# factorials

• May 9th 2010, 11:57 AM
alexandrabel90
factorials
may i know how to prove the theorem that factorials beat exponentials?
• May 9th 2010, 12:18 PM
Drexel28
Quote:

Originally Posted by alexandrabel90
may i know how to prove the theorem that factorials beat exponentials?

$\displaystyle n!\underset{n\to\infty}{\sim}\sqrt{2\pi n}n^ne^{-n}$
• May 9th 2010, 01:22 PM
alexandrabel90
sorry, i dont really get what you are trying to hint to me. how do i make use of this formula to show that

eg the fraction of an exponential over factorial will tend to 0?
• May 9th 2010, 01:34 PM
Drexel28
Quote:

Originally Posted by alexandrabel90
sorry, i dont really get what you are trying to hint to me. how do i make use of this formula to show that

eg the fraction of an exponential over factorial will tend to 0?

$\displaystyle \lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\ frac{a^n}{\sqrt{2\pi n}n^n e^{-n}}=\frac{1}{\sqrt{2\pi}}\lim_{n\to\infty}\frac{(a \cdot e)^n}{n^{n+\frac{1}{2}}}=0$

Why?
• May 9th 2010, 01:40 PM
Bruno J.
Here's a much simpler way. We know $\displaystyle e^a = \sum_{n=0}^\infty \frac{a^n}{n!}$ converges for every real number $\displaystyle a$, which immediately implies $\displaystyle \lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0$.
• May 9th 2010, 01:49 PM
Drexel28
Quote:

Originally Posted by Bruno J.
Here's a much simpler way. We know $\displaystyle e^a = \sum_{n=0}^\infty \frac{a^n}{n!}$ converges for every real number $\displaystyle a$, which immediately implies $\displaystyle \lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0$.

How do you know the OP has seen series? Really all you tacitly assumed was that if $\displaystyle \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1$ then it's dominated by a convergent geometric sequence.
• May 9th 2010, 02:00 PM
alexandrabel90
isit because n(n+0.5) can be taken as n^n thus the fraction becomes (a.e/n)^n where (a.e/n) is less than 1?
• May 9th 2010, 02:01 PM
alexandrabel90
in my course that im learning, i have yet to learn the formula for factorials. so assuming that i dont know that formula, is there another method to prove it?

thanks
• May 9th 2010, 02:32 PM
dwsmith
Quote:

Originally Posted by Drexel28
How do you know the OP has seen series?

What does OP stand for?
• May 9th 2010, 03:04 PM
Renji Rodrigo
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
$\displaystyle \lim \frac{a^n}{n!}=0$
• May 9th 2010, 03:06 PM
Bruno J.
Quote:

Originally Posted by Drexel28
How do you know the OP has seen series? Really all you tacitly assumed was that if $\displaystyle \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1$ then it's dominated by a convergent geometric sequence.

How do you know the OP has seen Stirling's formula? (Thinking)

You're right about all that I've assumed, which is comparatively quite little.
• May 9th 2010, 03:07 PM
Bruno J.
Quote:

Originally Posted by dwsmith
What does OP stand for?

"Original Poster"
• May 9th 2010, 03:21 PM
alexandrabel90
by the way, is there a way to use sandwich theorem to prove this?
• May 9th 2010, 03:45 PM
Renji Rodrigo
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
$\displaystyle \lim \frac{n!}{a^{n}}=\infty,\lim \frac{a^{n}}{n!}=0.$