Finding pole order and calculating residue of: $\frac1{z-\tan z} - \frac1{z}$ Part of the the problem I'm trying to do involves finding the residue of 
$\frac{1}{z-\tan z} - \frac{1}{z}$ at z=0
I am not sure of the order of the pole
Here's what I did to find the order of the pole
f(z) = $\frac{1}{z-\tan z} - \frac{1}{z}$
      = $\frac{\tan z}{z(z - \tan z)}$
Now $z - \tan z = \frac{1}{3}z^3 + \frac{2}{15}z^5 + ...\quad$
and $\quad z(z - \tan z) =  \frac{1}{3}z^4 + \frac{2}{15}z^6 + ... $
So f has pole of order 4 at z=0. 
Is this correct and to find the residue I use the formula
$\frac{1}{n!}\lim_{z\to 0} \frac{d^3}{dz^3}(z^4f(z))$ right?

You may write
$$
\begin{align}
\frac{1}{z-\tan z} - \frac{1}{z}&=\frac{1}{z-\left(z+\frac{1}{3}z^3 + \frac{2}{15}z^5 + \mathcal{O}(z^7)\right)} - \frac{1}{z}\\\\
&=\frac{1}{-\frac{1}{3}z^3 - \frac{2}{15}z^5 + \mathcal{O}(z^7)} - \frac{1}{z}\\\\
& =-\frac{3}{z^3}\frac{1}{1 + \frac{2}{5}z^2 +\mathcal{O}(z^4)} - \frac{1}{z}\\\\
& =-\frac{3}{z^3}\left(1- \frac{2}{5}z^2 + \mathcal{O}(z^4)\right) - \frac{1}{z}\\\\
& =-\frac{3}{z^3}+ \frac{6}{5z} - \frac{1}{z}+  \mathcal{O}(z)\\\\
& =-\frac{3}{z^3}+ \frac{1}{5z} +  \mathcal{O}(z)
\end{align}
$$  giving the value $\dfrac15$ for the desired residue.
