Factorial limits

**Mathstud28** · May 22nd 2008, 12:47 PM

Someone told me this wasn't correct?

Could anyone clarify please

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

**flyingsquirrel** · May 22nd 2008, 01:29 PM

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 $3^nn!<n^n$ .

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.

**wingless** · May 22nd 2008, 02:25 PM

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.

**wingless** · May 22nd 2008, 02:38 PM

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

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

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

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

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

**Mathstud28** · May 22nd 2008, 02:52 PM

Originally Posted by flyingsquirrel

Hello

You should show that $3^nn!<n^n$ .

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.

As for your proof, I thought about this but it wasn't as aesthetically pleasing.

**wingless** · May 22nd 2008, 03:18 PM

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.

As for your proof, I thought about this but it wasn't as aesthetically pleasing.

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 =)

**Mathstud28** · May 22nd 2008, 03:21 PM

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

**wingless** · May 22nd 2008, 03:35 PM

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?

**PaulRS** · May 22nd 2008, 04:21 PM

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

**Mathstud28** · May 22nd 2008, 04:25 PM

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

Thread: Factorial limits

LinkBack

Thread Tools

Display

Factorial limits

Similar Math Help Forum Discussions

factorial

Sum of Factorial

Factorial N

factorial!

factorial

Search Tags