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I was wondering if the following sum has a closed form: $$S_n=\sum_{k=1}^{n-1} \frac{n!}{k!}=n+n\cdot(n-1)+\cdots+n!$$

$S_n$ satisfies the following recursive relation

$$S_n=n\cdot(S_{n-1}+1)$$

Is there a simple closed form representation of these sums?

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2 Answers 2

up vote 4 down vote accepted

Edit: Initially I misread the summation, including an extra term.

This sequence is almost A007526 in OEIS, which has exponential generating function $$\frac{xe^x}{1-x}$$ and (more or less) closed form $a_n=\lfloor en!-1\rfloor$.

However, the sequence in OEIS is actually $$a_n=\sum_{k=0}^{n-1}\frac{n!}{k!}=S_n+n!\;,$$ which satisfies the same recurrence with $a_0=0$. For $S_n$ the recurrence starts at $S_1=0$ and holds thereafter. The $S_n$ are actually A038156 and satisfy $$S_n=\lfloor(e-1)n!)\rfloor-1\;.$$

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Well it depends what you mean by a closed form..

$$S_n= n! [\frac{1}{1!}+..+\frac{1}{(n-1)!}] \,.$$

Then

$$en!-S_n= n! [1-\frac{1}{n!}-\frac{1}{(n+1)!}-...]$$

$$en!-n!+2-S_n=1- n! [-\frac{1}{(n+1)!}-...]$$

Now, I think an easy computation shows that the RHS is between 0 and 1.

Thus

$$\lfloor en!-n!+2-S_n \rfloor =0$$

Hence

$$\lfloor en!-n!+2\rfloor =S_n$$

Is this a closed form or not? :)

P.S. Got to go, and typed in a hurry thus there might be mistakes in the computation. The basic idea should be right though...

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