Laurent series of $e^{z+1/z}$ What is the Laurent series  of $e^{z+1/z}$?
I had used $$a_k= \frac{1}{2\pi i}\int_c \frac{f(z)}{z^{k+1}}\,dz $$
for a curve $c$ in which we can use $e^z$ as an analytic func. and expanded the $e^{1/z}$ series expansion.

Assuming you are looking for the Laurent series around $z=0$: Since
$$
e^z = 1 + z + \frac{z^2}{2!} + \cdots = \sum_{k=0}^\infty \frac{z^k}{k!}
$$
for all $z \in \mathbb{C}$, we also have
$$
e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \cdots = \sum_{m=0}^\infty \frac{1}{m!z^m}
$$
for all $z\neq 0$. Multiplying the two series together we get the Laurent series we are looking for:
\begin{align}
e^{z+1/z} &= \sum_{n=-\infty}^\infty \left(\sum_{k-m=n} \frac{1}{k!m!} \right) z^n \\
&= \sum_{n=-\infty}^\infty \left(\sum_{m=0}^\infty \frac{1}{(m+n)!m!} \right) z^n
\end{align}
