Find $\ \limsup_{n\to \infty}(\frac{2^n}{n!}) $

Question

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!

Just strengthen the second fact a little, e.g. $LHS < r RHS$, for some $r<1$. Then you can bound the sequence by a geometric one. — vadim123, Oct 20 '15 at 18:07
See also: Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. — Martin Sleziak, Apr 16 '17 at 15:15

score 7 · Accepted Answer · answered Oct 20 '15 at 18:17

7

This might be a little simpler: For large $n,$ $$\frac{2^n}{n!} = \frac{2}{n}\frac{2}{n-1}\cdots \frac{2}{3}\frac{2}{2}\frac{2}{1}\le \frac{4}{n(n-1)}\cdot 2 = \frac{8}{n(n-1)}\to 0.$$

answered Oct 20 '15 at 18:17

zhw.

105,693

Well done! Concise and simple. +1 – Mark Viola Oct 20 '15 at 18:20
Out of curiosity, why stop at $\frac{\cdots}{n(n-1)}$? Couldn't you also ignore the $n-1$ factor? – user170231 Oct 20 '15 at 19:16
Yes, I could have, good point. I think I thought "series" for a moment and just wrote it down. So in fact the above shows $\sum 2^n/n!$ converges, which is stronger. – zhw. Oct 20 '15 at 19:53

score 3 · Answer 2 · answered Oct 20 '15 at 18:26

3

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, $\lim_{n \to \infty} \frac{2^n}{n!} = 0$.

answered Oct 20 '15 at 18:26

Michael Biro

13,757

Find $\ \limsup_{n\to \infty}(\frac{2^n}{n!}) $

2 Answers2