Residue of $\exp(z-z^{-1})$ at $z=0$ 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 $\sum_{n\ \text{odd}}\frac{(-1)^{\frac{n+1}{2}}}{\left(\frac{n+1}{2}\right)!\left(\frac{n+-1}{2}\right)!}$
 A: The residue is given by
$$ \frac{1}{2\pi i}\int_{|z|=1} e^{z-1/z} \, dz, $$
which is probably easier to evaluate than a series. Putting $z=e^{i\theta}$ gives
$$ \frac{1}{2\pi}\int_{-\pi}^{\pi} \exp{(e^{i\theta}-e^{-i\theta})} e^{i\theta} \, d\theta = \frac{1}{2\pi}\int_{0}^{2\pi} e^{2i\sin{\theta}} e^{i\theta} \, d\theta $$
This is a Fourier integral, of course. What is the Fourier expansion of $e^{2i\sin{\theta}}$? The answer is provided by the Jacobi–Anger expansion,
$$ e^{iz\sin{\theta}} = \sum_{n = -\infty}^{\infty} J_n(z) e^{in\theta}, $$
from which we see that the integral pulls out the $n=-1$ coefficient, $J_{-1}(2)=-J_1(2)$, and the answer can't be expressed more simply.
A: $$
\begin{align}
\left[z^{-1}\right]\sum_{k=0}^\infty\frac{\left(z-z^{-1}\right)^k}{k!}
&=\left[z^{-1}\right]\sum_{k=0}^\infty(-1)^k\frac{\left(1-z^2\right)^k}{z^k\,k!}\\
&=\sum_{\substack{k=0\\k\text{ odd}}}^\infty\frac{(-1)^k}{k!}(-1)^{\frac{k-1}2}\binom{k}{\frac{k-1}2}\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)!}\binom{2k+1}{k}\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{k!(k+1)!}
\end{align}
$$
