TITLE: Compute the residues of $\frac{e^{\frac{1}{z}}}{z+1}$
QUESTION [0 upvotes]: Compute the residues of $$\frac{e^{\frac{1}{z}}}{z+1}$$

The singular points I detected are the following:
$z=-1$ which is a pole of order 1.
$z=0$ which is essential
$z=\infty$ which is removable.
$\operatorname{res}_{z=-1}\frac{e^{\frac{1}{z}}}{z+1}=\frac{e^{\frac{1}{-1}}}{1}=\frac{1}{e}\\\operatorname{res}_{z=0}\frac{e^{\frac{1}{z}}}{z+1}=c_{-1}$
In order to compute $c_{-1}$(first negative index term of the Laurent series):
$\frac{e^{\frac{1}{z}}}{z+1}=\sum_\limits{n=0}^{\infty}(\frac{1}{z})^n\frac{1}{n!}\times \sum_\limits{n=0}^{\infty}(-1)^n{z}^n$
However I am not seeing how to compute $c_{-1}$. In the book solution $\operatorname{res}_{z=0}\frac{e^{\frac{1}{z}}}{z+1}=1-\frac{1}{e}$
Taking the limit: $\lim_{z\to\infty}\frac{e^{\frac{1}{z}}}{z+1}=0$ so $\operatorname{res}_{z=\infty}\frac{e^{\frac{1}{z}}}{z+1}=0$
Since $-\operatorname{res}_{z=\infty}\frac{e^{\frac{1}{z}}}{z+1}=\operatorname{res}_{z=-1}\frac{e^{\frac{1}{z}}}{z+1}+\operatorname{res}_{z=0}\frac{e^{\frac{1}{z}}}{z+1}$
However according to the solution $-\operatorname{res}_{z=\infty}\frac{e^{\frac{1}{z}}}{z+1}=\operatorname{res}_{z=-1}\frac{e^{\frac{1}{z}}}{z+1}+\operatorname{res}_{z=0}\frac{e^{\frac{1}{z}}}{z+1}\implies 0=1-\frac{1}{e}+\frac{1}{e}=1$
Questions:
How do I compute $\operatorname{res}_{z=0}\frac{e^{\frac{1}{z}}}{z+1}$?
Why does not the solution verify the equality 
$-\operatorname{res}_{z=\infty}\frac{e^{\frac{1}{z}}}{z+1}=\operatorname{res}_{z=-1}\frac{e^{\frac{1}{z}}}{z+1}+\operatorname{res}_{z=0}\frac{e^{\frac{1}{z}}}{z+1}$?
Thanks in advance!

REPLY [1 votes]: As you wrote, the Laurent series at $0$ is a product of two:
$$ \sum_{n=0}^\infty \frac{z^{-n}}{n!} \cdot \sum_{m=0}^\infty (-1)^m z^m $$
(it's better two use different letters for the indices).
The terms with $m-n=-1$, i.e. $n=m+1$, give you $z^{-1}$.  Thus the residue is
$$ \sum_{m=0}^\infty \frac{(-1)^m}{(m+1)!} = 1 -
 1/e $$ 
The singularity at $z=\infty$ is removable, but that doesn't make the residue there $0$: the residue at $\infty$ of a Laurent series $\sum_n a_n z^n$ is $-a_{-1}$.
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