Cauchy product of two series I‘m trying to understand this equality:
$\displaystyle e^{z+z^{-1}}=\sum_{n=-\infty}^\infty \Big(\sum_{m=0}^{\infty} \frac{1}{(m+n)!m!}\Big)z^n$
(The sum on the right side then turns out to be some kind of Bessel function)
I think the equality follows from the individual Taylor series:
$\displaystyle e^{z+z^{-1}}=\Big(\sum_{n=0}^\infty \frac{z^n}{n!}\Big) \Big(\sum_{n=0}^\infty \frac{1}{z^nn!}\Big)$
Then I tried, using the Cauchy product, to get the final representation above, but I somehow can‘t do it. Any help would be appreciated!

We have $$e^z e^{z^{-1}}=\sum_j \frac{z^j}{j!}\sum_k\frac{z^{-k}}{k!}=\sum_{j,k}\frac{z^{j-k}}{j!k!}=\sum_{n,k}\frac{z^n}{(n+k)!k!}=\sum_n\left(\sum_k\frac{1}{(n+k)!k!}\right)z^n$$
where $n=j-k$ so $j=n+k$. The sums are taken over $j$, $k$ and $n$ varying on $\Bbb Z$.
