Evaluation of $S_{k,j}=\sum_{n_1,\ldots,n_k=1}^\infty\frac{n_1\cdots n_j}{(n_1+\cdots+n_k)!}$ for $0\leqslant j\leqslant k>0$ For a positive integer $k$, and an integer $j$ with $0\leqslant j\leqslant k$, the problem of evaluating $$S_{k,j}=\sum_{n_1,\ldots,n_k=1}^\infty\frac{n_1\cdots n_j}{(n_1+\cdots+n_k)!}$$ appears as an extension of the problem 3.137 in the book "Limits, Series, and Fractional Part Integrals" by O. Furdui (which asks for evaluation of $S_{k,k}$). It's stated as an "open problem" there, and a quick search over the Internet reveals a few solutions, like this one.
In the answer below, I'm sharing a solution that looks straightforward to me. I wonder is there anything similar online. I didn't find it.
 A: I'm using the Cauchy integral formula ($n$ is a nonnegative integer):
$$\frac{1}{n!}=\frac{1}{2\pi\mathrm{i}}\oint_C\frac{e^z\,dz}{z^{n+1}},$$
where $C$ is, say, the circle $|z|=r$, with $r>1$ to ensure the convergence:
\begin{align*}
S_{k,j}&=\frac{1}{2\pi\mathrm{i}}\sum_{n_1,\ldots,n_k=1}^{\infty}n_1\cdots n_j\oint_C\frac{e^z\,dz}{z^{n_1+\cdots+n_k+1}}
\\&=\frac{1}{2\pi\mathrm{i}}\oint_C\frac{e^z}{z}\left(\sum_{n=1}^\infty\frac{n}{z^n}\right)^j\left(\sum_{n=1}^\infty\frac{1}{z^n}\right)^{k-j} dz
\\&=\frac{1}{2\pi\mathrm{i}}\oint_C\frac{e^z}{z}\left(\frac{z}{(z-1)^2}\right)^j\frac{dz}{(z-1)^{k-j}}
\\&=\frac{1}{2\pi\mathrm{i}}\oint_C\frac{z^{j-1}e^z}{(z-1)^{k+j}}\,dz.
\end{align*}
For $j>0$, using the "coefficient-of" notation, we get readily
$$S_{k,j}=e[z^{k+j-1}](1+z)^{j-1}e^z=e\sum_{r=0}^{j-1}\binom{j-1}{r}\frac{1}{(k+r)!}.$$
For $j=0$, the residue at $z=0$ enters the picture, resulting in
$$S_{k,0}=(-1)^k+e[z^{k-1}](1+z)^{-1}e^z=(-1)^k\left(1-e\sum_{r=0}^{k-1}\frac{(-1)^r}{r!}\right).$$
