Laurent Series of $(z^2 + 3z + 2)e^{\frac{1}{z+1}}$ I want to find the Laurent series of $(z^2 + 3z + 2)e^{\frac{1}{z+1}}$ around $z_0 = -1$. However, since this is not a fraction in the form $\frac{a}{z-b}$, I am not sure how to calculate it. 
 A: Note that in $\mathbb{C} \backslash \{ -1 \} $
$$
e^{\frac{1}{z+1}} = \sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n}
$$
Then
\begin{align*}
(z^2+3z+2)e^{\frac{1}{z+1}} & =\left[(z+1)^2+(z+1)\right] \sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n} \\
& =\left[(z+1)^2 \sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n} \right] + \left[(z+1)\sum_{n=0}^{\infty} \frac{1}{n! (z+1)^n}\right]  \\
& = \left[(z+1)^2 + (z+1) + \sum_{n=0}^{\infty}\frac{1}{(n+2)!(z+1)^n} \right] + \left[(z+1) + \sum_{n=0}^{\infty}\frac{1}{(n+1)!(z+1)^n}\right]  \\
& = 2(z+1)+(z+1)^2 + \sum_{n=0}^{\infty}\left(\frac{1}{(n+1)!}+\frac{1}{(n+2)!}\right)\frac{1}{(z+1)^n} \\
& = 2(z+1)+(z+1)^2 + \sum_{n=0}^{\infty}\left(\frac{n+3}{(n+2)!}\right)\frac{1}{(z+1)^n}
\end{align*}
A: Setting $\xi=z+1$, i.e. $z=\xi-1$, we have
\begin{eqnarray}
(z^2+3z+2)e^\frac{1}{z+1}&=&[(\xi-1)^2+3(\xi-1)+2]e^\frac1\xi=[(\xi^2-2\xi+1)+(3\xi-3)+2]e^\frac1\xi\\
&=&(\xi^2+\xi)e^\frac1\xi=(\xi^2+\xi)\sum_{n=0}^\infty\frac{1}{n!\xi^n}=\sum_{n=0}^\infty\left(\frac{1}{n!\xi^{n-2}}+\frac{1}{n!\xi^{n-1}}\right)\\
&=&\sum_{n=-2}^\infty\frac{1}{(n+2)!\xi^n}+\sum_{n=-1}^\infty\frac{1}{(n+1)!\xi^n}=2\xi+\xi^2+\sum_{n=0}^\infty\frac{n+3}{(n+2)!\xi^n}\\
&=&2(z-1)+(z-1)^2+\sum_{n=0}^\infty\frac{n+3}{(n+2)!(z-1)^n}.
\end{eqnarray}
