}
s0 $a_{-1}=\dfrac16$.
