TITLE: Residue of $\exp(z-z^{-1})$ at $z=0$
QUESTION [5 upvotes]: I'm interested in the residue of $\exp(z-z^{-1})$ at $z=0$.
We have $(z-z^{-1})^n=\sum_{k=0}^n\binom{n}{k}z^{n-k}(-z)^{-k}=\sum_{k=0}^n \binom{n}{k}(-1)^k z^{n-2k}$ so
$\exp(z-z^{-1})=\sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^n \binom{n}{k}(-1)^k z^{n-2k}$
But as I see it, I don't really have a chance to get the coefficient $a_{-1}$ since for every odd $n$ there is a $k$ such that $n-2k=-1$.
Edit: A non-closed form of the residue would be $\sum_{n\ \text{odd}}\frac{(-1)^{\frac{n+1}{2}}}{\left(\frac{n+1}{2}\right)!\left(\frac{n+-1}{2}\right)!}$

REPLY [3 votes]: $$
\begin{align}
\left[z^{-1}\right]\sum_{k=0}^\infty\frac{\left(z-z^{-1}\right)^k}{k!}
&=\left[z^{-1}\right]\sum_{k=0}^\infty(-1)^k\frac{\left(1-z^2\right)^k}{z^k\,k!}\\
&=\sum_{\substack{k=0\\k\text{ odd}}}^\infty\frac{(-1)^k}{k!}(-1)^{\frac{k-1}2}\binom{k}{\frac{k-1}2}\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)!}\binom{2k+1}{k}\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{k!(k+1)!}
\end{align}
$$