Jtdylktuy

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

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

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

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

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

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.

Another way to show $$lim_{ntoinfty}frac{ln(n!)}{n}=infty$$ is to consider the following property of logarithms: $$log(n!)=log(n)+log(n-1)+cdots+log(2)>frac{n}{2}logleft(frac n2right).$$ Now $$frac{log (n!)}{n}>frac{log(n/2)}{2}.$$ As $ntoinfty$, this clearly diverges to $+infty$.

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 $3n$.

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,

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

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

• Prove that $n ln(n) - n le ln(n!)$ without Stirling

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*}