3

What is the asymptotic behaviour of the function $e !n-n!$ , where $!n = n! \sum_{k=0}^n \frac{(-1)^k}{k!}$ is the subfactorial of $n$. I tried Wolfram Alpha but the series for n=$\infty$ is pretty complicated. There should be a simplier function doing the job.

The function arises from the integrals

$\int_{0}^{1}e^xx^ndx$ = $(-1)^n(e !n-n!)$ for every positive integer $n$.

Thomas Andrews
  • 177,126
Peter
  • 84,454

1 Answers1

1

It's best to work directly with the integral

$$ I(n) = \int_0^1 e^x x^n\, dx. $$

Make the substitution $x = e^{-t}$ to put it into the form

$$ I(n) = \int_0^\infty e^{e^{-t}-t} e^{-nt} \,dt. $$

Watson's lemma tells us that we can expand the integrand as a power series about $t=0$

$$ e^{e^{-t}-t} = e-2 e t+\frac{5 e t^2}{2} + \cdots $$

and integrate term-by-term to obtain an asymptotic expansion for the integral. Thus

$$ \begin{align} I(n) &\approx e \int_0^\infty e^{-nt}\,dt - 2e \int_0^\infty te^{-nt}\,dt + \frac{5e}{2} \int_0^\infty t^2 e^{-nt}\,dt + \cdots \\ &= \frac{e}{n} - \frac{2 e}{n^2} + \frac{5 e}{n^3} + \cdots. \end{align} $$

Antonio Vargas
  • 24,993
  • 1
    I found an even easier way which only gives $\frac{e}{n}$ : By integration by parts, it is easy to check I(k)=e-kI(k-1). I(k) is a positive and decreasing function. So 0<I(k)=e-kI(k-1)<I(k-1). It follows immediately $\frac{e}{k+1}$<I(k-1)<$\frac{e}{k}$ and therefore $\frac{e}{k+2}$<I(k)<$\frac{e}{k+1}$. So, $\frac{e}{k}$ is an asymptote. – Peter Nov 10 '13 at 13:47
  • But since your formula shows better bounds, I vote it up. – Peter Nov 10 '13 at 13:52