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

[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

 

[istevenson]

\frac{2^n}{n!}=\frac{2}{n}\frac{2}{n-1}...\frac{2}{2}<br /> \frac{2}{1}

so when n\rightarrow\infty

we get \frac{2^n}{n!}&lt;\frac{4}{n}

to any \varepsilon &gt; 0

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

when n>N

\frac{2^n}{n!}&lt;\frac{4}{n}<br /> &lt;\frac{4}{[\frac{4}{\varepsilon}]+1}&lt;\varepsilon

now we get the conclusion:

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