Which grows faster? $(ax)!$ or $x^x$?

Question

Messing around on wolphramalpha I found that:

$\lim_\limits{x \to\infty} \dfrac { x !}{x^x} = 0\;$ and $\;\lim_\limits{x \to\infty} \dfrac { (2x) !}{x^x} = \infty$

testing with a few more values I found:

$\lim_\limits{x \to\infty}\!\dfrac { (1.000001 x) !}{x^x}\!=\!0\;$ and $\;\lim_\limits{x \to\infty}\!\dfrac { (1.000002 x) !}{x^x}\!=\!\infty$

I was wandering: if it exists, what is the number $a$ such that $\lim_\limits{x \to\infty} \dfrac { (ax) !}{x^x} = 1\;?\;$ And if it doesn't exist, what is the number $a$ such that:

if $\,b > a\,,\;\;\lim_\limits{x \to\infty} \dfrac { (bx) !}{x^x} = 0$
if $\,c < a\,,\;\;\lim_\limits{x \to\infty} \dfrac { (cx) !}{x^x} = \infty$

and how could someone calculate this ?

Answer 1

We know that $$ \sqrt{2\pi}\; n^{n+1/2} e^{-n+1/(12n+1)} < n! < \sqrt{2\pi}\; n^{n+1/2} e^{-n+1/(12n)}. $$ (See formula $(26)$ on https://mathworld.wolfram.com/StirlingsApproximation.html). Therefore $$ \sqrt{2\pi}\; (ax)^{ax+1/2} e^{-ax+1/(12ax+1)} < (ax)! < \sqrt{2\pi}\; (ax)^{ax+1/2} e^{-ax+1/(12ax)}. $$ and $$ \sqrt{2\pi}\; a^{ax+1/2} x^{(a-1)x+1/2} e^{-ax+1/(12ax+1)} < \frac{(ax)!}{x^x} < \sqrt{2\pi}\; a^{ax+1/2} x^{(a-1)x+1/2} e^{-ax+1/(12ax)}. $$

Regrouping the terms,

$$ \sqrt{2\pi a}\; \left(\frac{a^a x^{a-1}}{e^a}\right)^x x^{1/2} e^{1/(12ax+1)} < \frac{(ax)!}{x^x} < \sqrt{2\pi a}\; \left(\frac{a^a x^{a-1}}{e^a}\right)^x x^{1/2} e^{1/(12ax)}. $$

Now we know that $\dfrac{(ax)!}{x^x} > 0$ for $x > 0,$ so if you can prove that $$ \lim_{x\to 0} \sqrt{2\pi a}\; \left(\frac{a^a x^{a-1}}{e^a}\right)^x x^{1/2} e^{1/(12ax)} = 0 $$ for a particular value of $a,$ then by the squeeze theorem you can prove that $\lim_{x\to 0} \dfrac{(ax)!}{x^x} = 0.$ I think you will find that this is an easy proof if $a = 1.$

On the other hand, if (for some value of $a$) you can prove that $$ \lim_{x\to 0} \sqrt{2\pi a}\; \left(\frac{a^a x^{a-1}}{e^a}\right)^x x^{1/2} e^{1/(12ax+1)} = \infty $$ then you can prove that $\lim_{x\to 0} \dfrac{(ax)!}{x^x} = \infty.$ In order to do this, you may find it helpful to consider whether there exists a number $x_0$ such that whenever $x > x_0,$ $$ \frac{a^a x^{a-1}}{e^a} > 2. $$ (There's nothing special about $2$ here; you could use any real constant greater than $1.$) Also note that if $a > 1$ then $a - 1$ is a positive number, although possibly a very small one. So $x_0$ might be very, very large indeed.......

Answer 2

Look at $\frac{(cx)!}{x^x}$.

If you let $c = 1+\frac{1}{n}$, and $x = \exp(dn)$, then the expression becomes very large very quick as soon as $d > 1$.

Now with $c=1.000001$ and $n = 1,000,000 x$ would be a number with several hundred thousand digits, and for $x^x$ the number of digits has several $100,000$ digits. So I think we can forgive wolfram for not getting the right result. You’d probably need to ask wolfram engineers about what exactly is going on.