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.
Best Answer
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, $$\frac{e^z}{z-1} = e \sum_{k=-\infty}^{-1} z^k+ \sum_{k=0}^\infty\frac{\gamma(k+1,1)}{k!} z^k.$$