different answer when developing Laurent series

Question

given: $$f(z) = \frac{2}{z^{2}-1}$$ and $$\sum_{n=\infty}^{\infty}{a_ {n}(z-2)^{2}}$$ will be the Laurent series of that function.

I've task to calculate the $a_{-3}+a_{2}$

what I've done so far is: $$\frac{2}{z^2-1}= \frac{1}{z-1}- \frac{1}{z+1}$$ and I got the following sums: $$\frac{1}{z-1} = \sum_{n=0}^{\infty}{(-1)^{n}\cdot (z-2)^{-n-1}}$$ $$\frac{1}{z+1} = \sum_{n=0}^{\infty}{(-3)^{n}\cdot (z-2)^{-n-1}}$$ now, not only I'm not sure how to find $a_{-3}$ and $a_{2}$, but in the official answer the last sum is: $$\frac{1}{z+1} = \sum_{n=0}^{\infty}{\frac{(-1)^{2}}{3^{n+1}}\cdot(z-2)^{n}}$$ is that matter?

Also, in the official answer didn't told me how to find $a_{-3}$ and $a_{2}$, and why I can use $\sum_{0}^{\infty}$ instead of $\sum_{-\infty}^{\infty}$?

the official answer is: $a_{-3}+a_{2}=1-\frac{1}{27}=\frac{26}{27}$

Answer 1

Actually $\frac 1{z-1}= \sum\limits_{n=0}^{\infty} (-1)^{n}(z-2)^{n}$.

Also, $\frac 1 {z+1}=\frac 1 {(z-2)+3}=\frac 1 3 \frac 1 {1+\frac {z-2} 3}=\frac 1 3\sum\limits_{n=0}^{\infty} (z-2)^{n} (-3)^{-n}$. $a_{-3}=0$ and you can read out $a_2$ by subtracting the second series from the first and the dividing by $2$.

Answer 2

I figure out my mistake: this is not true: $$\frac{1}{z+1} = \sum_{n=0}^{\infty}{(-3)^{n}\cdot (z-2)^{-n-1}}$$ since: $$\frac{1}{z+1} = \frac{1}{(z-2)}\cdot\frac{1}{1-(\frac{-3}{z-2})}=\sum_{n=0}^{\infty}{(-3)^{n}\cdot (z-2)^{-n-1}}$$ but: $$\left |\frac{-3}{z-2} \right| >1$$ and therefore, the way i made this series isn't correct.