<s>[INST] 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. 
 [/INST] Even though it has already been answered, I might aswell post my already typed solution:
Your expresion is
$$\frac{1}{1-z} \cdot \sum_{n=0}^\infty \left(-\frac{1-a}{1-z}z\right)^n$$
Using the geometric series, this is 
$$\frac{1}{1-z} \cdot \frac{1}{1+\frac{1-a}{1-z}z}$$
which is
$$\frac{1}{1-z} \cdot \frac{1-z}{1-z+(1-a)z} = \frac{1}{1-az}$$
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