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$$S_1=1\\ S_n=n!+S_{n-1}$$ Is there a simple way to express $S_n$ without summing up all the previous terms? Sorry I haven't put any effort in the problem but I don't know where to start.

So this means that $$S_1=1\\ S_2=3\\ S_3=9\\ S_4=33$$ and so on. Thanks in advance.

Joao
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  • Even if you (understandably) want to avoid summing up a series like this, it's still a good idea to compute the first few terms in order to get a feel for the sequence. (And it's an easy thing to include in your question.) – Semiclassical Aug 26 '14 at 03:29
  • @Semiclassical Done! – Joao Aug 26 '14 at 03:32

3 Answers3

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According to a CAS, there is effectively a closed form $$S_n=-(-1)^n \Gamma (n+2)~ \text{Subfactorial}[-n-2]-2 ~\text{Subfactorial}[-3]+1$$ in which function $\text{Subfactorial}[k]$ gives the number of permutations of $n$ objects that leave no object fixed.

I bet that this will not be very practical.

Claude Leibovici
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$S_n=1!+2!+\cdots + n!$. Now, a detailed discussion of this sum can be obtained here.

Community
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voldemort
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OEIS sequence A007489. No "closed form" is listed.

Robert Israel
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