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Using Wolfram alpha, I'm pretty sure that if $k\in\mathbb{Z}^+$, then $\displaystyle\sum_{n=0}^\infty \dfrac{n!}{(n+k)!}=\dfrac{1}{(k-1)(k-1)!}$. How do you prove this?

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 What about the case $k=1$? Try partial fraction decomposition. This question would be better suited on math.stackexchange.com – András Bátkai Aug 2 2011 at 19:24
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 «Using Wolfram alpha, I'm pretty sure that …» is a strange way to phrase things! :) – Mariano Suárez-Alvarez Aug 2 2011 at 19:33
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 Here is how you can prove this. Using repeated integration you can see that $\sum_{n=0}^\infty \frac{x^{n+k}}{(n+1)\cdots(n+k)}$ equals $c_k (1-x)^{k-1}\ln(1-x)+p_k(1-x)+q_k(x)$, where $c_k$ is a constant, $p_k$ and $q_k$ are polynomials of degree $k-1$. Keeping track of $c_k,p_k,q_k$ and plugging $x=1$ should yield the equation. – GH Aug 2 2011 at 19:55

closed as too localized by Andreas Blass, Felipe Voloch, Mariano Suárez-Alvarez, quid, S. Carnahan♦ Aug 2 2011 at 19:48

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