# Calculating the limit of $\sqrt[n]{n!}$ as $n \rightarrow \infty$ [duplicate]

Some may consider this a duplicate, but the only similar question I have found make use of Stirling's identity and then conclude the result. I would like to try and avoid this and so would like a more elementary approach of solving the problem.

This is my progress so far:

I think the following is true (and the limit I'm trying to calculate would follow immediately):

For every $k\in\mathbb{N}$, for sufficiently large $n$ we have $n! > k^{n}$

I'm not sure how to prove this result, but it would be equivalent to proving that

For sufficiently large $n$, $\sum_{i=1}^{n} \log_{k}i > n$ for any $k \in \mathbb{N}$.

Would somebody be able to provide a hint on how to proceed, and whether or not my method could be fruitful?

## marked as duplicate by Hans Lundmark, Misha Lavrov, Xander Henderson, Ethan Bolker, user99914 Jun 13 '18 at 20:50

• there is an easier proof, among the $n$ numbers $1,\ldots, n$, at least half of them is greater or equal to $\frac{n}{2}$. so $\sqrt[n]{n!} \ge \sqrt{n/2}$. – achille hui Jun 13 '18 at 9:37
• I know this is short but it's a perfect answer, so if you submit it as an actual answer I would accept it! – Adam Higgins Jun 13 '18 at 9:38

Among the $n$ numbers $1,\ldots, n$, at least half of them is $\ge \frac{n}{2}$ while the rest is $\ge 1$.
When $n \ge 2$, this leads to

$$n! \ge \left(\frac{n}{2}\right)^{\# \{ k\,:\,\frac{n}{2} \le k \le n\} } \ge \left(\frac{n}{2}\right)^{\frac{n}{2}}\quad \implies\quad \sqrt[n]{n!} \ge \sqrt{\frac{n}{2}}$$

As a result, $\sqrt[n]{n!}$ diverges to $\infty$ as $n \to \infty$.

By the ratio test, the power series $\sum_{n=0}^{\infty}\frac{x^n}{n!}$ has radius of convergence $= \infty$. Thus

$\lim \sup \frac{1}{n!^{1/n}}=0$, hence $\lim_{n\to\infty}n!^{1/n}=\infty$.

Correct me if wrong .

$e^n \ge n^k/k!,$ $k \in \mathbb{Z^+}.$

Set $k=n:$

$n!\ge n^n/e^n = (n/e)^n.$

$\sqrt[n]{n!} \ge (n/e).$

Hence?

• Whilst this is correct, I'm assuming that the first inequality that you stated comes from the power series of $e^{x}$. I prefer Achille's answer since it is, in some sense, more elementary, although this is a nice slick answer. – Adam Higgins Jun 13 '18 at 9:55
• Adam.Yes, power series of $e^x$.Yes, again, Achille's solution is more elementary . – Peter Szilas Jun 13 '18 at 10:01

In this answer, Riemann Sums are used to show that \begin{align} \lim_{n\to\infty}\frac1n\log\left(\frac{n!}{n^n}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac kn\right)\frac1n\\ &=\int_0^1\log(x)\,\mathrm{d}x\\[9pt] &=-1 \end{align} Thus, $$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}n=\frac1e$$ This implies that your limit is $\infty$.