# Math Help - Factorial limits

1. ## Factorial limits

Someone told me this wasn't correct?

I wanted to show that $\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=0$
two ways

Method one

Since $\frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!},\forall{n}> 0$

and also since as n gets sufficiently large there $\exists\text{ an }N\backepsilon\forall{n>N}\frac{3^nn!}{(2n)!}\leq\ frac{n^n}{(2n)!}$

So from there I set up my inequalities

$\frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!}\leq\frac{n^ n}{(2n)!}$

and $\therefore\lim_{n\to\infty}\frac{1}{(2n)!}\leq\lim _{n\to\infty}\frac{3^nn!}{(2n)!}\leq\lim_{n\to\inf ty}\frac{n^n}{(2n)!}$

Knowing the left and right integrals we see that

$0\leq\lim_{n\to\infty}\frac{3^nn!}{(2n)!}\leq{0}$

$\therefore\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=0$

by the squeeze theorem

Method two

Since as $n\to\infty$

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

and $(2n)!\sim{\sqrt{4\pi{n}}(2n)^{2n}e^{-2n}}=2\sqrt{\pi{n}}4^n(n^n)^2(e^{-n})^2$

$\therefore\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=\li m_{n\to\infty}\frac{3^n\sqrt{2}\sqrt{\pi{n}}n^ne^{-n}}{2\sqrt{\pi{n}}4^n(n^n)^2(e^{-n})^2}=\lim_{n\to\infty}\frac{3^ne^n}{\sqrt{2}4^nn ^n}$

and just to completely be thorough I included this step

$\lim_{n\to\infty}\frac{3^ne^n}{\sqrt{2}4^nn^n}=\fr ac{1}{\sqrt{2}}\cdot\lim_{n\to\infty}\frac{3^n}{4^ n}\cdot\lim_{n\to\infty}\frac{e^n}{n^n}=\frac{1}{\ sqrt{2}}\cdot{0}\cdot{0}=0$

could somone point me in the direction of my mistake? I am sure it is a wrong assumption

Thanks...Mathstud

2. Hello

Method one

Since $\frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!},\forall{n}> 0$

and also since as n gets sufficiently large there $\exists\text{ an }N\backepsilon\forall{n>N}\frac{3^nn!}{(2n)!}\leq\ frac{n^n}{(2n)!}$
You should show that $3^nn!.
So from there I set up my inequalities

$\frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!}\leq\frac{n^ n}{(2n)!}$

and $\therefore\lim_{n\to\infty}\frac{1}{(2n)!}\leq\lim _{n\to\infty}\frac{3^nn!}{(2n)!}\leq\lim_{n\to\inf ty}\frac{n^n}{(2n)!}$

Knowing the left and right integrals
You should also show that $\lim_{\infty}\frac{n^n}{(2n)!}=0$

we see that

$0\leq\lim_{n\to\infty}\frac{3^nn!}{(2n)!}\leq{0}$

$\therefore\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=0$

by the squeeze theorem
I did not notice anything wrong... (by the way, you don't need $\frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!}$ to apply the squeeze theorem since $\frac{3^nn!}{(2n)!}>0$)

Method two

[...]
It is correct too.

3. Originally Posted by Mathstud28
Someone told me this wasn't correct?
and also since as n gets sufficiently large there $\exists\text{ an }N\backepsilon\forall{n>N}\frac{3^nn!}{(2n)!}\leq\ frac{n^n}{(2n)!}$
This is not true.

$3^n n! \geq n^n$ for all n>0.

4. $\lim_{n\to\infty} \frac{3^n n!}{(2n)!}$

$\lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}\cdot \overbrace{(\not 1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n)}^{\text{n terms}}}{\underbrace{[\not1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n \cdot (n+1) \cdot (n+2) \cdot ...\cdot (2n)]}_{\text{2n terms}}}$

$\lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}}{\underbrace{[(n+1)\cdot(n+2)\cdot(n+3)\cdot...\cdot(2n)]}_{\text{n terms}}}$

$\lim_{n\to\infty} \underbrace{\frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\f rac{3}{n+3}\cdot...\cdot \frac{3}{2n}}_{\text{n terms}}$

Each term will approach 0 while n approaches infinity. So we can say that $\lim_{n\to\infty} \frac{3^n n!}{(2n)!} = 0$

5. Originally Posted by flyingsquirrel
Hello

You should show that $3^nn!.

You should also show that $\lim_{\infty}\frac{n^n}{(2n)!}=0$

I did not notice anything wrong... (by the way, you don't need $\frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!}$ to apply the squeeze theorem since $\frac{3^nn!}{(2n)!}>0$)

It is correct too.
Originally Posted by wingless
$\lim_{n\to\infty} \frac{3^n n!}{(2n)!}$

$\lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}\cdot \overbrace{(\not 1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n)}^{\text{n terms}}}{\underbrace{[\not1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n \cdot (n+1) \cdot (n+2) \cdot ...\cdot (2n)]}_{\text{2n terms}}}$

$\lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}}{\underbrace{[(n+1)\cdot(n+2)\cdot(n+3)\cdot...\cdot(2n)]}_{\text{n terms}}}$

$\lim_{n\to\infty} \underbrace{\frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\f rac{3}{n+3}\cdot...\cdot \frac{3}{2n}}_{\text{n terms}}$

Each term will approach 0 while n approaches infinity. So we can say that $\lim_{n\to\infty} \frac{3^n n!}{(2n)!} = 0$
To Flyingsquirrel
Thank you very much!

