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

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- May 9th 2010, 11:57 AMalexandrabel90factorials
may i know how to prove the theorem that factorials beat exponentials?

- May 9th 2010, 12:18 PMDrexel28
- May 9th 2010, 01:22 PMalexandrabel90
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 PMDrexel28
- May 9th 2010, 01:40 PMBruno 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 PMDrexel28
- May 9th 2010, 02:00 PMalexandrabel90
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 PMalexandrabel90
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 PMdwsmith
- May 9th 2010, 03:04 PMRenji 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 PMBruno J.
- May 9th 2010, 03:07 PMBruno J.
- May 9th 2010, 03:21 PMalexandrabel90
by the way, is there a way to use sandwich theorem to prove this?

- May 9th 2010, 03:45 PMRenji 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. $