may i know how to prove the theorem that factorials beat exponentials?
Proposition[ratio teste for sequences]
Iff (x_n) is a sequence with x_n >0 forall n in N
and $\displaystyle \lim \frac{x_{n+1}}{x_{n}} <1 $
then $\displaystyle \lim x_{n} =0 $.
Take $\displaystyle x_n =\frac{a^n}{n!} $ a>0
then
$\displaystyle \frac{x_{n+1}}{x_{n}} = \frac{a^{n+1} n!}{(n+1)! a^n}=\frac{a}{n+1} $
so $\displaystyle \lim \frac{x_{n+1}}{x_{n}}=0 <1 $
then
$\displaystyle \lim \frac{a^n}{n!}=0 $
Given a>0 exists $\displaystyle n_{0}\in N $ with $\displaystyle n_{0}>2a $,
$\displaystyle \frac{n_{0}}{a}>2 $ so for $\displaystyle n\geq n_{0}$ we have
$\displaystyle \frac{n}{a}\geq \frac{n_{0}}{a}>2$
taking the product in $\displaystyle \frac{t}{a}>2 $ from $\displaystyle t=n_{0}$ to $\displaystyle n$ we have
$\displaystyle \prod^{n}_{t=n_{0}}\frac{t}{a}>\prod^{n}_{t=n_{0}} 2=2^{n+1-n_{0}} $
taking the product with $\displaystyle \prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}=p$ on the both sides off the inequality
$\displaystyle \prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}\prod^{n}_{t=n_{0}}\frac{t}{a}= \prod^{n}_{t=1}\frac{t}{a}=\frac{n!}{a^{n}}>p.2^{n +1-n_{0}} $
$\displaystyle \frac{n!}{a^{n}}>p.2^{n+1-n_{0}}$
the limit on the right goes to infinity so the limit on the left too
, so
$\displaystyle \lim \frac{n!}{a^{n}}=\infty,\lim \frac{a^{n}}{n!}=0. $