I'm meant to find the laurent series for this function;
$$f(z) = \frac{1}{z^2-1}, 1<|z-2|<3 $$
I started with substituting $$ t=z-2 \implies f(t+2) = \frac{1}{(t+1)(t+3)}$$
And then I did used partial fraction so I have 2 series to work with. My two series looked like this $$\frac{1}{2}\sum_{n=0}^\infty(-1)^n(z-2)^n - \frac{1}{2}\sum_{n=0}^\infty(-\frac{1}{3})^n(z-2)^n$$
Is it correct? When I add them together, I get a new sum which is the wrong answer. The correct answer also involves a sum from negative to positive infinity. I don't understand how, since both of the sums I got start from n=0.