As for your comment on the limit, I know we both know it is 0; but to show it see below

Using the same trick of asymptotic equivlaence

$\lim_{n\to\infty}\frac{n^n}{(2n)!}=\lim_{n\to\inft y}\frac{e^{2n}}{2\sqrt{\pi{n}}4^nn^n}=0$

To Wingless

Thank you very much. As for your first comment on the falseness of that inequality, I really appreciate it, its weird I am like inequality impaired . I will be more careful next time.

6. Originally Posted by Mathstud28
To Wingless

Thank you very much. As for your first comment on the falseness of that inequality, I really appreciate it, its weird I am like inequality impaired . I will be more careful next time.

You may be right about that. OK, then let me say:

$\left(\frac{3}{2n}\right)^n \leq \left ( \frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\frac{3}{n+3}\ cdot...\cdot \frac{3}{2n} \right ) \leq \left(\frac{3}{n}\right)^n$

$\lim_{n\to\infty} \left(\frac{3}{n}\right)^n = \lim_{n\to\infty} \left(\frac{3}{2n}\right)^n = 0$
( You can find this easily by a cheap l'hopital trick, turn (3/n)^n into e^(n ln(3/n)... )

I hope it's clear now =)

7. Originally Posted by wingless
You may be right about that. OK, then let me say:

$\left(\frac{3}{2n}\right)^n \leq \left ( \frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\frac{3}{n+3}\ cdot...\cdot \frac{3}{2n} \right ) \leq \left(\frac{3}{n}\right)^n$

$\lim_{n\to\infty} \left(\frac{3}{n}\right)^n = \lim_{n\to\infty} \left(\frac{3}{2n}\right)^n = 0$
( You can find this easily by a cheap l'hopital trick, turn (3/n)^n into e^(n ln(3/n)... )

I hope it's clear now =)
Yeah thanks! I would never use L'hopital's on something like this, its probably easiest with the stirlings formula, it isnt as fun anyways

8. Originally Posted by Mathstud28
Yeah thanks! I would never use L'hopital's on something like this, its probably easiest with the stirlings formula, it isnt as fun anyways
Ain't this the spirit of a true mathematician or what?

9. Ok, note that $\sqrt[n]{n!}\sim{\frac{n}{e}}$ (*) ( use Stirling's formula if you want, here you have a proof of (*): http://www.mathhelpforum.com/math-he...-sequence.html )

We have: $\lim \frac{\sqrt[n]{n!}}{\sqrt[n]{(2n)!}}=\lim \frac{\frac{n}{e}}{\left(\frac{2n}{e}\right)^2}=0$

Thus: $\lim \sqrt[n]{\frac{3^n\cdot{n!}}{(2n)!}}=0$ and that can only happen if $\lim \frac{3^n\cdot{n!}}{(2n)!}=0$

Try proving that $
\sum\limits_{n = 0}^\infty {\tfrac{{\left( {n!} \right)}}
{{\left( {2n} \right)!}} \cdot x^n } = 2 \cdot x \cdot \int_0^1 {\left( {t - t^2 } \right) \cdot e^{x \cdot \left( {t - t^2 } \right)} dt} + \int_0^1 {e^{x \cdot \left( {t - t^2 } \right)} dt}
$

Hint: Beta Function

10. Originally Posted by PaulRS
Ok, note that $\sqrt[n]{n!}\sim{\frac{n}{e}}$ (*) ( use Stirling's formula if you want, here you have a proof of (*): http://www.mathhelpforum.com/math-he...-sequence.html )

We have: $\lim \frac{\sqrt[n]{n!}}{\sqrt[n]{(2n)!}}=\lim \frac{\frac{n}{e}}{\left(\frac{2n}{e}\right)^2}=0$

Thus: $\lim \sqrt[n]{\frac{3^n\cdot{n!}}{(2n)!}}=0$ and that can only happen if $\lim \frac{3^n\cdot{n!}}{(2n)!}=0$

Try proving that $
\sum\limits_{n = 0}^\infty {\tfrac{{\left( {n!} \right)}}
{{\left( {2n} \right)!}} \cdot x^n } = 2 \cdot x \cdot \int_0^1 {\left( {t - t^2 } \right) \cdot e^{x \cdot \left( {t - t^2 } \right)} dt} + \int_0^1 {e^{x \cdot \left( {t - t^2 } \right)} dt}
$

Hint: Beta Function
Where is the hint?

And it appears to be a very convoluted form of the gamma or beta function..one of the two