Simplification of a series so that it converges to a given function  I am trying to rearrange the series 
$ \frac{1}{1-z} - \frac{(1-a)z}{(1-z)^2} +  \frac{(1-a)^2z^2}{(1-z)^3} - \cdots$ 
In such a way that I can show it converges to 
$\frac{1}{1-az} $


 What I have so far 
 Let $ w = \frac{z}{1-z} $, we can then write the series as

$ \frac{w}{z} \left ( 1 - (1-a)w + (1-a)^2 w^2 - \cdots \right ) $
 Which is the taylor series expansion about $0$ of 
$ \frac{w}{z} \frac{1}{1+(a-1)w} $ 
 Which I can simplify down to 
$ \frac{1}{1+(a-2)z}$  as follows 
$ \frac{w}{z} \frac{1}{1+(a-1)w} = \frac{1}{1-z} \left ( \frac{1}{1 + (a-1)\frac{z}{1-z}}\right) $ 
$= \frac{1}{1-z} \left ( {\frac{1-z + az - z}{1-z}}\right)^{-1}  $ 
$ = \frac{1}{1+(a-2)z} $

Obviously this is not the same as $\frac{1}{1-az} $. I would love any guidance people can give as to where I went wrong. 

The mistake lies in this step:
$$\frac{w}{z}\left(1-(1-a)w+(1-a)^2w^2-\cdots\right)=\frac{w}{z}\cdot\frac{1}{1+(a-1)w}$$
Recall that $1+x+x^2+\cdots=\frac{1}{1-x}$. If we let $x=-(1-a)w$, then your series actually becomes:
$$\frac{w}{z}\left(1+x+x^2+\cdots\right)=\frac{w}{z}\cdot\frac{1}{1-x}=\frac{w}{z}\cdot\frac{1}{1+(1-a)w}$$
which is slightly different to what you wrote. If you continue from here, you will reach the desired answer:
\begin{align*}
\frac{w}{z}\cdot\frac{1}{1+(1-a)w} &= \frac{1}{1-z}\cdot\frac{1}{1+(1-a)\frac{z}{1-z}} \\
&= \frac{1}{(1-z)+(1-a)z} \\
&= \frac{1}{1-az}.
\end{align*}
