Finding the Laurent Series of $f(z)=\displaystyle\frac{z^2 e^{1/z}}{z-1}$ for $0<|z|<1$

Question

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 $$ \begin{array}{ccl} f(z)&=&\displaystyle\frac{z^2}{z-1}\cdot e^{1/z}\\ &=&\displaystyle\left(-\sum_{n=0}^\infty z^{n+2}\right)\cdot\left(\sum_{n=0}^\infty \frac{1}{n! z^n}\right)\\ &=&-\displaystyle\sum_{n=0}^\infty\left(\sum_{k=0}^nz^{k+2}\cdot \frac{1}{(n-k)!z^{n-k}}\right)\\ &=&-\displaystyle\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{z^{2k-n+2}}{(n-k)!}\right) \end{array} $$

Answer 1

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 $n\leq-2$.