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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.

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    $\begingroup$ Cesaro-Stolz is the key here. $\endgroup$ – Paramanand Singh Dec 8 '18 at 7:34
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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 $+\infty$.

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    $\begingroup$ Thank you for the answer, could you please elaborate on $\ln(n!) > \frac{n}{2}\ln\left(\frac{n}{2}\right)$? $\endgroup$ – roman Dec 7 '18 at 15:59
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    $\begingroup$ @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.) $\endgroup$ – Clayton Dec 7 '18 at 16:05
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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 $3n$.

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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

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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,

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

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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*}

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