How to show that lim 2^n/n! = 0 with n--> infinity

1. Sep 21, 2006

kreil

Using the definition of a limit, show that

$$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$

If someone could get me started that would be great.

thanks
josh

Last edited: Sep 21, 2006
2. Sep 21, 2006

istevenson

$$\frac{2^n}{n!}=\frac{2}{n}\frac{2}{n-1}.....\frac{2}{2} \frac{2}{1}$$

so when $$n\rightarrow\infty$$

we get $$\frac{2^n}{n!}<\frac{4}{n}$$

to any $$\varepsilon > 0$$

$$\exists N=[\frac{4}{\varepsilon}]+1$$

when n>N

$$\frac{2^n}{n!}<\frac{4}{n} <\frac{4}{[\frac{4}{\varepsilon}]+1}<\varepsilon$$

now we get the conclusion:

$$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$

Last edited: Sep 21, 2006