Finding the laurent expansion of $\frac{z}{(z-1)(z-2)}$ I want to  find the Laurent series for $\frac{z}{(z-1)(z-2)}$ in the region $1 < |z| < 2$. This implies that $\frac{1}{|z|} < 1$, so noticing that $(z-1) = z(1 - \frac 1 z)$ I can rewrite the desired function as
$$\frac{z}{z(1- \frac 1 z)(z-2)} = \frac{1}{z-2} \cdot \frac{1}{1 - \frac 1 z}.$$
Now using the definition of the geometric series I rewrite it as
$$\sum_{k=0}^{\infty} \frac{1}{z^k(z-2)}$$
Is this the right Laurent series, and if not, where did I go wrong? Please note I am trying to understand where I made a mistake, not simply finding any solution.
I've read a similar question at Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ and one answer uses the fact that $\frac{|z|}2 < 1$, but I am not sure if my method is also valid.

For $1 < |z| < 2$,
\begin{align*}
 \frac{1}{(z-1)(z-2)} &=-\frac{1}{z-1} + \frac{1}{z-2}\\
&=-\frac{1}{z}\left[\frac{1}{1-(1/z)}\right]-\frac{1}{2}\left[\frac{1}{1-(z/2)}\right]\\
&=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{z^n}-\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\\
\end{align*}
\begin{align*}
 \frac{z}{(z-1)(z-2)}
&=-\sum_{n=0}^{\infty} \frac{1}{z^n}-\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^{n+1}\\
&=-1-\sum_{n=1}^{\infty} \frac{1}{z^n}-\sum_{n=1}^{\infty} \left(\frac{z}{2}\right)^{n}\\
&=-\sum_{n=1}^{\infty} \frac{1}{z^n}-\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^{n}\\
\end{align*}
