0
$\begingroup$

This question already has an answer here:

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?

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ 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 $\sqrt[n]{n!} \ge \sqrt{n/2}$. $\endgroup$ – achille hui Jun 13 '18 at 9:37
  • $\begingroup$ I know this is short but it's a perfect answer, so if you submit it as an actual answer I would accept it! $\endgroup$ – Adam Higgins Jun 13 '18 at 9:38
5
$\begingroup$

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 $n \to \infty$.

$\endgroup$
1
$\begingroup$

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 $\lim_{n\to\infty}n!^{1/n}=\infty$.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Adam Higgins Jun 13 '18 at 9:55
  • $\begingroup$ Adam.Yes, power series of $e^x$.Yes, again, Achille's solution is more elementary . $\endgroup$ – Peter Szilas Jun 13 '18 at 10:01
0
$\begingroup$

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 $\infty$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.