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. 
 A: Laurent series and partial fractions go hand in hand, we can find the partial fraction decomposition using the singular parts of the Laurent expansions, while the partial fraction decomposition will yield the complete Laurent expansion. The sum of all the singular parts of the Laurent expansions around all the poles must equal to rational function, because the difference between the rational function and this sum only has removable singularities, therefore it must be a polynomial. But this polynomial tends to zero (this is the case if the numerator is of lower degree than the denominator), therefore the difference is zero. This means that the sum of the singular parts of the Laurent expansions is in fact the partial fraction decomposition.
I this case, we can easily see that the singular part of the Laurent expansion around $z = 0$ is given by
$$\frac{-1}{8z}\tag{1}$$
The singular part of the Laurent expansion around $z = 2$ can be found by noting that the rational function asymptotically tends to zero as a constant times $z^{-4}$ for large $|z|$. If we put $z = 2+t$, then we have that the function for large $|t|$ behaves like $t^{-4}$, so if we expand (1) in negative powers of $t$ to order $t^{-3}$ then this must yield minus the contribution to the partial fraction expansion from the singularity at $z = 2$, because terms of lower order than $t^{-4}$ must cancel. Expanding minus Eq. (1) for large $|t|$ yields:
$$\frac{1}{8(2+t)} = \frac{1}{8t} - \frac{1}{4t^2} + \frac{1}{2t^3}+\cdots$$
Therefore the partial fraction expansion is:
$$\frac{1}{8}\frac{1}{z-2}- \frac{1}{4}\frac{1}{(z-2)^2} + \frac{1}{2}\frac{1}{(z-2)^3} - \frac{1}{8z}$$
The complete Laurent expansion is now easily obtained by expanding this in powers of $z$ using the formulas for the geometric series and its derivatives.
A: 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
$$\frac{1}{w(1-w)^N} = \frac{1}{w}+\sum_{n=0}^\infty \sum_{k=1}^{N} \binom{n+1}{k}w^n$$
$$\frac{1}{w(1-w)^N} = \sum_{n=-N}^{\infty}(-1)^{2n+1}(w-1)^N$$
