Factorial limit convergence on e

Question

I'm working with some particular infinite products. Each infinite product applies to a range of real numbers, in much the same way as a Riemann sum applies to a range of real numbers. The specifics of the products aren't relevant, but I've managed to reduce each infinite product to a limit. The first infinite product (for the range 0->1) reduces to the following form:

$$ \lim_{x \to \infty} \sqrt[x]{\frac{x!}{x^x}} \approx 0.367882 $$

It's a classical result that the limit of the inner fraction is 0. However, adding the $\sqrt[x]{}$ makes this limit converge to a positive value. A bit of Googling later, and I was surprised to discover that this value is $1/e$.

First question: why does this limit converge on $1/e$?

The logical "next" infinite product would be for the range 1->2, which reduces to:

$$ \lim_{x \to \infty} \sqrt[x]{\frac{(2x)!/x!}{x^x}} $$

This converges to $\approx 1.471517$.

Or, we could look at the infinite product for the whole range 0->2, which reduces to:

$$ \lim_{x \to \infty} \sqrt[x]{\frac{(2x)!}{x^{2x}}} $$

This converges to $\approx 0.54134$. (The result for this range 0->2 equals the product of the results for ranges 0->1 and 1->2, which makes sense).

Second question: is there an exact representation for these second two limits? I assume it would involve $e$. How could I have determined it?

Answer 1

The first result follows from Stirling's formula: $n! \sim \sqrt{2\pi n} (\frac ne)^n$, and the $\sqrt{2\pi n}$ becomes insignificant after taking the $n^{\text{th}}$ root.

We can also recover this by taking logs. Rewriting $\sqrt[x]{\frac{x!}{x^x}}$ as $\left(\frac1x \cdot \frac2x \cdots \frac xx\right)^{1/x}$, we can take the log and get $\frac1x\left(\log \frac1x + \log \frac2x + \dots + \log \frac xx\right)$, which is exactly a Riemann sum for $\int_{0^+}^1 \log x\,dx = -1$. Therefore the original limit must be $e^{-1} = \frac1e$.

From the first limit, we expect $$\lim_{x \to \infty} \sqrt[2x]{\frac{(2x)!}{(2x)^{2x}}} = \frac1e$$ as well, and getting the third limit from here is just a matter of pulling out some factors. We can write $\sqrt[2x]{\frac{(2x)!}{(2x)^{2x}}}$ as $\frac12 \sqrt[2x]{\frac{(2x)!}{x^{2x}}}$, or $\frac12 \sqrt{ \sqrt[x]{\frac{(2x)!}{x^{2x}}}}$. So your third limit converges to some value $L$ such that $\frac12 \sqrt{L} = \frac1e$, or $L = \frac{4}{e^2}$.

It follows that the second limit is $\frac 4e$.

Answer 2

$$A=\sqrt[x]{\frac{x!}{x^x}} \implies\log(A)=\frac 1x \big[\log(x!)-x\log(x)\big]$$ Using Stirling approximation, you then obtain $$\log(A)=-1+\frac{\log (2 \pi x)}{2 x}+\frac{1}{12 x^2}+O\left(\frac{1}{x^4}\right)$$ $$A=e^{\log(A)}=\frac 1 e\Bigg[1+ \frac{\log (2 \pi x)}{2 x}\Bigg]+O\left(\frac{1}{x^2}\right)$$

Using the same procedure $$B= \sqrt[x]{\frac{(2x)!}{x^x x!}}=\frac 4 e\Bigg[1+ \frac{\log (2 )}{2 x}\Bigg]+O\left(\frac{1}{x^2}\right)$$

$$C=\sqrt[x]{\frac{(2x)!}{x^{2x}}}=\frac 4 {e^2}\Bigg[1+ \frac{\log (4 \pi x)}{2 x}\Bigg]+O\left(\frac{1}{x^2}\right)$$

Notice that you have the limit, how it is approached and moreover a shortcut method for good approximations (you can use your phone for that).