factorials

**alexandrabel90** · May 9th 2010, 11:57 AM

may i know how to prove the theorem that factorials beat exponentials?

**Drexel28** · May 9th 2010, 12:18 PM

Originally Posted by alexandrabel90

may i know how to prove the theorem that factorials beat exponentials?

$n!\underset{n\to\infty}{\sim}\sqrt{2\pi n}n^ne^{-n}$

**alexandrabel90** · May 9th 2010, 01:22 PM

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?

**Drexel28** · May 9th 2010, 01:34 PM

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?

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

Why?

**Bruno J.** · May 9th 2010, 01:40 PM

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

**Drexel28** · May 9th 2010, 01:49 PM

Originally Posted by Bruno J.

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

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

**alexandrabel90** · May 9th 2010, 02:00 PM

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?

**alexandrabel90** · May 9th 2010, 02:01 PM

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

**dwsmith** · May 9th 2010, 02:32 PM

Originally Posted by Drexel28

How do you know the OP has seen series?

What does OP stand for?

**Renji Rodrigo** · May 9th 2010, 03:04 PM

Proposition[ratio teste for sequences]
Iff (x_n) is a sequence with x_n >0 forall n in N
and $\lim \frac{x_{n+1}}{x_{n}} <1$
then $\lim x_{n} =0$ .

Take $x_n =\frac{a^n}{n!}$ a>0

then

$\frac{x_{n+1}}{x_{n}} = \frac{a^{n+1} n!}{(n+1)! a^n}=\frac{a}{n+1}$
so $\lim \frac{x_{n+1}}{x_{n}}=0 <1$
then
$\lim \frac{a^n}{n!}=0$

**Bruno J.** · May 9th 2010, 03:06 PM

Originally Posted by Drexel28

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

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

You're right about all that I've assumed, which is comparatively quite little.

**Bruno J.** · May 9th 2010, 03:07 PM

Originally Posted by dwsmith

What does OP stand for?

"Original Poster"

**alexandrabel90** · May 9th 2010, 03:21 PM

by the way, is there a way to use sandwich theorem to prove this?

**Renji Rodrigo** · May 9th 2010, 03:45 PM

Given a>0 exists $n_{0}\in N$ with $n_{0}>2a$ ,

$\frac{n_{0}}{a}>2$ so for $n\geq n_{0}$ we have
$\frac{n}{a}\geq \frac{n_{0}}{a}>2$

taking the product in $\frac{t}{a}>2$ from $t=n_{0}$ to $n$ we have

$\prod^{n}_{t=n_{0}}\frac{t}{a}>\prod^{n}_{t=n_{0}} 2=2^{n+1-n_{0}}$

taking the product with $\prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}=p$ on the both sides off the inequality

$\prod\limits^{n_{0}-1}_{t=1}\frac{t}{a}\prod^{n}_{t=n_{0}}\frac{t}{a}= \prod^{n}_{t=1}\frac{t}{a}=\frac{n!}{a^{n}}>p.2^{n +1-n_{0}}$

$\frac{n!}{a^{n}}>p.2^{n+1-n_{0}}$

the limit on the right goes to infinity so the limit on the left too
, so
$\lim \frac{n!}{a^{n}}=\infty,\lim \frac{a^{n}}{n!}=0.$

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