How to find the Laurent expansion for $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$? I want to find the Laurent expansion for  $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$,
I've tried to apply this formula $\frac{1}{1-\omega}=\sum_{n=0}^{\infty }\omega^{n}$ and the usual Taylor series of the exponential function, but I don't know how to continue:
$$\begin{align}f(z)&=\frac{1}{z-1}\exp\left(\frac{1}{z^{2}}\right)\\
&=-\frac{1}{1-z}\exp\left(\frac{1}{z^{2}}\right)\\&=-\left (\sum_{n=0}^{\infty }z^{n}  \right )\left ( \sum_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )\end{align}$$
Thanks in advance.
Ps: I tried applying a Cauchy product, but I think this is not appropriate.
Edit 1: If it is useful at the end of the text, the authors say that the Laurent expansion is:
$\sum_{k=-\infty }^{\infty }a_{k}z^{k}$ with $a_{k}=-e$ if $k\geq 0$ and
$a_{k}=-e+1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(j-1)!}$if $k=-2$ or $k=-2j+1$ where $j=1,2,...$
 A: Starting with your $=-\left (\sum\limits_{m=0}^{\infty }z^{m}  \right )\left ( \sum\limits_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )$ changing one of the $n$ to $m$, you can say the coefficient of $z^k$ is

*

*$-\sum\limits_{n=0}^{\infty} \frac1{n!} =-e$ when $k\le 0$

*$-\sum\limits_{n=k/2}^{\infty} \frac1{n!} =\sum\limits_{n=0}^{n=(k-2)/2} \frac1{n!}-e$ when $k\gt 0$ and even

*$-\sum\limits_{n=(k+1)/2}^{\infty} \frac1{n!} =\sum\limits_{n=0}^{(k-1)/2} \frac1{n!}-e$ when $k\gt 0$ and even

But that looks wrong to me:  I do not think $$\cdots -e z^{-5} -e z^{-4} -e z^{-3} -e z^{-2} -e z^{-1} -e z^{0}+ \\(1-e)z^1 +(1-e)z^2 +(2-e)z^3 +(2-e)z^4+\left(\frac52-e\right)z^5+\cdots$$ converges when $|z| \le 1$.
Meanwhile for the same question asked elsewhere, a suggested answer was in effect $$z^{-1}+z^{-2}+2 z^{-3}+2 z^{-4}+\frac{5 }{2}z^{-5}+\frac{5}{2}z^{-6}+\cdots$$ but I do not think that converges either when $|z|\le 1$
A: First, we can write two series for $\frac1{z-1}$ in the two regions $|z|<1$ and $|z|>1$ as
$$\frac1{z-1}=\begin{cases}
-\sum_{n=0}^\infty z^n&,|z|<1\\\\
\sum_{n=1}^\infty z^{-n}&,|z|>1\tag1
\end{cases}$$

Second, the Laurent series for $e^{1/z^2}$ for $0<|z|$ is given by
$$e^{1/z^2}=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\tag2$$
where $a_n$ the sequence such hat
$$a_n=\begin{cases}
1&,n\,\text{even}\\\\
0&,n\,\text{odd}
\end{cases}$$

Putting $(1)$ and $(2)$ together reveals
$$\frac{e^{1/z^2}}{z-1}=
\begin{cases}
-\sum_{m=0}^\infty z^m \sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,0<|z|<1\tag3\\\\
\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,1<|z|
\end{cases}
$$


For $|z|>1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written
$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,\sum_{m=1}^\infty z^{-(n+m)}\\\\
&\overbrace{=}^{p=n+m}\sum_{n=0}^\infty  \frac{a_n}{(n/2)!}\sum_{p=n+1}^\infty\,z^{-p}\\\\
&=\sum_{p=1}^\infty\left(\sum_{n=0}^{p-1}  \frac{a_n}{(n/2)!}\right)\,z^{-p}
\end{align}$$


For $0<|z|<1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written
$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=-\sum_{m=0}^\infty z^{m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{m=0}^\infty z^{m-n}\\\\
&\overbrace{=}^{p=m-n}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=-n}^\infty z^{p}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\left(\sum_{p=-n}^{0} z^{p}+\sum_{p=1}^\infty z^{p}\right)\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=0}^{n} z^{-p}\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{p=0}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}\\\\
&=-e \sum_{p=0}^\infty z^{p}-\sum_{p=1}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}
\end{align}$$

