TITLE: Find Taylor series of $f(z)=\frac{1}{z^2-1}$ for $1<|z+2|<3$
QUESTION [5 upvotes]: find Taylor series of $f(z)=\frac{1}{z^2-1}$ for $1<|z+2|<3$,
No matter what I do I reach the conclusion this series does not exist unless
$|z+2|<1$ but I don't understand why...
I can write  $f(z)=\frac{1}{(z-1)(z+1)}$ and then try to develop the series for each factor separately $f_1(z)=\frac{1}{(z+1)},f_2(z)=\frac{1}{(z-1)}$.
Both analytic  on any $1 \neq z \in \mathbb{C}$, so it must be analytic on the disc 1<|z+2|<3 since $z_0=\pm1$ is not in the disc...
So the disc should be an appropriate Domain to develop a Taylor Series...
But any series I find does not converge in it...
Can someone please explain to me  why is this?

REPLY [2 votes]: $$\frac1{z^2-1} = \frac12 \left (\frac1{z-1} - \frac1{z+1} \right )  = \frac12 \left (\frac1{z+2-3} - \frac1{z+2-1} \right )$$
$|z+2| \gt 1$ and $|z+2| \lt 3$ implies that we approach the expansions of the two terms differently:
$$\frac1{z+2-3} = -\frac13 \frac1{1-\frac{z+2}{3}}  = -\frac13 \sum_{k=0}^{\infty} \left (\frac{z+2}{3} \right )^k$$
$$\frac1{z+2-1} = \frac1{z+2} \frac1{1-\frac1{z+2}} =  \sum_{k=1}^{\infty} \frac1{(z+2)^k} $$
Combine the expansions and you are there.