Find the first three Laurent expansion terms of $\frac{1-z}{z^2} e^z$ I've this function $f(z)=\frac{1-z}{z^2} e^z$ and I've to find the first three Laurent expansion terms in $z=0$.
I've proceeded in this way:
First of all I've considered the expansion series of $e^z = \sum_{n=0}\frac{z^n}{n!}$ and I found the first three terms, so $$e^z=1+z+\frac{z^2}{2}+o(z^3)$$
At this point my function would be
$$f(z)=\frac{1}{z^2}(1-z)(1+z+\frac{z^2}{2}+o(z^3))$$
$$f(z)= \frac{1}{z^2}-\frac{1}{2}-\frac{z}{2}+o(z^3)$$
and it seems right, but using Wolfram Alpha, I've discovered that should be
$f(z)=\frac{1}{z^2}-\frac{1}{2}-\frac{z}{3}-\frac{z^2}{8}-\frac{z^3}{30}+o(z^4)$.
As you can see, the third term is different. What I've to do?

$$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
So:
$$\frac{e^z}{z^2}=\sum_{n=0}^{\infty}\frac{z^{n-2}}{n!}=\sum_{n=-2}^{\infty}\frac{z^n}{(n+2)!}=\frac{1}{z^2}+\sum_{n=-1}^{\infty}\frac{z^n}{(n+2)!}$$
and
$$\frac{e^z}{z}=\sum_{n=0}^{\infty}\frac{z^{n-1}}{n!}=\sum_{n=-1}^{\infty}\frac{z^n}{(n+1)!}$$
Thus, since both converges we can subtract them
$$e^z\frac{1-z}{z^2}=\frac{e^z}{z^2}-\frac{e^z}{z}=\frac{1}{z^2}+\sum_{n=-1}^{\infty}\frac{z^n}{(n+2)!}-\sum_{n=-1}^{\infty}\frac{z^n}{(n+1)!}=\\\frac{1}{z^2}-\sum_{n=-1}^{\infty}\frac{z^n(n+1)}{(n+2)!}=\frac{1}{z^2}-\frac{1}{2}-\frac{z}{3}-\frac{z^2}{8}+O(z^4)$$
