Proving lim (as n -> infinity) 2^n/n = 0

[vikkisut88]

Homework Statement

Prove that lim n \rightarrow\infty 2^{}n/n! = 0

Homework Equations

This implies that 2^{}n/n! is a null sequence and so therefore this must hold:
(\forall E >0)(\existsN E N^{}+)(\foralln E N^{}+)[(n > N) \Rightarrow (|a_{}n| < E)

The Attempt at a Solution

Whenever I have proved these before I have tried to eliminate n from the top and then use Archimedian Principle to finsih off the proof. However I don't know how to do this in this case. I have the idea that 2/k \leq2/3 for any k\geq3

 

[HallsofIvy]

It is very easy to see that
\frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}= \frac{2}{n+1}
so that each term is less than
\frac{2}{n+1}
times the previous term.

It is also very easy to see that, for n= 4,
\frac{2^n}{n!}= \frac{16}{24}&lt; 1[/itex]<br /> <br /> That is, if n&gt; 4, <br /> \frac{2^n}{n!}&amp;lt; \left(\frac{2}{5}\right)^n\frac{2}{3}<br /> and that last goes to 0.

 

[vikkisut88]

i see where you're going but where does that final 2/3 come from?