Find the coefficient of $z$ in the Laurent series expansion of $\frac{e^z}{z-1}$ in ${|z| > 1}$ I've found this duplicate from '15: Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$, but I think it's wrong, since it looks for the Laurent expansion in ${|z-1|>1}$. 
After developing it by myself, I've reached the following:
$$f(z)=\frac{e^z}{z-1}=\frac{1}{z-1}\cdot e^z$$
Then, $e^z=\sum_{n\geq 0} \frac{z^n}{n!}$ and $$\frac{1}{z-1}=\sum_{n \geq 0} \frac{1}{z^{n+1}}$$.
Here I thought of using the Cauchy Product, but I end up with
$$f(z)=\sum_{n \geq 0} \sum_{k=0}^n \frac{z^{2k-n-1}}{k!}$$ (How can I make this font bigger? The exponent is too small. Sorry about that. The exponent would be $2k-n-1$).
So I have to sum whenever $2k-n-1 = 1$
If I'm not getting it wrong, this means that I have to calculate $$\sum_{n \geq 0} \frac{1}{(1+2n)!}$$ (I'm not sure how to show it better, but I've put the first 3 terms and were $\frac{1}{1!}, \frac{1}{3!}$ and $\frac{1}{5!}$. I'm not sure how to calculate that. The nearest thing I thought was that $$\sum_{n \geq 0} \frac{1}{n!}=e $$. Maybe what I found before could be considered as an answer, but I think I can go one step further.
Thanks 
 A: We denote with $[z^n]$ the coefficient of $z^n$ of a series.

We obtain for $n\in\mathbb{Z}$
  \begin{align*}
[z^n]\frac{e^z}{z-1}&=[z^{n}]\frac{e^z}{z}\cdot\frac{1}{1-\frac{1}{z}}\tag{1}\\
&=[z^{n+1}]\sum_{k=0}^\infty\frac{z^k}{k!}\sum_{j=0}^\infty\frac{1}{z^j}\tag{2}\\
&=\sum_{k=0}^\infty \frac{1}{k!}[z^{n+1-k}]\sum_{j=0}^\infty\frac{1}{z^j}\tag{3}\\
&=\left\{
\begin{array}{ll}
\sum_{k=n+1}^\infty\frac{1}{k!}&\qquad n\geq 0\\
\sum_{k=0}^\infty\frac{1}{k!}&\qquad n< 0\\
\end{array}
\right.\tag{4}\\
&\color{blue}{=}\left\{
\begin{array}{ll}
\color{blue}{e-\sum_{k=0}^n\frac{1}{k!}}&\color{blue}{\quad\, n\geq 0}\\
\color{blue}{e}&\color{blue}{\quad\, n< 0}\\
\end{array}
\right.
\end{align*}

Comment:


*

*In (1) we factor out $z$.

*In (2) and (3) we use $[z^p]z^qA(z)=[z^{p-q}]A(z)$.

*In (4) we select the coefficient of $z^{n+1-k}$.

We conclude the Laurent expansion of $\frac{e^z}{z-1}$ in $|z|>1$ is
  \begin{align*}
\color{blue}{\frac{e^z}{z-1}=e\sum_{n=-\infty}^{-1}z^n + \sum_{n=0}^\infty\left(e-\sum_{k=0}^n\frac{1}{k!}\right)z^n}
\end{align*}

A: We have
$$\frac{e^z}{z-1} = e^z \cdot \frac{1}z \cdot \frac1{1-\frac1z} = \left(\sum_{n=0}^\infty \frac{z^{n}}{n!}\right)\left(\sum_{n=0}^\infty \frac{1}{z^{n+1}}\right)$$
so the coefficient of $z$ is equal to
$$\sum_{n=2}^\infty \frac1{n!} = e-2$$
It is because $z$ is obtained when multiplying the term $\frac{z^{k+2}}{(k+2)!}$ from the first sum, with the term $\frac1{z^{k+1}}$ from the second sum:
$$\frac{z^2}{2!}\cdot \frac1z + \frac{z^3}{3!}\cdot \frac1{z^2} + \frac{z^4}{4!}\cdot \frac1{z^3} + \cdots = z \left(\frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots \right)$$
