The binomial expansion of $\frac{1}{z-2}$ is $-\frac{1}{2}\sum^\infty_{n=0} (-1)^n \left(\frac{z}{2}\right)$.
Does this converge for $\left|\frac{z}{2}\right|<2$?
Does $\frac{1}{z} \sum^\infty_{n=0} (-1)^n \left(\frac{2}{z}\right)^n$, converge for $|z|>1$?
I don't know where your expanding this series at but here is a good rule to help you:
$|z-\alpha|<|\alpha-\beta|$
Where $\beta$ is the nearest singularity which in this case is $2$. So, if $\alpha$ is the point in which you expanded your binomial series at a point $\alpha$ then the series converges for:
$|z-\alpha|<|\alpha-2|$
Please comment on my post and tell me where you are expanding your binomial series at, then I'll solve it further.