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I was trying to think how I'd prove the asymptotic formula $n! \sim \sqrt{2\pi n} n^n e^{-n}$ to a student who could think like a mathematician but lacked a background in Taylor series, Riemann sums & integration, and analysis beyond this. Of course, to even parse what an asymptotic equivalence is they would have to know about limits, so let's assume this and also that they understand the concepts required to prove that, say, a bounded monotone sequence converges.

Anyhow, I got stuck. Even bounds that are not as sharp as this seem to require a bit more (e.g., here) than I'm assuming these hypothetical students know. Obviously Euler-Maclauren summation and the gamma function are out of the question, but I've found a geometric proof of Wallis' product (e.g., here), and so proving what I want reduces to showing that $n! \sim C\sqrt{n}n^n e^{-n}$. Hence my question is:

Do you know of a way to obtain this latter asymptotic formula under the above constraints? (Or at least assuming as little extra as possible)

I realize this question is vague, but I'm interested in seeing some of the different approaches out there that one might take. Geometric or combinatorial methods would be especially appreciated.

Perhaps this should be community wiki?

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  • $\begingroup$ Perhaps $n!=(-1)^n\displaystyle\int_0^1\ln^nx~dx,~$ and $~\displaystyle\int\ln x~dx=x\ln\dfrac xe=\ln\bigg(\dfrac xe\bigg)^x$ ? $\endgroup$ – Lucian Feb 3 '15 at 16:17
  • $\begingroup$ It's been a long time since I read it, and I don't have access right now to check, but I seem to recall that Mermin's book Boojums All the Way Through has a chapter on Stirling's approximation that might be worth a look. $\endgroup$ – Barry Cipra Feb 6 '15 at 18:13
  • $\begingroup$ @BarryCipra Thank you for the interesting reference! $\endgroup$ – sourisse Mar 17 '15 at 10:35

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