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

1. Jan 18, 2009

### vikkisut88

1. The problem statement, all variables and given/known data

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

2. Relevant equations
This implies that 2$$^{}n$$/n! is a null sequence and so therefore this must hold:
($$\forall$$ E >0)($$\exists$$N E N$$^{}+$$)($$\forall$$n E N$$^{}+$$)[(n > N) $$\Rightarrow$$ (|a$$_{}n$$| < E)

3. 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 $$\leq$$2/3 for any k$$\geq$$3

2. Jan 18, 2009

### HallsofIvy

Staff Emeritus
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,
$$\frac{2^n}{n!}= \frac{16}{24}< 1[/itex] That is, if n> 4, [tex]\frac{2^n}{n!}< \left(\frac{2}{5}\right)^n\frac{2}{3}$$
and that last goes to 0.

3. Jan 18, 2009

### vikkisut88

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