<s>[INST] Results for values of the residues in the Residue Theorem 
*

*If the sum of the residue is $0$, what can I conclude: that the value of the integral is $0$ or that the integral diverges?  

*If the residue tends to $\infty$, should I conclude that the integral diverges? 
 [/INST] If you REALLY want to use the Laurent series expansion, this is how it goes:
\begin{align}
\dfrac{1}{1+z} (z e^{1/z}) &= (1-z+z^2-z^3+\dots)\cdot z (1+\dfrac{1}{z}+\dfrac{1}{2!z^2} + \dfrac{1}{3!z^3}+\dots)\\
& = z^{-1}\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dfrac{1}{5!}+\dots\right)+\text{other terms we don't care about}
\end{align}
Now, if you look at the coefficient, 
$$\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dfrac{1}{5!}+\dots\right)$$ is exactly 
\begin{align}
e^z & = 1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!}+\dfrac{z^4}{4!}+\dfrac{z^5}{5!}+\dots\\
e^{-1}&=1+(-1)+\dfrac{(-1)^2}{2!} + \dfrac{(-1)^3}{3!}+\dfrac{(-1)^4}{4!}+\dfrac{(-1)^5}{5!}+\dots\\
e^{-1}&=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dfrac{1}{5!}+\dots
\end{align}
</s>