<s>[INST] Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stuck. The general idea I know is to get each of the 3 terms above into the form $D\cdot \frac{1}{1-z}$, where $D$ is a constant. But for the first term, it would be $\frac{1}{4}\cdot\frac{1}{(0-z)}$. I can't get rid of the zero. 
 [/INST] HINT:
Write $\frac1z$ as
$$\frac1z=\frac{1/2}{1+(z-2)/2}$$
Then, recall the sum of a geometric series.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

$$\begin{align}\frac1z&=\frac{1}{2+(z-2)}\\\\&=\frac{1/2}{1+(z-2)/2}\\\\&=\frac12 \sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n\end{align}$$for $|z-2|<2$.  Therefore, the Laurent series is given by $$\frac12 (z-2)^{-2}-\frac14 (z-2)^{-1}+\frac18 \sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n=\frac18\sum_{n=-2}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n$$

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