explicit formula for coefficients of Laurent series Give an explicit formula for the coefficients of the Laurent series on $A:=\{z: |z|>1\}$ for the function $g(z)=\frac{e^z}{z-1}$.
I know how to  go about finding the Laurent series, using the geometric series for $\frac{1}{1-z}$. But how do I obtain an explicit formula for ALL the coefficients?.
I have seen some earlier posts,but I am not entirely clear on their work. Can we use the coefficient formula in the Laurent series expansion and then use Cauchy's formula?.
Any help is appreciated.
 A: You can at first determine the Laurent series for
$$ e^{z} = \sum_{m=0}^\infty \frac{z^m}{m!}$$
and
$$ \frac1{z-1} = \sum_{n=0}^\infty z^{-(n+1)} , \qquad |z|>1$$
independently. 
To multiply two Laurent series, we can use this formula or simply calculate
$$\begin{align}\frac{e^{z}}{z-1}= &\left(\sum_{m=0}^\infty \frac{z^m}{m!} \right) 
 \left(\sum_{n=0}^\infty z^{-(n+1)}\right)\\
&=\sum_{m,n\in\mathbb{Z}}[0\leq m][0\leq n] \frac{z^{m-n-1}}{m!}\\
&=\sum_{k,m} [0\leq m][0\leq m-k-1] \frac{z^{k}}{m!}\\
&= \sum_{k=-\infty}^{-1}  z^k \underbrace{\sum_{m=0}^\infty \frac1{m!}}_{e}
+ \sum_{k=0}^\infty z^k \sum_{m=1+k}^\infty \frac1{m!}.
\end{align}$$
with $k=m-n-1$ and where I used Iverson's bracket.
Edit: As Robert Israel pointed out, we can still express the last sum using the incomplete $\gamma$-function
$$\gamma(n,x) = \int_0^x t^{n-1} e^{-t} dt = x^n (n-1)! e^{-x} \sum_{k=0}^\infty \frac{x^k}{(n+k)!} .$$
Setting $x=1$, we have with $n\geq 1$
$$\gamma(n,1) = (n-1)! \sum_{i=n}^\infty \frac{1}{i!}.  $$
Thus,
$$\frac{e^z}{z-1} = e \sum_{k=-\infty}^{-1} z^k+ \sum_{k=0}^\infty\frac{\gamma(k+1,1)}{k!} z^k.$$
A: First observe
$$\frac{1}{1-z}=-\frac1z\frac{1}{1-\frac1z}=\sum_{-\infty}^{n=-1}-z^n$$
and$$e^z=\sum_{k=0}^{\infty}\frac{z^k}{n!}$$
The general term of the first series is $a_n=0$, $n\ge 0$ and $a_n=-1$, $n<0$ while for the second $b_n=\frac1{n!}$ for $n\ge 0$ and $b_n=0$ for $n<0$.
$$\sum_{n=-\infty}^{\infty}a_n\sum_{n=-\infty}^{\infty}b_n=\sum_{n=-\infty}^{\infty}c_n$$
where $$c_n=\sum_{k=-\infty}^{\infty}a_kb_{n-k}$$
Can you continue?
