<s>[INST] Find the laurent series of $\frac{1}{z(z-2)^3} $ Find the laurent series of $$\frac{1}{z(z-2)^3} $$ about  $$ z_0 = 0, z_0 = 2 $$
I am having trouble with this question. 
I have worked out that with some manipulation I can get f(z) into the form:
$$ \frac{-1}{8z(1-\frac{z}{2})^3} $$
However the solution to the question uses the following geometric series to find residue at $$z_0 = 0 $$
$$\sum_{n=0}^\infty n(n-1)z^{n-2} = \frac{2}{(1-z)^3} $$
I have tried repeatedly to manipulate the standard geometric series for 1/(1-z) however I can't seem to replicate this geometric series. This is also a recurring problem I am having when trying to find the laurent series of a function f(z) when it has a pole of order n where n > 1. 
Any help about how to solve this and similar questions of the same form in future would be much appreciated. 
 [/INST] Let $z=2w$.
$$\frac{1}{z(z-2)^3} = -\frac{1}{16}\frac{1}{w(1-w)^3}  $$
First comes partial fraction decomposition
\begin{align}
\frac{1}{w(1-w)^3} &= \frac{1}{w}+\frac{1}{1-w}+\frac{1}{(1-w)^2}+\frac{1}{(1-w)^3}
\end{align}
If you're seeking the residue, the problem is almost over. The latter three terms can be expressed as a geometric series and its derivatives, which are all Taylor series. The residue must be coming from the first term $-\frac{1}{16w} = -\frac{1}{8z}$. If you want to actually write down the Laurent series, consider
\begin{align}
\frac{1}{1-w} &= \sum_{n=0}^\infty w^n
\\
\\
\frac{1}{(1-w)^2} &= \sum_{n=1}^\infty nw^{n-1} = \sum_{n=0}^\infty (n+1)w^{n}
\\
\\
\frac{1}{(1-w)^3} &= \frac{1}{2}\sum_{n=2}^\infty n(n-1)w^{n-2} = \frac{1}{2}\sum_{n=0}^\infty (n+2)(n+1)w^{n}
\\
\\
\frac{1}{(1-w)^{N+1}} &= \frac{1}{N!}\sum_{n=N}^\infty \frac{n!}{(n-N)!}w^{n-N}
= \frac{1}{N!}\sum_{n=0}^\infty \frac{(n+N)!}{n!}w^{n}
= \sum_{n=0}^\infty \binom{n+N}{N}w^{n}
\end{align}
Then 
\begin{align}
 -\frac{1}{16}\frac{1}{w(1-w)^3} 
&=
-\frac{1}{16}\left\{ \frac{1}{w}+\sum_{n=0}^\infty\left(w^n+(n+1)w^n+\frac{(n+2)(n+1)}{2}w^n\right)\right\}
\\
&=
-\frac{1}{16}\left\{ \frac{1}{w}+\sum_{n=0}^\infty\left(1+(n+1)+\frac{(n+2)(n+1)}{2}\right)w^n\right\}
\\
\\
&=
-\frac{1}{16}\left\{ \frac{1}{w}+\sum_{n=0}^\infty\frac{(n+2)(n+3)}{2}w^n\right\}
\\
\\
\frac{1}{z(z-2)^3} &=
-\frac{1}{16}\left\{ \frac{2}{z}+\sum_{n=0}^\infty\frac{(n+2)(n+3)}{2^{n+1}}z^n\right\}
\end{align}
Finally, let's discuss the approach for the expansion about $z_0=2$. We already have the partial fraction decomposition.
\begin{align}
\frac{-1}{16}\left(\frac{2}{z(1-z/2)^3}\right) &= \frac{-1}{16}\left(\frac{2}{z}+\frac{1}{1-z/2}+\frac{1}{(1-z/2)^2}+\frac{1}{(1-z/2)^3}\right)
\\
&= \frac{2}{(z-2)^3}-\frac{1}{4(z-2)^2}+\frac{1}{8(z-2)}-\frac{1}{8z}
\end{align}
The first few terms are already terms in the Laurent series. This will come down to expanding $1/z$ about $z_0=2$. You should find that this expansion is given by a Taylor series.

Looking over this, the general series for $\frac{1}{w(1-w)^N}$ about $w_0=0$ and $w_0=1$ look fairly easily. From that you can transform over to $\frac{1}{z(z-a)^N}$. Anyway, the two solutions respectively appear to be
$$\frac{1}{w(1-w)^N} = \frac{1}{w}+\sum_{n=0}^\infty \sum_{k=1}^{N} \binom{n+1}{k}w^n$$
$$\frac{1}{w(1-w)^N} = \sum_{n=-N}^{\infty}(-1)^{2n+1}(w-1)^N$$
</s>