The closed form of $\sum_{n=1}^{x}n!$

Asked Aug 13 '14 at 14:17

Active Aug 20 '14 at 00:15

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Let $$y=\sum_{n=1}^{x}n!$$ be the sum of consecutive factorials.

What is closed form for $y$ in terms of $x$? Wolfram Alpha says that $$y=-(-1)^x\Gamma(x+2)(!(-x-2))-!(-1)-1$$ where $!x$ is subfactorial function of $x$, but there isn't any step-by-step method available. Can you explain how solve this?

edited Aug 20 '14 at 00:15

asked Aug 13 '14 at 14:17

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Have you tried induction on $x$? – Frunobulax Aug 13 '14 at 14:39
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For others reading this, note that $!x$ is not a typo but rather the so-called subfactorial. (It would be helpful to put that into your question.) – Semiclassical Aug 13 '14 at 14:59
see this and this. – Aug 13 '14 at 15:12
Given @aziiri's first link, this is essentially a duplicate question. But I don't find myself terribly satisfied by the answers in those questions. – Semiclassical Aug 13 '14 at 15:15
Me too, the result is quite ugly. But this the case of many finite sums, for example the harmonic sum also have no beautiful closed form (some will consider $H_n$ a closed form). – Aug 13 '14 at 15:19
You may want to look at http://www.math.upenn.edu/~wilf/AeqB.html – Max Aug 13 '14 at 23:34

The closed form of $\sum_{n=1}^{x}n!$

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