Residue of $z_0=1$ for $f(z)=\frac{z^3+5}{z(z-1)^3}$ 
Consider $$f(z)=\frac{z^3+5}{z(z-1)^3}.$$ I am trying to find the residue of the pole of order $3$, $\ z_0=1$.

I know from calculations that $$\text{Res}(f,1)=\frac{1}{2}\lim_{z\to 1}\frac{\partial^2}{\partial z^2}\left(\frac{z^3+5}{z}\right)=6.$$
I wish to express $f$ as a Laurent series, to find the coefficient of $(z-1)^{-1}$ and hence confirm that Res$(f,1)=6$. I believe the radius of convergence is $0<|z-1|<1$.
I start by taking out a mulitplicative factor of $\frac{1}{(z-1)^2}$, so
\begin{align}
f(z)&=\frac{1}{(z-1)^2}\left(\frac{z^3+5}{z(z-1)}\right) \\
&=\frac{1}{(z-1)^2}\left(-\frac{5}{z}+\frac{6}{z-1}\right) \\
&=\frac{6}{(z-1)^3}-\frac{5}{(z-1)^2}\left(\frac{1}{z}\right)
\end{align}
But I am having difficulty completing the Laurent series as I don't know how to manipulate $\frac{1}{z}$ into a series in terms of $z-1$ that is convergent in $0<|z-1|<1$. Thank you in advanced.

First of all, note that$$\frac{z^3+5}{z(z-1)^3}=-\frac5z+\frac6{z-1}-\frac3{(z-1)^2}+\frac6{(z-1)^3}.$$So, what can we do with $-\frac5z$? We have\begin{align}-\frac5z&=-\frac5{1+(z-1)}\\&=-5\left(1-(z-1)+(z-1)^2-(z-1)^3+\cdots\right)\end{align}As you can see, this doesn't matter for compution of the residue, which is $6$, as you wrote.
