1
$\begingroup$

This question already has an answer here:

$$S_1=1\\ S_n=n!+S_{n-1}$$ 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.

$\endgroup$

marked as duplicate by Thomas Andrews, voldemort, Jonas Meyer, Antonio Vargas, Adam Hughes Aug 26 '14 at 5:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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.) $\endgroup$ – Semiclassical Aug 26 '14 at 3:29
  • $\begingroup$ @Semiclassical Done! $\endgroup$ – Joao Aug 26 '14 at 3:32
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.