Improve the inequality

In the induction proof for the inequality $\left( \frac{n}{3} \right)^n < n!$ one inevitably finds the known expression $\left( 1 + \frac{1}{n} \right)^n$ and uses the fact that this is between $2$ and $3$ to conclude. But having more experience, we know that $\left( 1 + \frac{1}{n} \right)^n$ is related to the famous constant $e$ . In particular, $\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e$ so one way of getting a better inequality is finding a stronger way to relate $e$ and $\left( \frac{n}{3} \right)^n$ . Consider the following: Let $u = n-1$ . Then

$\left( \frac{n}{3} \right)^n = \left( \frac{u+1}{3} \right)^{u+1} = \frac{1}{3^{u+1}} (u+1)^{(u+1)} = \frac{1}{3^{u+1}} \sum_{k=0}^{u+1} {u+1 \choose k} u^k \leq \frac{1}{3^{u+1}} \sum_{k=0}^{u+1} \frac{(u+1)!}{k!} u^k < \frac{(u+1)!}{3^{u+1}} \sum_{k=0}^{\infty} \frac{u^k}{k!} = \frac{(u+1)!}{3^{u+1}} e^{u} = \frac{n!}{3^{n}} e^{n-1}$

This implies

$\frac{n!}{\left( \frac{n}{3} \right)^n} > \frac{n!}{ \frac{n!}{3^{n}} e^{n-1}} = \frac{3^n}{e^{n-1}}$

And notice that this last expression goes to infinity so to pick our $n_0$ simply choose one such that $\frac{3^{n_0}}{e^{n_0 -1}} > 3^a$ and then that will imply that for $n \geq n_0$ , $\frac{n!}{\left( \frac{n}{3} \right)^n} > 3^a \implies n! > 3^a \left( \frac{n}{3} \right)^n = \frac{n^n}{3^{n-a}}$

And with just this, we were able to obtain some very nice inequalities:

Since $\ln(x)$ is monotonically increasing for $x>0$ , we have
$\ln(n!)= \sum_{r=1}^{n} \ln(r) \geq \int_{1}^{n} \ln(x) dx > n\ln(n/e).$ Exponentiating both sides, we have $n! > (\frac{n}{e})^n.$ Since $(3/e)^n \to \infty$ , as $n \to \infty$ , for any $a$ , there exists a finite $n_0$ , such that $3^{n-a} > e^n$ for all $n\geq n_0$ . This gives, for all $n \geq n_0$ , $n! > \frac{n^n}{3^{n-a}} .$