Find $\ \limsup_{n\to \infty}(\frac{2^n}{n!}) $ Find$\ \limsup_{n\to \infty}(\frac{2^n}{n!}) $
The following 2 facts have already been proven.


*

*$\ 2^n < n! $ for $n \ge 4$ (Proof by induction)

*$\ \frac{2^{n+1}}{(n+1)!} < \frac{2^n}{n!} $ for $\ n\ge 4 $
It would be sufficient to show that $\ \lim_{n\to \infty} \frac{2^n}{n!} = 0 $
However, what are some ways to show this? 
I have found a proof using the definition of convergence. 
(For all $\ \epsilon > 0 $, Find N such that if n >= N, $\ |\frac{2^n}{n!} - 0| < \epsilon $)
However, I was wondering if there was a simpler proof.
Thanks!

Here's an alternative idea: 
$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$ converges for all $x$ by the ratio test so, in particular, $\lim_{n \to \infty} \frac{2^n}{n!} = 0$.
