Let the residue of the function $f(z)=\frac{z^ne^{\frac{1}{z}}}{1+z} $ at the points $z=0$ and $z=\infty$ be $A$ and $B$ respectively. Show that if $n$ is a non-negative integer, then $$A=(-1)^{n+1}\frac{1}{e}+\frac{1}{n!}-\frac{1}{(n-1)!}+\ldots+(-1)^{n}\frac{1}{1!};$$
$$A+B+(-1)^n\frac{1}{e}=0$$
To find the residue at point $z=0$, I think I need to write the Laurent series around point $z=0$. $$f(z)=z^n(\sum_{m=0}^{\infty}(-z)^m)(\sum_{j=0}^{\infty}\frac{z^{-j}}{j!}).$$ To find the resduie at point $z=0$, I guess it's enough to count the coeffiecnt of $z^{-1}.$
However, I don't know how to look at this sum to get the desired $A$. I appreciate any help.
First, we expand $e^{1/z}$ and $\frac1{1+z}$ in Laurent and Taylor series to write
$$\frac{e^{1/z}}{1+z}=\sum_{m=0}^\infty \sum_{j=0}^\infty \frac{(-1)^m}{j!} z^{m-j}$$
Next, letting $k=m-j$ reveals
$$\begin{align} \sum_{m=0}^\infty \sum_{j=0}^\infty \frac{(-1)^m}{j!} z^{m-j}&=\sum_{m=0}^\infty \sum_{k=-\infty}^m \frac{(-1)^mz^k}{(m-k)!}\\\\ &=\sum_{m=0}^\infty \sum_{k=-\infty}^{-1} \frac{(-1)^mz^k}{(m-k)!}+\sum_{m=0}^\infty \sum_{k=0}^m \frac{(-1)^mz^k}{(m-k)!}\\\\ \end{align}$$
Finally, the residue of $z^n \frac{e^{1/z}}{1+z}$ at $z=0$ is, therefore, simply the coefficient on the $z^{-1}$ term. This is given by
$$\begin{align} \text{Res}\left(z^n \frac{e^{1/z}}{1+z},z=0\right)&=\sum_{m=0}^\infty \frac{(-1)^m}{(m+n+1)!}\\\\ &=\sum_{m=n+1}^\infty \frac{(-1)^{m-n-1}}{m!} \\\\ &=\sum_{m=0}^\infty \frac{(-1)^{m-n-1}}{m!}-\sum_{m=0}^n \frac{(-1)^{m-n-1}}{m!} \\\\ &=(-1)^{n+1}\left(e^{-1}-\sum_{m=0}^{n}\frac{(-1)^m}{m!}\right) \end{align}$$
as was to be shown!