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?
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$\begingroup$ $\sum_{k=0}^n C_n^k k!=\int_0^\infty C_n^k x^k e^{-x}dx=\int_0^\infty (1+x)^n e^{-x}dx$ expressible via the incomplete Gamma function $\endgroup$– Peter KravchukCommented Jun 6, 2016 at 12:13
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2$\begingroup$ Useful summary here: oeis.org/wiki/Subfactorial. $\endgroup$– Todd TrimbleCommented Jun 6, 2016 at 14:30
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See https://oeis.org/A000522 . It is known (and not too hard to prove) that it is $\lfloor e \cdot n! \rfloor$ for $n \geq 1$.
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$\begingroup$ Jeffrey - Thank you so much for the answer. Btw, do you have a reference for the proof? $\endgroup$– Arun SenCommented Jun 6, 2016 at 14:09
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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 $(e-\frac{1}{n!},e)$. $\endgroup$ Commented Jun 6, 2016 at 16:17