I have an exercise where I'm supposed to find convergence if any and if there are show how it's... constituted. The series:

$\displaystyle \[

\displaystyle \sum_{n=0}^{\infty} \left(\frac{z-1}{1-2z}\right)^n

\]

$

I'd like help when it comes to ways of solving it. I solve it as a real problem, getting the zeros on the real axis. Then I use these to find the zeros in the img-plane.

$\displaystyle \[

\displaystyle \frac{\left|(\frac{z-1}{1-2z})^{n+1}\right|}{\left|(\frac{z-1}{1-2z})^n\right|}

\]

$ = $\displaystyle \[

\left|(\frac{z-1}{1-2z})^{1}\right|

\]

$ = $\displaystyle \[

\displaystyle \frac{\left|z-1\right|}{\left|1-2z\right|}

$

For what z is the fraction smaller than 1?

$\displaystyle \[

\displaystyle \frac{\left|z-1\right|}{\left|1-2z\right|} < 1

\]$

I get $\displaystyle 0 < z < \frac{2}{3}$ - treating z as a real variable, so centre on x-axis should be $\displaystyle \frac{1}{3}$

Now to the imaginary part.

$\displaystyle \[

\displaystyle \frac{\left|(x+iy)-1\right|}{\left|1-2(x+iy)\right|} < 1

\]$ which gives $\displaystyle \[

\displaystyle \frac{\left|(\frac{1}{3}+iy)-1\right|}{\left|1-2(\frac{1}{3}+iy)\right|} < 1

\]$

I get $\displaystyle y = \pm \frac{i}{3}$ - which goes nicely with the real values.

Centre = $\displaystyle \frac{1}{3}$

Radius = $\displaystyle \frac{1}{3}$

First on all, is this right? And is there a more solid way of doing this?