$\frac{e^{\frac{1}{z}}}{z^2-1}$ for $|z|>1$
I factored out the denominator and rewrote it to a geometric series and got the following expression:
$$e^{1/z}\sum_{n=0}^{\infty}\frac{1}{(z^2)^{n+1}}=\sum_{n=0}^{\infty}\frac{1}{z^nn!}\sum_{n=0}^{\infty}\frac{1}{(z^2)^{n+1}}$$
Now to evaluate the first terms I figured I'd use the Cauchy product which give me:
$$\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}$$
But this isn't correct. What am I doing wrong?
$$\left(1+\frac1z+\frac1{2z^2}+\cdots\right)\left(\frac1{z^2}+\frac1{z^4}+\cdots\right)= \cdots+\left(1\frac1{z^4} + \frac1{2z^2}\frac1{z^2}\right)+\cdots$$