Laurent Series  $\exp(1/z)/(1-z)$ I need some help finding the Laurent expansion and residue of
$$\dfrac{\exp \left(\frac1z \right)}{(1-z)}$$
So far I've done
$$\sum_{j=0}^\infty \frac{z^{-j}}{j!} \sum_{k=0}^\infty z^k = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}$$
but don't know where to go from here. And is it also possible to use Cauchy product when one of the powers is $<0$ and the other is $>0$?
 A: As for the residue: going "naive" may be a good idea$$\frac{1}{1-z}\,e^{\frac{1}{z}}=\left(1+z+z^2+z^3+...+z^n+...\right)\left(1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+...+\frac{1}{n!z^n}+...\right)$$
Well, it seems fairly easy to see what products are going to give us the coefficient of $\,z^{-1}\,$:
(first term left) times (second term right), (second left) times (third right),...,(n-th left) times ((n+1)-th right),..., so:$$\frac{1}{z}+\frac{1}{2z}+\frac{1}{6z}+...+\frac{1}{n!z}+...=\frac{1}{z}\sum_{n=1}^\infty\frac{1}{n!}=\frac{1}{z}(e-1)$$
A: You can rewrite the series in the form: 
$$\sum_{j=0}^\infty \sum_{k=0}^\infty \frac{z^{k-j}}{j!}=\sum_{n=1}^{\infty}\left(\sum_{j=n}^{\infty}\frac{1}{j!}\right)z^{-n}+\sum_{n=0}^{\infty}\left(\sum_{j=0}^{\infty}\frac{1}{j!}\right)z^n=\sum_{n=1}^{\infty}\left(e-\sum_{j=0}^{n-1}\frac{1}{j!}\right)z^{-n}+\sum_{n=0}^{\infty}ez^n$$
So the residue at $0$ is the coefficient of $z^{-1}$, which is $\sum_{j=1}^\infty\frac{1}{j!}=e-1$.
A: Another approach is to take the circle $\gamma=\partial B_\rho(0)$ with $\rho\in(0,1)$ and get the residue using the integral definition which yields
$$
\begin{align*}
\operatorname{Res}_{z=0}\left(\frac{\exp\left(\frac{1}{z}\right)}{1-z}\right)
&= \frac{1}{2\pi\mathrm i}\oint_\gamma \exp\left(\frac{1}{z}\right)(1+z+z^2+\ldots)\,\mathrm dz\\
&=\frac{1}{2\pi\mathrm i}\oint_\gamma \exp\left(\frac{1}{z}\right)\,\mathrm dz + \frac{1}{2\pi\mathrm i}\oint_\gamma z\exp\left(\frac{1}{z}\right)\,\mathrm dz+\ldots\\
&=\frac{1}{1!}+\frac{1}{2!}+\ldots=\mathrm e-1.
\end{align*}
$$
