Can a Laurent series be found for $f(z)=\frac{1}{(z+1)(z+2)}$ in the region $0<|z+1|<2$?

Question

I know that a Laurent series can be found for $\frac{1}{(z+1)(z+2)}$ in the region $0<|z+1|<1$, but can a Laurent series be found for $0<|z+1|<2$?

I am confused because in the region $0<|z+1|<1$ the Laurent series is

$$ \frac{1}{(z+1)(z+2)}=\frac{1}{(z+1)}\frac{1}{(1+(1+z))} $$ $$ \frac{1}{(z+1)(z+2)}=\frac{1}{(z+1)}\sum_{n=0}^{\infty}(-1)^n(z+1)^n $$ $$ \frac{1}{(z+1)(z+2)}=\sum_{n=0}^{\infty}(-1)^{n}(z+1)^{n-1} $$

but I am not sure how this will differ for $0<|z+1|<2$.

Answer 1

Hint: Consider the partial fraction decomposition.

Answer 2

Every holomorphic function $f$ on an annulus (or a punctured disk) has a (unique) Laurent series. By uniqueness, however you find the coefficients, if it actually converges to $f$ it is $f$'s Laurent series.

The thing is: is $f$ actually holormophic in that annulus?