$\displaystyle{\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{n}{(k+n)!} \frac{n}{k!}} ~= ~?$
From numerical calculation, I believe the answer is $e^2$. However, I have no idea how to show this.
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Are you familiar with the Cauchy product? Hence I guess you could use it to rewrite the product of the sum $\sum_{k=0}^{\infty}\frac{1}{k!}$, which defines $e$, with itselft to somehow end up by your double sum. – mrtaurho Aug 09 '18 at 11:26
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Thank you for a comment! It was an important step to change the indices as like the Cauchy product. @mrtaurho – rudgns55 Aug 09 '18 at 12:26
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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 $2S=2e^2$.
Angina Seng
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Thank you for your answer. The crucial step was using symmetry! Your answer is very clear and impressive :) – rudgns55 Aug 09 '18 at 12:13