Find the Laurent expansion of $\dfrac{e^\frac{1}{z}}{z+1}$ in the domain $|z|>0$.
My attempt:
Since we have to work on domain $|z|>0$, let $z=\dfrac{1}{t}$, so I have to find the Laurent expansion of $\dfrac{te^t}{t+1}$ at $t=0$. Particularly, I have to find the Laurent expansion of $\dfrac{e^t}{t+1}$ at $t=0$.
I already knew that $$e^t=\sum\limits_{0}^{\infty}\dfrac{t^n}{n!},\quad\forall t\in\mathbb{C}$$
$$\dfrac{1}{t+1}=\sum\limits_{n=0}^{\infty}(-1)^nt^n,\quad\forall |t|<1.$$
But I don't know how to find the product of these two series. Could someone help me pr have another way to deal with this problem? Thanks in advance!
The product of those two series – and they are really Taylor rather than Laurent swifties in $t$ – is the convolution: $$\frac{e^t}{1+t}=\sum_{n=0}^\infty t^n\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}$$ Thus the Laurent series of the original function is $$\frac{e^{1/z}}{1+z}=\sum_{n=0}^\infty(1/z)^{n+1}\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}$$