Expand $f(z) = \frac{1}{(z+1)(z+2)}$ in the region $1<|z|<3$.
Here's my working: By partial fractions,
$f(z) = \frac{1}{z+1}-\frac{1}{z+2}=\frac{1}{z(1+1/z)}-\frac{1}{z(1+2/z)}$
Now I can expand the first term since |1/z|<1, but according to the domain of convergence |2/z|>1, so how do I deal with the second term? (Because $(1+x)^{-1}=1-x+x^2-x^3+...$ only when $|x|<1\ $)
For $|a/z|<1$ or equivalently $|z|>|a|$,$$\frac{1}{z-a} = \frac{1}{z}\cdot\frac{1}{1-\frac{a}{z}}=\frac{1}{z}\left(1+\frac{a}{z}+\frac{a^2}{z^2}+ \cdots\right)=\frac{1}{z}+\frac{a}{z^2}+\frac{a^2}{z^3}+ \cdots$$
For $|a/z|>1$ or equivalently $|z|<|a|$,$$\frac{1}{z-a} = -\frac{1}{a}\cdot\frac{1}{1-\frac{z}{a}}=-\frac{1}{a}\left(1+\frac{z}{a}+\frac{z^2}{a^2}+ \cdots\right)=-\frac{1}{a}-\frac{z}{a^2} -\frac{z^2}{a^3}+ \cdots$$