1

For positive functions $f$ and $g$ on real domains, define $f(n) \sim g(n)$ to mean $\displaystyle\lim_{n\to\infty}\frac {f(n)}{g(n)}=1$.

Given that $$\frac{n^{n+\frac12}}{e^{n-1}n!}\sim\frac e{\sqrt{2\pi}},$$ I would like to show that $$n!\sim \sqrt{2\pi n}(\frac ne)^n.$$

It is simple enough to shuffle the terms around to reach the desired conclusion, but for the non-constant terms, how is this justified? Don't I need to show that each non-constant term has a (non-infinite) limit before I'm allowed to tear it away and shuffle it around? OR do we simply treat the $\sim$ like an equality (asymptotic equality) with the caveat that we are working in the region of sufficiently large $n$?

ryang
  • 38,879
  • 14
  • 81
  • 179
  • Write it as a limit then shuffle. – Antonio Vargas Feb 28 '13 at 16:09
  • @AntonioVargas That's what I'm doing, and hence my question. Normally when working with limits, I don't shuffle the terms around unless I have shown that the extracted terms themselves each has a non-infinite limit. I.e. If $\lim f =l$ and $\lim g = m$, then $\lim (f+g)=l+m$. However, $\lim (f+g)=l+m$ does not mean that $\lim f=l$ or that $\lim g =m$, since we don't know that either f or g has a limit. Am I right? – ryang Feb 28 '13 at 16:10
  • Use \sim. $ $ – Did Feb 28 '13 at 16:39
  • @Did Thanks for the tip. – ryang Feb 28 '13 at 16:42

1 Answers1

1

There's nothing mysterious happening here. For all $n$ it is true that

$$ \frac{\left(\frac{n^{n+1/2}}{e^{n-1} n!}\right)}{\left(\frac{e}{\sqrt{2\pi}}\right)} = \frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n!}, $$

So if

$$ \lim_{n \to \infty} \frac{\left(\frac{n^{n+1/2}}{e^{n-1} n!}\right)}{\left(\frac{e}{\sqrt{2\pi}}\right)} = 1 $$

then

$$ \lim_{n \to \infty} \frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n!} = 1 $$

because you are taking the limit of the same quantity both times. It may sometimes be the case that you have extra terms/factors left over, and there you'll have to make some arguments about the limits existing. Here, however, there are no such complications.

Antonio Vargas
  • 24,993