Alternative proof about divergence of of {n!/(a^n)}

Question

Can we find a more elegant or special proof of $\frac {n!}{a^n}\to \infty, a>0$?

For example I suggest this:

$\forall a>0\forall n\in \Bbb N:\frac {n!}{a^n}=\int _{0}^{\infty}(\frac xa)^n e^{-x}dx=\int _0^a (\frac xa)^n e^{-x}dx +\int_a^\infty (\frac xa)^n e^{-x}dx\geq \int_a^\infty (\frac xa)^n e^{-x}dx=-(\frac xa)^n e^{-x}|_a^\infty +\frac na\int_a^\infty (\frac xa)^{n-1} e^{-x}dx=...=e^{-a}(1+\frac {n}{a}+\frac {n(n-1)}{a^2}+...+\frac {n!}{a^n})\geq \frac {e^{-a}}{a}n\to \infty.$

P.S. I am asking about a non trivial answer.

Answer 1

I can give a proof by using only the definition and an elementary result. The statement is equivalent to: $\lim\dfrac{a^{n}}{n!}=0$.

Let $N\in\mathbb{N}$ such that $N>a$. For $n>N$ we have

$\dfrac{a^{n}}{n!}=\dfrac{a}{1}\cdot \dfrac{a}{2}\cdot \ldots\cdot \dfrac{a}{N-1}\cdot \dfrac{a}{N}\cdot \ldots\cdot \dfrac{a}{n}$$\leq\,J\cdot \dfrac{a}{N}\cdot \dfrac{a}{N}\cdot \ldots\cdot \dfrac{a}{N}=J(\dfrac{a}{N})^{n-N+1}$

where $J=\dfrac{a^{N}}{(N-1)!}$ is a constant. (Because $a, N$ are constants.)

$\lim(\dfrac{a}{N})^{n-N+1}=(\dfrac{a}{N})^{-N+1}\lim(\dfrac{a}{N})^{n}$.

But by our choice of $N$ we have $\dfrac{a}{N}<1$ and hence $\lim(\dfrac{a}{N})^{n}=0$.

Thus $\lim\dfrac{a^{n}}{n!}=0$, and we are done!

Answer 2

Let $$m<a<m+1, m<n.$$ Then $$\frac{a^n}{n!}=\frac{(a\cdot a\cdot a\cdot a\cdot a\cdot \ldots\cdot a)(a\cdot a\cdot a\cdot a\cdot \ldots\cdot a)}{[(1\cdot 2\cdot 3\cdot 4\cdot\ldots \cdot m)(m+1)(m+2)(m+3)\ldots n]}=\frac{a^m}{m!}\frac{a}{m+1}\frac{a}{m+2}\frac{a}{m+3}\ldots\frac{a}{n}.$$ Next, $$\frac{a}{m+1}>\frac{a}{m+2}>\frac{a}{m+3}>\ldots>\frac{a}{n}.$$ Then we have $$0 <\frac{a^n}{n!}\le \frac{a^m}{m!}\left(\frac{a}{m+1}\right)^{n-m}.$$ Also, $$\frac{a}{m+1}<1\implies \lim_{n\to \infty}\left( \frac{a}{m+1}\right)^{n-m}\to 0.$$ Finally, by the sandwich theorem, $$\lim_{n\to \infty}\frac{a^n}{n!}\to 0,\quad a>0.$$