Computing $\text{Res}_{z=\infty} \frac{e^{1/z}z^n}{z+1}$, Let $n$ be a nonnegative integer and define $\displaystyle f(z)=\frac{e^{1/z}z^n}{z+1}$. I need to compute $\text{Res}_{z=\infty}f(z)$. I did the following:
$$\text{Res}_{z=\infty}f(z) = - \frac{1}{z^2} \text{Res}_{z=0}f(1/z) = - \text{Res}_{z=0}  \frac{e^z}{z^{n+1}(z+1)}.$$
Now, $\displaystyle \frac{e^z}{z^{n+1}(z+1)}$, has a pole of order $n+1$ at $0$, so
$$\text{Res}_{z=\infty}f(z) = -\frac{1}{n!} \lim_{z \to 0} \frac{d^n}{dz^n} \left( \frac{e^z}{z+1} \right).$$
This last derivative is giving me a headache, I'd like to know if there's an easy way to compute it, or a complete different way to compute the residue which is easier (I tried with the integral definition of the coefficent $a_{-1}$ of the Laurent series and I arrived to the same derivative).
 A: The residue of $\frac{e^z}{z^{n+1} (z+1)}$ at $0$ is the coefficient of $z^{-1}$ in the Laurent series of that function about $0$, which is the coefficient of $z^{n}$ in the Maclaurin series of $e^z/(z+1)$.  Now $e^z = \sum_{k=0}^\infty z^k/k!$ while $1/(z+1) = \sum_{j=0}^\infty (-1)^j z^j$, so that coefficient is
$$ \sum_{k=0}^n \frac{(-1)^{n-k}}{k!}$$
That can be written using an incomplete Gamma function as
$$ {\frac { \left( -1 \right) ^{n}\Gamma \left( n+1,-1 \right) {{\rm e}^{
-1}}}{n!}}
$$
A: I thought it might be instructive to present an approach that relies on application of Leibniz's Product Rule.  To that end we proceed.
Applying Leibniz's Rule to the function $\frac{e^z}{z+1}$ reveals
$$\begin{align}
\frac{d^n}{dz^n}\left(\frac{e^z}{z+1}\right)&=\sum_{k=0}^n\binom{n}{k}\underbrace{\frac{d^{n-k}e^z}{dz^{n-k}}}_{e^z}\,\,\underbrace{\frac{d^k(z+1)^{-1}}{dz^k}}_{(-1)^kk!(z+1)^{-(k+1)}}\\\\
&=e^z\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^kk!}{(z+1)^{k+1}}
\end{align}$$
Multiplying by $-\frac1{n!}$ and letting $z\to 0$, we find that
$$\begin{align}
\text{Res}(f(z), z=0)&=-\sum_{k=0}^n \frac{(-1)^n}{(n-k)!}\\\\
&\overbrace{=}^{k\mapsto n-k}-\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}
\end{align}$$
And we are done!
