TITLE: $\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{n}{(k+n)!} \frac{n}{k!}$
QUESTION [6 upvotes]: $\displaystyle{\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} 
\frac{n}{(k+n)!} \frac{n}{k!}} ~= ~?$

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

REPLY [4 votes]: 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 $2S=2e^2$.