1
$\begingroup$

It is known that $ \sum_{k = 0}^{n} {n \choose k} = 2^n$ and $ \sum_{k = 0}^{n} {n \choose k} (!k)= n!$. But is it known what $ \sum_{k = 0}^{n } {n \choose k}(k!)$ is equal to?

$\endgroup$
2
  • $\begingroup$ $\sum_{k=0}^n C_n^k k!=\int_0^\infty C_n^k x^k e^{-x}dx=\int_0^\infty (1+x)^n e^{-x}dx$ expressible via the incomplete Gamma function $\endgroup$ – Peter Kravchuk Jun 6 '16 at 12:13
  • 2
    $\begingroup$ Useful summary here: oeis.org/wiki/Subfactorial. $\endgroup$ – Todd Trimble Jun 6 '16 at 14:30
9
$\begingroup$

See https://oeis.org/A000522 . It is known (and not too hard to prove) that it is $\lfloor e \cdot n! \rfloor$ for $n \geq 1$.

$\endgroup$
2
  • $\begingroup$ Jeffrey - Thank you so much for the answer. Btw, do you have a reference for the proof? $\endgroup$ – Arun Sen Jun 6 '16 at 14:09
  • 2
    $\begingroup$ Note that the sum expands into $n!(\frac{1}{0!}+\dots+\frac{1}{n!})$; the sum in parentheses converges to $e$ from below, and can easily be shown to be in $(e-\frac{1}{n!},e)$. $\endgroup$ – Klaus Draeger Jun 6 '16 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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