Find Laurent series Let $$f(z):=\frac{e^{\frac{1}{z}}}{z^2+1}$$ and let $$\sum_{k}a_kz^k$$ with k in Z the Laurent series of $f(z)$ for $0<|z|<1$. I have to find a formula for $a_k$. I've tried a lot, but I'm stuck. Can somebody help me? I want to do it with ordening of absolute series.

$$\frac{e^{1/z}}{1+z^2}=\left(1+\frac1z+\frac1{2z^2}+\ldots\right)\left(1-z^2+z^4-z^6+\ldots\right)=$$
$$=\sum_{k=0}^\infty\frac{z^{-k}}{k!}\sum_{k=0}^\infty(-1)^kz^{2k}$$
Now multiply within the common convergence radius...
