/sci/ - Science & Math

I need to find the Lauren Series of [math]\frac{\mathrm{exp(\frac{1}{z^2})}}{z-1} \text{ about } z=0[/math].

[math] \frac{\mathrm{exp(\frac{1}{z^2})}}{z-1}=\frac{1}{z-1}\mathrm{exp(\frac{1}{z^2})} = (-1-z-z^2-z^3-...)(1+\frac{1}{z^2}+\frac{1}{2z^4}+...)= \\ \sum\limits_{k=0}^{\infty}{-z^k}\sum\limits_{j=0}^{\infty}{\frac{1}{j!z^{2j}}} = \sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{\infty}{\frac{z^{k-2j}}{j!}}[/math]

now what do?


Answer they give is

[math]\sum\limits_{k=-\infty}^{\infty}{a_{k}z^{k}}[/math] where
[math]\\ a_{k}=\begin{cases}
-e \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ if } \ k\ge0 \\
-e+1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(j-1)!} \ \ \ \ \ \ \ \ \ \ \text{ if } k=-2j \text{ or } k=-2j+1, j=1,2,...\
\end{cases}[/math]