$\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{n}{(k+n)!} \frac{n}{k!}$

Question

$\displaystyle{\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{n}{(k+n)!} \frac{n}{k!}} ~= ~?$

From numerical calculation, I believe the answer is $e^2$. However, I have no idea how to show this.

Answer 1

Set $m=n+k$. Then the sum is $$S=\sum_{k,m:0\le k\le m}\frac{(m-k)^2}{k!m!} =\sum_{k,m:0\le k<m}\frac{(m-k)^2}{k!m!}.$$ By the symmetry of the summand in $(m,k)$, and the fact it vanishes for $m=k$, this is half the sum over all $m$ and $k$: $$2S=\sum_{k=0}^\infty\sum_{m=0}^\infty\frac{(m-k)^2}{m!k!} =S_2S_0-2S_1^2+S_0S_2$$ where $$S_j=\sum_{m=0}^\infty\frac{m^j}{m!}$$ Obviously, $S_0=e$, but it's not hard to prove $S_1=e$ and $S_2=2e$. We get $2S=2e^2$.