Results for values of the residues in the Residue Theorem 
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*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? 
 A: Note: This was too long to post as a comment, but it is written in response to a comment by OP.
Here is a nice way to compute the residue at 0 that avoids having to fall back to series.
$$2\pi i\, \mathrm{Res}_{z=0}(f)= \int_C \dfrac{ze^{1/z}}{1+z} dz$$
Let's just say that $C$ is the circle $C = \{z\in \mathbb{C}: |z|=0.5\}$
Finding the Laurent series of this function centered at 0 can be a bit of a pain.
But what if we applied a change of variable $z = 1/w$?
\begin{align}
\int\limits_{C} \dfrac{ze^{1/z}}{1+z} dz& = \int\limits_{-\Gamma} \dfrac{e^w/w}{1+1/w} \left(-\frac{1}{w^2}\right)dw\\
& = \int\limits_{\Gamma} \dfrac{e^w/w}{1+1/w} \left(\frac{1}{w^2}\right)dw\\
& = \int\limits_{\Gamma} \dfrac{e^w}{w+1} \left(\frac{1}{w^2}\right)dw\\
\end{align}
Where $\Gamma = \{ w\in \mathbb{C}: |w|=2\}$ is a circle of radius $2$ centered at 0. It encloses two singular points, $w=0$ and $w=-1$.
$$\displaystyle\int\limits_{\Gamma} \dfrac{e^w}{w+1} \left(\frac{1}{w^2}\right)dw$$ can be computed by looking at $w=0$ (pole of order 2) and $w=-1$ (pole of order 1).

To calculate the residue of a pole of order $(n+1)$, 
$$\mathrm{Res}_{z=z_0}(f) = \dfrac{1}{n!}\dfrac{d^{n}(z-z_0)^{n+1}f}{dz^n}(z_0)$$
To calculate the residue at $w=0$ of $\dfrac{e^w}{w+1} \frac{1}{w^2}$, first note that this is a pole of order $2 = 1+1$, $n=1$. So
$$\mathrm{Res}_{w=0}(f)= \left.\dfrac{1}{1!}\dfrac{d}{dw}\dfrac{e^w}{w+1}\right|_{w=0}$$
A: 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}
