Laurent series for $f(z) = \exp(z+\frac{1}{z})$ around $0$ I need to find the Laurent series of the following function around $0$ - $$f(z) = \exp(z+\frac{1}{z})$$
Now by power series expansion, I got $$f(z) = \sum_{m=0}^{\infty} \frac{z^m}{m!} \sum_{k=0}^{\infty}\frac{1}{k! z^k}$$ 
Now I know I am supposed to use the concept of Cauchy product of power series but I am not much comfortable with that. So, can someone explain to me how to perform the product here ?
 A: The function $f(z)=\exp(z+1/z)$ is analytic in the annular region
$0<|z|<\infty$.
We can represent the factors $\exp(z)$ and $\exp(1/z)$ as Laurent
series over the annulus $0<|z|<\infty$, viz.,
$\exp(z)=\sum_{k=-\infty}^\infty a_k z^k$ and
$\exp(1/z)=\sum_{k=-\infty}^\infty b_k z^k$, where
$$a_k=\begin{cases} 1/k!&\text{for $k\ge0$}\\
                       0&\text{for $k<0$}\,,\end{cases}$$ and
$$b_k=\begin{cases} 0&\text{for $k>0$}\\
             1/(-k)!&\text{for $k\le0$}\,.\end{cases}$$
The Laurent series of the product is 
$f(z)=\sum_{k=-\infty}^\infty c_k z^k$, where
$c_n=\sum_{k=-\infty}^\infty a_k b_{n-k}$.
The latter simplifies to $c_n=\sum_{k=0}^\infty\frac1{k!} \frac1{(n+k)!}$.
Further, by comparison with the series expression for the modified Bessel 
function of the first kind,
$$I_n(z)=\sum_{k=0}^\infty{(\frac12 z)^{n+2k}\over k!\,\Gamma(n+k+1)}\,,$$
it follows that $c_n=I_n(2)$.
Thus we have that
$$\exp\left(z+\frac1z\right)=
I_0(2)+\sum_{n=1}^\infty I_n(2)\left(z^n+{1\over z^n}\right).$$
