How do I compute the following limit: $ \lim_{x \to \infty} \frac{x!}{\left( \frac{x}{e} \right)^{x}}$?

Question

In class we proved that $$ \lim_{x \to \infty} \frac{x!}{2^{x}} = \infty$$

This got me thinking for what value $n$ $$ \lim_{x \to \infty} \frac{x!}{n^{x}}$$ would the limit be $= 0$.

So clearly $n = x$ makes the bottom part of the fraction go to infinity much faster than the top part, and this is the case for $n = \frac{x}{2}$ as well. However, the limit for $n = \frac{x}{3}$ is $\infty$. I immediately became suspicious that the "turning point" would be for $n = \frac{x}{e}$. Due to calculator approximation errors, a normal TI-89 says the limit is $\infty$, but I'm not really sure if that's correct.

In any case, how would one compute the limit for when $n = \frac{x}{e}$?

Answer 1

Hint

When you have to manipulate factorials, Stirling approximation is very often the trick to be used.

As a first approximation, you have $$x!\approx \sqrt{2\pi\, x}\,\Big(\frac x e\Big)^x$$ which than makes for your problem $$\frac{x!}{\left( \frac{x}{e} \right)^{x}}\approx \sqrt{2\pi\,x} $$ Please, remember it : it is very useful and you will often need it !

Answer 2

Maybe a nice way of solving your problem:

$$\lim_{x\to\infty}\frac{x!}{\left(\frac{x}{e}\right)^x}=\lim_{x\to\infty}\frac{e^xx!}{x^x}=\lim_{x\to\infty}e^xx^{-x}x!=$$ $$\lim_{x\to\infty}\exp\left[x-x\ln(x)+\ln(x!)\right]=\exp\left[\lim_{x\to\infty}\left(x-x\ln(x)+\ln(x!)\right)\right]=$$ $$\exp\left[\lim_{x\to\infty}\ln\left(e^xx^{-x}!\right)\right]=\infty$$

Answer 3

Let $r(x) =\frac{x!}{(x/e)^x} =\frac{x!e^x}{x^x} $. Then

$\begin{array}\\ s(x) &=\frac{r(x+1)}{r(x)}\\ &=\frac{\frac{(x+1)!e^{x+1}}{(x+1)^{x+1}}}{\frac{x!e^x}{x^x}}\\ &=\frac{(x+1)ex^x}{(x+1)^{x+1}}\\ &=\frac{ex^x}{(x+1)^{x}}\\ &=\frac{e}{(1+1/x)^{x}}\\ \text{so that}\\ \ln(s(x)) &=1-x\ln(1+1/x)\\ \end{array} $

We now use the expansion $z-z^2/2 <\ln(1+z) < z-z^2/2+z^3/3 $ for $0 < z < 1$.

Then $x\ln(1+1/x) \lt x(1/x-1/(2x^2)+1/(3x^3)) =1-1/(2x)+1/(3x^2) $ and $x\ln(1+1/x) \gt x(1/x-1/(2x^2)) =1-1/(2x) $.

Therefore $1/(2x)-1/(3x^2) \lt s(x) \lt 1/(2x) $. This implies that $\sum_{x=n}^m \ln(s(x))$ diverges as $m \to \infty$.

But

$\begin{array}\\ \sum_{x=n}^m \ln(s(x)) &=\sum_{x=n}^m \ln(r(x+1)/r(x))\\ &=\sum_{x=n}^m (\ln(r(x+1))-\ln(r(x)))\\ &=\ln(r(m+1))-\ln(r(n))\\ \end{array} $

and this implies that $\ln(r(m+1))$ diverges to $\infty$.

If we look more closely at this, the argument shows that, since

$\begin{array}\\ \sum_{x=n}^m \ln(s(x)) &\sim \sum_{x=n}^m 1/(2x)\\ &\sim \ln(m)/2\\ &=\ln(\sqrt{m})\\ \end{array} $

that $\ln(r(x)) \sim \ln(\sqrt{x}) $.

As is often the case, nothing here is original.