How to prove that $\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0$ [duplicate]
It recently came to my mind, how to prove that the factorial grows faster than the exponential, or that the linear grows faster than the logarithmic, etc...
I thought about writing: $$ a(n) = \frac{k^n}{n!} = \frac{ k \times k \times \dots \times k}{1\times 2\times\dots\times n} = \frac k1 \times \frac k2 \times \dots \times \frac kn = \frac k1 \times \frac k2 \times \dots \times \frac kk \times \frac k{k+1} \times \dots \times \frac kn $$ It's obvious that after k/k, every factor is smaller than 1, and by increasing n, k/n gets closer to 0, like if we had $\lim_{n \to \infty} (k/n) = 0$, for any constant $k$.
But, I think this is not a clear proof... so any hint is accepted. Thank you for consideration.
The series for $e^k$ $$ \sum_{n=0}^\infty\frac{k^n}{n!} $$ converges by the ratio test. The terms of a convergent series must tend to $0$.
For $n\ge2k$, the ratio of terms is $\frac{k^{n+1}/(n+1)!}{k^n/n!}=\frac{k}{n+1}<\frac{1}{2}$. We can remove the reference to series (which seems to have bothered someone) with the following sandwich, valid for $n\ge2k$: $$ 0\le\frac{k^n}{n!}\le\frac{k^{2k}}{(2k)!}\left(\frac{1}{2}\right)^{n-2k} $$ which shows that $\displaystyle\lim_{n\to\infty}\frac{k^n}{n!}=0$.
If you know that this limit exists, you have $$ \lim_{n \to \infty} \frac{k^n}{n!} = \lim_{n \to \infty} \frac{k^{n+1}}{(n+1)!} = \lim_{n \to \infty} \frac k{n+1} \frac{k^n}{n!} = \left(\lim_{n \to \infty} \frac k{n+1} \right) \left( \lim_{n \to \infty} \frac {k^n}{n!} \right) = 0. $$ Can you think of a short way to show the limit exists? (You need existence to justify my factoring of the limits at the end. If you don't have that then there's no reason for equality to hold.)