Simplification of a series so that it converges to a given function

Question

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.

Answer 1

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*}

Answer 2

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

$$\frac{1}{1-z} \cdot \frac{1-z}{1-z+(1-a)z} = \frac{1}{1-az}$$