Elementary proof of lower bound for factorial

Question

Let $\xi$ be a fixed but arbitrary real number. Prove that $n! > \xi^n$ for sufficiently large $n$ in as elementary, short and elegant a way as possible. In particular, you are not allowed to use Stirling's formula, transcendental functions or transcendental numbers (over $\mathbb{Q}[\xi]$); integrals and derivatives of polynomials are allowed, though. You do not have to provide an optimal bound, or even an explicit one.

Bonus problem: If you are feeling adventurous, prove that there is a positive real constant $C$ such that $n! > (Cn)^n$ for sufficiently large $n$ under the same conditions as above.

Answer 1

If $\xi$ is non-positive, there is nothing to prove (similar if $\xi \in (0, 1]$). So let's assume $\xi > 1$. Let $n_0 = \lceil \xi \rceil$. Since for each $i \in \{1, \ldots, n_0\}$, $$\frac{n_0^2 + i}{\xi} \times \frac{i}{\xi} \geq \frac{n_0^2}{\xi^2} \geq 1,$$ and $\frac{m}{\xi} > 1$ for all $m > n_0$, it follows that for all $n \geq n_0^2 + n_0$, $$\frac{n!}{\xi^n} = \frac{n}{\xi} \times \cdots \times \frac{n_0^2 + n_0}{\xi} \times \cdots \times \frac{n_0^2 + 1}{\xi} \times \cdots \times \frac{n_0 + 1}{\xi} \times \frac{n_0}{\xi} \times \cdots \times \frac{1}{\xi} > 1.$$ Therefore the result follows.

Answer 2

Let $N$ be the largest integer such that $N < \xi$. For $n > N$ consider the fraction $\frac{n!}{\xi^n}$ ans split into three parts:

$$\frac{n!}{\xi^n} = \left( \frac{1}{\xi} \dotsb \frac{N}{\xi} \right) \cdot \left( \frac{N+1}{\xi} \dotsb \frac{n-1}{\xi} \right) \cdot \frac{n}{\xi}$$

The first part is a fixed number $S(\xi)$. The second part is clearly $\geq 1$. The third part can be chosen arbitrary large, concretely let $n$ be large enough, such that $\frac{S(\xi)n}{\xi} > 1$ holds.

At the (now stated correctly) bonus problem: The inequality $n! > \left(\frac{n}{3} \right)^n$, which solves the bonus problem, is easily shown by induction, with the induction step needing the inequality $\left(1+\frac{1}{n}\right)^n < 3$. (Instead of $3$ we could use any upper bound of this series)

This inequality is well known (since its fundamental to the existence of $e$ and hence fundamental to mathematics) and there is a one-line proof:

$$\left(1+\frac{1}{n}\right)^n = \sum_{k=0}^n \frac{1}{k!} \frac{n!}{(n-k)!n^k} \leq \sum_{k=0}^n \frac{1}{k!} < \sum_{k=0}^\infty \frac{1}{k!} \leq 1+\sum_{k=1}^\infty \frac{1}{2^{k-1}} = 3.$$ We used $k! \geq 2^{k-1}$, which is trivial, since $k!$ is made up of $k-1$ factors not lower than $2$.

I am pretty sure that there is no way to prove that statement without using (implicitly or explicitly) that the series $\left(1+\frac{1}{n}\right)^n$ is bounded above. Note that you implicitly use this boundedness, once you speak about $\exp$ or $\ln$.

As an argument, note that with $a_n = \frac{n!C^n}{n^n}$, we obtain $\frac{a_{n+1}}{a_n} = \frac{C}{\left(1+\frac{1}{n}\right)^n}$, which shows that the behavior of $\left(1+\frac{1}{n}\right)^n$ for large $n$ is crucial for the behavior of $a_n$ for large $n$. And in particular we see: If we can show boundedness of this series, we have that $\frac{a_{n+1}}{a_n}$ is as large as we want if $C$ is chosen large enough. And of course this implies $a_n > 1$ for $n \gg 0$.

On the other hand, if the series would turn out to be unbounded, we would obtain $a_n \to 0$ for any fixed $C > 0$.

Consequently one could say that the easiest and most elegant proof to the bonus problem comes down to the easiest and most elegant proof the the boundedness of $\left(1+\frac{1}{n}\right)^n$. Since the well-known standard proof has only one line, there is not much to work on this, I guess :P