<s>[INST] Expand $\frac{1}{z-2}$ on $|z|>2$ 
Expand $\frac{1}{z-2}$ around $|z|>2$

So there is no singularity points 
$$\frac{1}{(z-2)}=-\frac{1}{2-z}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{z^n}{2^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{n+1}}$$
but we can also write:
$$\frac{1}{(z-2)}=\frac{1}{z}\frac{1}{1-\frac{2}{z}}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{2^n}{z^n}=\sum_{n=0}^{\infty} \frac{2^n}{z^{n+1}}$$
So which one is true? the second one as for the first we get $|\frac{z}{2}|<1\Rightarrow |z|<2$ and we are looking at $|z|>2$?
 [/INST] You already answered your own question:
To expand
$$\frac{1}{1-q}$$
as you did in both cases, you need $|q|<1$ for the series to converge. Thus, if $|z|<2$, the first one is right, whereas for $|z|>2$, the second one is right.
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