Find laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}. Find the Laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}.
So far I have the following; I'm not 100% sure if it is right.
$\frac{z-1}{(z-2)(z-3)}$ = $\frac{2}{(z-3)}$-$\frac{1}{(z-2)}$
For $\frac{1}{(z-2)}$=$\frac{1}{z}$ $\frac{1}{1-(\frac{2}{z})}$=$\frac{1}{z}$$\sum_{k=1}^n\frac{2^k}{z^k}$=$\sum_{k=1}^n\frac{2^k}{z^{k+1}}$ for $|z|<1$.
I am having trouble with the other fraction.  I have seen similar questions asked, but I cannot seem to get the information I need.  Any input would be much appreciated!

$$\frac{z-1}{(z-2)(z-3)}=\frac{2}{z-3}-\frac{1}{z-2}=\frac{2}{z-3}-\frac{1}{z}\frac{1}{1-\frac{2}{z}}=$$
$$=\frac{2}{z-3}-\frac{1}{z}\left[1+\frac{2}{z}+\left(\frac{2}{z}\right)^2+...\right]$$
Note that the first summand above is analytic in the given annulus, and the above expansion of $\,\dfrac{1}{z-2}\,$ applies whenever 
$$\left|\frac{2}{z}\right|<1\Longleftrightarrow |z|>2$$
