$\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}$In this post I’m going to evaluate
$$
\lim_{n\to\infty} (n!)^{1/n}
$$
by connecting it to the harmonic series. I’ll use $a_n = (n!)^{1/n}$ when I want a more compact notation.
I’ll begin with some intuitions and outlines of other proofs for why this limit is $\infty$ before I move to the one I particularly want to spend time on.
First of all, if $p > 0$ and I instead considered $\lim_{n} (p^n)^{1/n}$, this is just $p$ so this means that the power of $1/n$ is enough to balance out exponential growth. But factorial growth dominates exponential growth: now taking $p>1$,
$$
\begin{aligned}
&\frac{n!}{p^n} = \frac{1\cdot2\cdot \ldots \cdot \lfloor p \rfloor}{p^{\lfloor p \rfloor}} \cdot \frac{(\lfloor p \rfloor+1)(\lfloor p \rfloor+2) \cdot \ldots \cdot (n-1)n}{ p^{n-\lfloor p \rfloor}} \\
&= \frac{\lfloor p \rfloor!}{p^{\lfloor p \rfloor}} \cdot \prod_{j=1}^{n-\lfloor p \rfloor} \left(\frac{\lfloor p \rfloor+j}{p}\right).
\end{aligned}
$$
$\frac{\lfloor p \rfloor!}{p^{\lfloor p \rfloor}}$ is fixed so as $n$ grows, the only thing that changes is the product of terms like $\frac{\lfloor p\rfloor +j}p$. $j \geq 1$ so all of these terms are strictly positive, and as $n$ grows they get bigger and bigger, so $\frac{n!}{p^n}\to\infty$. This suggests that $n!$ will grow faster than the powers of $1/n$ can handle.
Another way to get a sense of what happens with this limit comes from Stirling’s approximation, which says that
$$
\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n} (n/e)^n} = 1
$$
so $n!$ is well approximated by $\sqrt{2\pi n} (n/e)^n$ as $n\to\infty$. Plugging in this approximation to the original limit yields
$$
\lim_{n\to\infty} \left[\sqrt{2\pi n}(n/e)^n\right]^{1/n} = e^{-1} \lim_n\; (2\pi)^{\frac 1{2n}} n^{1 + \frac 1{2n}} = \infty.
$$
Yet another way is that I can just compute the ratio of successive terms:
$$
\begin{aligned}
&\frac{a_{n+1}}{a_n} = \frac{[(n+1)!]^{1/(n+1)}}{(n!)^{1/n}} = \frac{\{[n!(n+1)]^{1/n}\}^{\frac{n}{n+1}}}{(n!)^{1/n}} \\&
= \frac{[(n+1)^{1/n} a_n]^{\frac n{n+1}}}{a_n} = (n+1)^\frac{1}{n+1} a_n ^{-\frac 1{n+1}}
= \left(\frac{n+1}{a_n}\right)^{\frac 1{n+1}}.
\end{aligned}
$$
For $n >1$ I know
$$
n^n > n!
$$
since the left hand side is a product of larger numbers than the right hand side (expect for one factor of $n$ that they both share). This means
$$
\begin{aligned}
&(n+1)^n > n^n > n! \\
&\implies n+1 > (n!)^{1/n} = a_n \\
&\implies \frac{n+1}{a_n} > 1 \\
&\implies \left(\frac{n+1}{a_n}\right)^{\frac 1{n+1}} > 1.
\end{aligned}
$$
This shows that the terms in $a_n$ are getting larger and larger so $\lim a_n = \infty$.
The final idea I’ll mention, which leads to my particular line of attack for this problem, is to note that $(n!)^{1/n}$ is the geometric mean of the first $n$ natural numbers, so this limit is like evaluating the geometric mean of $\mathbb N$. Any measure of central tendancy worth its salt should give an infinite value for this.
I’ll use the famous AM-GM inequality, which states that for non-negative $x_1, \dots, x_n\in\R$
$$
\bar x_n \geq \left(\prod_i x_i\right)^{1/n}.
$$
The proof for this uses the concavity of the logarithm and Jensen’s inequality:
$$
\log \bar x_n \geq \frac 1n \sum_i \log x_i = \log \left(\prod_i x_i\right)^{1/n}
$$
so taking the exponential of each side finishes it.
I’ll apply this to prove the HM-GM inequality, which I’ll then use to get a sequence that bounds $(a_n)$ from below.
Result 1: HM-GM inequality. For any non-negative $x_1,\dots,x_n\in\R$
$$
\frac{n}{\sum_i x_i^{-1}} \leq \left(\prod_i x_i\right)^{1/n}.
$$
Pf: I’ll plug $x_i^{-1}$ in to the AM-GM inequality so
$$
\frac 1n \sum_i x_i^{-1} \geq \left(\prod_i x_i^{-1}\right)^{1/n} = \frac 1{\left(\prod_i x_i\right)^{1/n}}.
$$
Inverting and reversing the direction of the inequality, I have
$$
\frac{n}{\sum_i x_i^{-1}} \leq \left(\prod_i x_i\right)^{1/n}
$$
as desired.
$\square$
This means that
$$
(n!)^{1/n} \geq \frac{n}{\sum_{k=1}^n k^{-1}}
$$
and the denominator of the right hand side can be recognized as $H_n$, the $n$th harmonic number.
Result 2: $H_n\to\infty$ as $n\to\infty$. In other words, the harmonic series $\sum_{k=1}^\infty \frac 1k$ diverges.
Pf: I’ll follow a standard proof which notes that the terms of the series are monotonically decreasing. This means I can bound the sum from below by collecting the terms in groups (sequentially, not rearranging the sum, so I’m properly bounding the partial sums) and bounding each group from below. (effectively the Cauchy condensation test). This lower bound will diverge which proves the result.
I’ll collect the terms into groups of size $1$, $2$, $4$, $8$, and etc, and then each group’s size is estimated from below by the minimum value of the group times the group size. This works out to
$$
\begin{aligned}
&1 + \underbrace{\frac 12 + \frac 13}_{\geq 2 \cdot \frac 13} + \underbrace{\frac 14 + \frac 15 + \frac 16 + \frac 17}_{\geq 4 \cdot \frac 17} + \dots \\
&\geq \sum_{k=0}^\infty2^k \frac 1{2^{k+1}-1} = \frac 12 \sum_{k=0}^\infty \frac {2^k}{2^{k}-1/2} \geq \frac 12 \sum_{k=0}^\infty 1 = \infty
\end{aligned}
$$
so $\sum_{k=1}^\infty \frac 1k = \lim_n H_n = \infty$.
$\square$
This shows that $\frac{n}{H_n}$ is a $\frac{\infty}{\infty}$ indeterminate. But $H_n$ grows sublinearly so I will bound it from above so that I can show that $\frac n{H_n}\to\infty$.
Result 3: $H_n \leq 1 + \log n$.
Pf: the main idea is shown in the figure below. I can view $H_n$ as the integral of the step function shown in blue, and then I can note that this step function is bounded above by $y = 1/x$.
As the figure shows, I have
$$
\int_1^n x^{-1}\,\text dx = \log n \geq \frac 12 + \frac 13 + \dots + \frac 1n = H_n -1
$$
so
$$
H_n \leq 1 + \log n
$$
$\square$
I’ve now shown that
$$
(n!)^{1/n} \geq \frac{n}{H_n} \geq \frac n{1+\log n} \to \infty
$$
so
$$
\lim_{n\to\infty} (n!)^{1/n} = \infty
$$
as expected.