# Show that $\lim_{n\to\infty}\frac{\ln(n!)}{n} = +\infty$

Show that $$\lim_{n\to\infty}\frac{\ln(n!)}{n} = +\infty$$

The only way i've been able to show that is using Stirling's approximation: $$n! \sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$

Let: $$\begin{cases} x_n = \frac{\ln(n!)}{n}\\ n \in \Bbb N \end{cases}$$

So we may rewrite $$x_n$$ as: $$x_n \sim \frac{\ln(2\pi n)}{2n} + \frac{n\ln(\frac{n}{e})}{n}$$

Now using the fact that $$\lim(x_n + y_n) = \lim x_n + \lim y_n$$ : $$\lim_{n\to\infty}x_n = \lim_{n\to\infty}\frac{\ln(2\pi n)}{2n} + \lim_{n\to\infty}\frac{n\ln(\frac{n}{e})}{n} = 0 + \infty$$

I'm looking for another way to show this, since Stirling's approximation has not been introduced at the point where i took the exercise from yet.

• Cesaro-Stolz is the key here. – Paramanand Singh Dec 8 '18 at 7:34

Another way to show $$\lim_{n\to\infty}\frac{\ln(n!)}{n}=\infty$$ is to consider the following property of logarithms: $$\log(n!)=\log(n)+\log(n-1)+\cdots+\log(2)>\frac{n}{2}\log\left(\frac n2\right).$$ Now $$\frac{\log (n!)}{n}>\frac{\log(n/2)}{2}.$$ As $$n\to\infty$$, this clearly diverges to $$+\infty$$.

• Thank you for the answer, could you please elaborate on $\ln(n!) > \frac{n}{2}\ln\left(\frac{n}{2}\right)$? – roman Dec 7 '18 at 15:59
• @roman: There are $n-1$ terms in the sum after expanding the logarithm (therefore, there are at least $n/2$ terms for $n>3$...this is where the first $n/2$ comes from). Now, if we take the first $n/2$ terms in the sum, the argument of the logarithms in all of these terms is at least $n/2$ (that is, $\log(n)>\log(n/2)$, $\log(n-1)>\log(n/2)$, etc.) – Clayton Dec 7 '18 at 16:05

Hint: $$\ln(n!)=\ln 1+\ln2+\cdots+\ln n$$For all but the smallest $$n$$, most of those terms are larger than $$1$$, so the sum is larger than $$n$$.

For somewhat large $$n$$, most of those terms are larger than $$2$$, so the sum is larger than $$2n$$.

For even larger $$n$$, most of those terms are larger than $$3$$, so the sum is larger than $$3n$$.

We have that

$$\ln(n!)\ge n \ln n - n$$

then

$$\frac{\ln(n!)}{n}\ge \ln n -1 \to \infty$$

For the proof of the first inequality refer to

The Cesaro-Stolz criterion is your easiest and cleanest way out of this. It states that given sequences $$x, y \in \mathbb{R}^{\mathbb{N}}$$ such that $$y$$ is strictly increasing and unbounded and the sequence of successive increments converges in the extended real line

$$\lim_{n \to \infty} \frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=t \in \overline{\mathbb{R}}$$

then $$\lim_{n \to \infty}\frac{x_{n}}{y_{n}}=t$$

You can apply this to $$x=(\mathrm{ln}(n!))_{n \in \mathbb{N}}$$ and $$y=(n)_{n \in \mathbb{N}}$$.

A similar argument would rely on a version of the ratio criterion: if $$x \in (0, \infty)^{\mathbb{N}}$$ is a sequence of strictly positive reals such that

$$\lim_{n \to \infty} \frac{x_{n+1}}{x_n}=a \in [0, \infty]$$

then

$$\lim_{n \to \infty} \sqrt [n]{x_{n}}=a$$

This criterion itself can be proved by the Cesaro-Stolz criterion (there are also other methods) and you can apply it to conclude that

$$\sqrt[n]{n!} \xrightarrow{n \to \infty} \infty$$

as $$\frac{(n+1)!}{n!}=n+1 \xrightarrow{n \to \infty} \infty$$.

In the same vein of employing ratios, one can settle the convergence of the sequence $$\left(\frac{\sqrt[n]{n!}}{n}\right)_{n \in \mathbb{N}^{*}}=\left(\sqrt[n]{\frac{n!}{n^n}}\right)_{n \in \mathbb{N}^{*}}$$

by studying the sequence of successive ratios:

$$\frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^{n} \xrightarrow{n \to \infty} \frac{1}{\mathrm{e}}$$

Hence,

$$\sqrt[n]{n!}=n \cdot \frac{\sqrt[n]{n!}}{n} \xrightarrow{n \to \infty} \infty \cdot \frac{1}{\mathrm{e}}=\infty$$

Here is another way considering

• $$e^{\frac{\ln n!}{n}}$$

$$\begin{eqnarray*} e^{\frac{\ln n!}{n}} & = & \sqrt[n]{n!}\\ & \stackrel{GM-HM}{\geq} & \frac{n}{\frac{1}{1}+\cdots \frac{1}{n}} \\ & \stackrel{\sum_{k=1}^n \frac{1}{k} < \ln n + 1}{>} & \frac{n}{\ln n +1} \\ & \stackrel{n \to \infty}{\longrightarrow} & +\infty \end{eqnarray*}$$