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.
$$ \begin{align} e^ze^{\frac1z} &=\sum_{k=0}^\infty\frac{z^k}{k!}\sum_{j=0}^\infty\frac1{j!z^j}\\ &=\sum_{k=-\infty}^\infty z^k\color{#0000F0}{\sum_{j=0}^\infty\frac1{(k+j)!j!}}\\ &=\sum_{k=-\infty}^\infty\color{#0000F0}{I_{|k|}(2)}\,z^k \end{align} $$ where it is convention that $\frac1{k!}=0$ when $k<0$.
$I_n$ is the Modified Bessel Function of the Second Kind.
Why is the Sum in $\boldsymbol{j}$ Even in $\boldsymbol{k}$?
In the preceding, it is stated that $$ e^{z+\frac1z}=\sum_{k=-\infty}^\infty I_{|k|}(2)z^k $$ which implies that $\sum\limits_{j=0}^\infty\frac1{(k+j)!j!}$ is even in $k$; and in fact it is, because of the convention, mentioned above, that $\frac1{k!}=0$ when $k<0$.
For $k\ge0$, we have
$$
\begin{align}
\sum_{j=0}^\infty\frac1{(-k+j)!j!}
&=\sum_{j=k}^\infty\frac1{(-k+j)!j!}\tag{1a}\\
&=\sum_{j=0}^\infty\frac1{j!(j+k)!}\tag{1b}
\end{align}
$$
Explanation:
$\text{(1a)}$: the terms with $j\lt k$ are $0$ due to the convention
$\text{(1b)}$: substitute $j\mapsto j+k$