What is the result of $ \lim_{n \rightarrow \infty} \frac{ 2^n }{ n! } $ and why?

Question

What is the result of the next limit: $ \displaystyle\lim_{n \rightarrow \infty} \frac{ 2^n }{ n! } $

Why (theorem)?

You could of course use stirling, but its not really necessary. — Vincent Pfenninger, Aug 26 '13 at 08:40
@Prism Looked into it. It's definately a duplicate; flagged as such. — AlexR, Aug 26 '13 at 08:55
http://math.stackexchange.com/a/77561/8581, I think, could be the best. — Mikasa, Aug 26 '13 at 09:03

score 5 · Answer 1 · answered Aug 26 '13 at 08:37

$$\frac{2^n}{n!} = \frac{2}{n} \cdot \frac{2}{n - 1}\cdot \frac{2}{n - 2}\cdot \dots \cdot \frac{2}{2}\cdot \frac{2}{1} \leq \frac{2}{n} \frac{2}{2} \frac{2}{2} ... \frac{2}{2} \frac{2}{1} = \frac{4}{n} \to 0$$

as $n \to \infty$.

What is the result of $ \lim_{n \rightarrow \infty} \frac{ 2^n }{ n! } $ and why?

1 Answers1