Why $n!^{(1/n)}$ behaves like an arithmetic progression?

Question

The sequence $(n!)^{(1/n)}$ is an increasing sequence. Here are some terms,

$$ 1,1.4142, 1.8171,2.2133, 2.6051,2.9937,....$$

Could someone explain why this sequence is so close to an arithmetic progression?

Answer 1

Stirling's formula says that asymptotically, $$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n.$$ Taking $n$th roots, we get that your expression is asymptotic to $n/e,$ which is an arithmetic progression.

Answer 2

Stirling's approximation tells you that $n! \approx e^{n\log n - n}$. So $(n!)^{1/n}\approx e^{\log n - 1} = n\cdot e^{-1}$. Here are some terms of that sequence: $1.10364, 1.47152, 1.8394, 2.20728, 2.57516, 2.94304, 3.31091$.

Answer 3

The first step in an explanation is Stirling's formula:

$$n!\approx \sqrt{2\pi n}\left(\frac ne\right)^n$$

so that

$$n!^{1/n}\approx (\sqrt{2\pi n})^{1/n}\frac ne.$$

The $(\sqrt{2\pi n})^{1/n}$ part rapidly approaches 1, so can be disregarded, and the common difference of your terms is simply $\frac1e\approx 0.367$.

This blog post of Terry Tao explains where Stirling's formula comes from.