For what value of n does $((x!)^n) / (x^x)$ approach neither 0 or infinity?

Question

I like messing around with numbers and I had a stupid thought. When $n = 2$ the equation approaches infinity but when $n = 1$ it approaches $0$. This isn't that weird but shouldn't there be a number of n that neither approaches 0 nor infinity, if so what would it be?

Answer 1

There is no such value of $n$, but it'll be hard to see that by just numerical experimentation. Let's define

$$f_n(x) = {(x!)^n \over x^x}$$

and then you're asking what $\lim_{x \to \infty} f_n(x)$ is as a function of $n$. You can take logs to get

$$\log f_n(x) = n \log (x!) - x \log x$$

At this point it helps to use "Stirling's approximation" $$x! \approx \sqrt{2\pi x} \left( {x \over e} \right)^x$$ and so $$\log (x!) \approx {1 \over 2} \log (2\pi) + x \log x - x.$$

Therefore you have

$$\log f_n(x) = n \left( {1 \over 2} \log (2\pi) + x \log x - x \right) - x \log x$$

or with some rearrangement

$$ \log f_n(x) = (n-1) x \log x + n (C-x) $$

where I've also defined $C = {1 \over 2} \log (2\pi)$. Now if $n \le 1$ this goes to $-\infty$ as $x \to \infty$. and so $f_n(x) \to 0$. If $n > 1$ then $\lim_{n \to \infty} \log f_n(x) = \infty$ and so $\lim_{n \to \infty} f_n(x) \to \infty$ as well.

But if $n$ is not much more than 1 then you basically have no chance of seeing this if you're just experimenting by sticking numbers into a calculator. The way to compute this when $x$ gets large is to work in terms of the logarithm $\log (x!)$, which can be computed directly in some mathematical software. This shows, for example, that $f_{1.1}(20000) \approx 2.4 \times 10^{-950}$. Sure looks like it's going to zero!

But as $x$ increases it turns around: $f_{1.1}(30000) \approx 3.5 \times 10^{-897}$. And eventually it goes off to positive infinity: $f_{1.1}(100000) \approx 6.3 \times 10^{+2230}$.