Laurent series of $e^{e^{\frac{1}{z}}}$ around $z=0$

Question

Actually I need only the $res(f;0)$ where $f = e^{e^{\frac{1}{z}}}$
I thought of finding the Laurent series of $e^{e^{\frac{1}{z}}}$ around $z=0$
Any other Ideas if you have ?

Answer 1

$$\exp{\left ( e^{1/z} \right )} = \sum_{k=0}^{\infty} \frac{e^{k/z}}{k!} = \sum_{k=0}^{\infty} \frac1{k!} \sum_{m=0}^{\infty}\frac{k^m}{m!} \frac1{z^m}=\sum_{m=0}^{\infty} \frac1{m! z^m} \sum_{k=0}^{\infty} \frac{k^m}{k!}$$

The residue of $\exp{\left ( e^{1/z} \right )}$ at $z=0$ is simply the coefficient of $1/z$ in the Laurent expansion about $z=0$, which is simply read off the from the $m=1$ term of the last sum on the right:

$$\frac1{1!} \sum_{k=0}^{\infty}\frac{k^1}{k!} = e$$

Answer 2

Using $f(w) = e^{w} = \sum_{n=0}^{\infty} \frac{w^{n}}{n!}$, We have

$$f(1/z) = e^{1/z} = \sum_{n=0}^{\infty} \frac{\frac{1}{z^{n}}}{n!} = \sum_{n=0}^{\infty}\frac{1}{z^{n}n!}.$$

Then

$$e^{e^{\frac{1}{z}}} = \sum_{m=0}^{\infty} \frac{1}{m!}\left(\sum_{n=0}^{\infty}\frac{1}{z^{n}n!}\right)^{m}.$$

We have

$$\begin{cases} 1 + \frac{1}{z} + \mathcal{O}(z^{-2}) &m=1\\ \frac{1}{2!}\left(1 + \frac{2}{z}\right) + \mathcal{O}(z^{-2}) & m=2\\ \frac{1}{3!}(1+\frac{3}{z}) + \mathcal{O}(z^{-2}) & m=3 \end{cases}$$

and in general, collecting the coefficients of $z^{-1}$, we arrive at

$$\operatorname{Res}(f,0) = \sum_{k=1}^{\infty} \frac{k}{k!} = \sum_{k=1}^{\infty}\frac{1}{(k-1)!} = e^{1} = e$$