1. factorials

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

2. Originally Posted by alexandrabel90
may i know how to prove the theorem that factorials beat exponentials?
$\displaystyle n!\underset{n\to\infty}{\sim}\sqrt{2\pi n}n^ne^{-n}$

3. 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?

4. 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?

5. 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 $\displaystyle \lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0$.

6. 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.

7. 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?

8. 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

9. Originally Posted by Drexel28
How do you know the OP has seen series?
What does OP stand for?

10. 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
$\displaystyle \lim \frac{a^n}{n!}=0$

11. 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?

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

12. Originally Posted by dwsmith
What does OP stand for?
"Original Poster"

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

14. 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
$\displaystyle \lim \frac{n!}{a^{n}}=\infty,\lim \frac{a^{n}}{n!}=0.$