Expand the function $$f(z) = \frac{1}{z^2(z-2)}$$ into a Laurent series which converges on a region $0<|z|<R$, and determine the greatest possible value of $R$.
We can see that we need a Laurent Series about the point $z=0$.
What I tried was \begin{align}\frac{1}{z^2(z-2)} &= \frac{1}{z^3 -2z^2} \\ &= \frac{1}{z^3} \bigg[ \frac{1}{1-\frac{2}{z}}\bigg] \\ &= \frac{1}{z^3} \sum_{n=0}^\infty\bigg( \frac{2}{z}\bigg)^n \\ &= \sum_{n=0}^\infty\frac{2^n}{z^{n+3}}\end{align}
When I use the ratio test, I get \begin{align}\lim_{n \to \infty} \bigg| \frac{2^{n+1} \cdot z^{n}}{z^{n+1}\cdot2^n}\bigg| &= \lim_{n \to \infty} \bigg|\frac{2}{z}\bigg| \\ &=\bigg|\frac{2}{z}\bigg| \end{align} Now for the radius of convergence, we require $$\bigg|\frac{2}{z}\bigg| <1$$
This does not seem to be correct.
When I use WolframAlpha, I get a completely different (negative) Laurent Series with radius of convergence $|z| < 2$, which makes a lot more sense.
Where did I go wrong?
EDIT: HERE IS MY RE-ATTEMPT
I retried it using the following
\begin{align}f(z) &= \frac{1}{z^2}\cdot \frac{1}{z-2} \\ &= \frac{1}{z^2} \cdot \frac{-1}{2} \cdot \frac{1}{1-\frac{z}{2}} \\ &= \frac{1}{z^2} \cdot \frac{-1}{2} \sum_{n=0}^\infty \bigg( \frac{z}{2} \bigg)^n \\ &= - \sum_{n=0}^\infty \frac{z^{n-2}}{2^{n+1}}\end{align}
This seems to have radius of convergence $|z|<2$. Is this correct??