As the question title states, I'm tasked with finding the Laurent series expansion for $e^{z+\frac{1}{z}}$ about $z_0=0$. My approach is as follows.
$$ \begin{align*} e^{z+\frac{1}{z}}&=e^ze^{\frac{1}{z}}\\ &=\left(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots\right)\left(1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}\right)\\ &=\cdots+\left(1+\frac{1}{2}+\frac{1}{12}+\cdots\right)z^{-1}+\left(1+1+\frac{1}{4}+\frac{1}{36}+\cdots\right)+\left(1+\frac{1}{2}+\frac{1}{12}+\cdots\right)z+\cdots\\ &=\cdots+\left(\frac{1}{0!1!}+\frac{1}{1!2!}+\frac{1}{2!3!}+\cdots\right) z^{-1} + \left(\frac{1}{0!0!}+\frac{1}{1!1!}+\frac{1}{2!2!}+\cdots\right)+\left(\frac{1}{0!1!}+\frac{1}{1!2!}+\frac{1}{2!3!}+\cdots\right)z+\cdots\\ &=\sum_{n=0}^{\infty}\left(\left[\sum_{m=0}^{\infty}\frac{1}{m!(m+n!)}\right]z^n\right)+\sum_{n=0}^{\infty}\left(\left[\sum_{m=0}^{\infty}\frac{1}{m!(m+n!)}\right]z^{-n}\right) \end{align*} $$
Is this valid? It was the only answer I could settle on, and when I check on Wolfram alpha, what comes out is that the coefficients are related to the modified Bessel function of the first kind (which I'm not familiar with).