Finding the Laurent Series of $f(z)=\displaystyle\frac{z^2 e^{1/z}}{z-1}$ for $0<|z|<1$ I have to calculate the Laurent series at the origin of $f(z)=\displaystyle\frac{z^2 e^{1/z}}{z-1}$ for $0<|z|<1$, my idea was to use the Cauchy product, but I don't know if it is correct.
$$\frac{1}{z-1}=-\sum_{n=0}^\infty z^n\Rightarrow \frac{z^2}{z-1}=-\sum_{n=0}^\infty z^{n+2}$$
and also $e^{1/z}=\displaystyle\sum_{n=0}^\infty \frac{1}{n! z^n}$, then we have
$$
\begin{array}{ccl}
f(z)&=&\displaystyle\frac{z^2}{z-1}\cdot e^{1/z}\\
&=&\displaystyle\left(-\sum_{n=0}^\infty z^{n+2}\right)\cdot\left(\sum_{n=0}^\infty \frac{1}{n! z^n}\right)\\
&=&-\displaystyle\sum_{n=0}^\infty\left(\sum_{k=0}^nz^{k+2}\cdot \frac{1}{(n-k)!z^{n-k}}\right)\\
&=&-\displaystyle\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{z^{2k-n+2}}{(n-k)!}\right)
\end{array}
$$
 A: Using the Cauchy-product is fine, but  we should also aim at the  Laurent-series representation $$\sum_{n=-\infty}^{\infty}a_nz^n$$

We obtain
\begin{align*}
\color{blue}{f(z)}&=\frac{z^2e^{1/z}}{z-1}\\
&=-\sum_{k=0}^\infty z^{k+2}\sum_{l=0}^\infty \frac{1}{l!z^l}\\
&=-\sum_{k=2}^\infty z^{k}\sum_{l=0}^\infty \frac{1}{l!z^l}\tag{1}\\
&=-\sum_{n=-\infty}^0\sum_{l=-n+2}^{\infty}\frac{1}{l!}z^n-\sum_{l=1}^\infty\frac{1}{l!}z
-\sum_{n=2}^\infty\sum_{l=0}^{\infty}\frac{1}{l!}z^n\tag{2}\\
&\,\,\color{blue}{=-\sum_{n=-\infty}^0\left(e-\sum_{l=0}^{-n+1}\frac{1}{l!}\right)z^n-(e-1)z-e\sum_{n=2}^\infty z^n}
\end{align*}

Comment: We denote with $[z^n]$ the coefficient of $z^n$  of a series. We calculate the  line (2) by considering $[z^n]f(z)$ in (1) for specific $n$ till we see what's going on.

*

*$[z^0]$: We look at (1) noting that $[z^0]f(z)=\sum_{l=2}^\infty \frac{1}{l!}$.


*$[z^1]$: We obtain $\sum_{l=1}^\infty \frac{1}{l!}$.


*$[z^2]$: We obtain $\sum_{l=0}^\infty \frac{1}{l!}$ and the same results for $[z^n]f(z)$ where $n\geq2$.


*$[z^{-1}]$: We get $\sum_{l=3}^{\infty} \frac{1}{l!}$.


*$[z^{-2}]$: We get $\sum_{l=4}^{\infty} \frac{1}{l!}$ and in general we obtain $[z^n]f(z)=\sum_{l=-n+2}^{\infty} \frac{1}{l!}$ where $n\leq-2$.
