Limit of (2n)!/(4^n (n!)^2)

1. Mar 13, 2012

looserlama

1. The problem statement, all variables and given/known data

This is basically just a step in a problem I'm doing, but I can't find a solution for it anywhere.

So, limit as n$\rightarrow$$\infty$ (2n)!/(4n(n!)2)

3. The attempt at a solution

I tried writing out the factorials of the top and bottom to see if anything would cancel out, but it doesn't work, at least I don't see how.

I know that it goes to 0, but I don't know how to actually compute it.

Any help at all would be much appreciated.

2. Mar 13, 2012

Dick

Try a ratio test.

3. Mar 13, 2012

looserlama

Isn't the ratio test just for series though?

Also, if we are talking about the same ratio test, then that equals 1, so it's inconclusive. That's kind of how I got to this point in my problem.

Last edited: Mar 13, 2012
4. Mar 13, 2012

scurty

I did a bunch of factoring and rearranging and ended up with 0. I'm almost certain there is an easier way but I didn't see one. I'm hesitant to tell you what I did because there's a good chance I made an error. If you are still stuck I can guide you along with what I did.

Well, I'm pretty sure my first step is valid, I canceled out a n! from top and bottom.

5. Mar 13, 2012

Dick

Yeah, you are right. If you can use the ratio test to show a series converges then you know that the terms of the series go to zero. But you are right, it's inconclusive.

6. Mar 13, 2012

scurty

Disregard my first post. The squeeze theorem works in this case, I'm fairly sure. I found two functions that bound the function in the first post that both converge to 0. I think that's how you would solve this problem?

7. Mar 13, 2012

Dick

My next approach would be to use Stirling's approximation. The leading terms will cancel. You could just keep adding subleading terms until one doesn't cancel. That's not very easy either. What's your idea?

8. Mar 13, 2012

scurty

My idea didn't turn out correct, I made a few errors. After you cancel out the n! everything just gets messy.

9. Mar 13, 2012

Dick

My first idea didn't work either. Try using Stirling's approximation in the form $n! \approx \sqrt{2 \pi n} (\frac{n}{e})^n$. The limit is a little delicate. It's the square root term that make the difference.

Last edited: Mar 13, 2012
10. Mar 13, 2012

scurty

That works pretty well, I simplified it down to $\displaystyle\frac{1}{\sqrt{\pi}} \cdot \displaystyle\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} = 0.$

But as I mentioned above, I think the squeeze theorem works as well. I like your way though! Sterling is an upper bound approximation, right? So if all the terms are positive and the approximation approaches 0, the original limit must approach 0, right?

11. Mar 13, 2012

Dick

Stirling is an asymptotic approximation. The limit of the approximation over n! goes to one as n goes to infinity. I'm not sure there is any guarantee it's greater or less than n!. I still think this way might be too complicated. What's your idea? I thought you decided it didn't work?

12. Mar 13, 2012

scurty

My first idea didn't work. My second one I think is pretty conclusive:

Notice:

$\displaystyle\frac{(2n)!}{4^{n}((2n)!)^{2}} \leq \displaystyle\frac{(2n)!}{4^{n}(n!)^{2}} \leq \displaystyle\frac{2^{n}(n!)^{2}}{4^{n}(n!)^{2}}$.

I think it's pretty obvious why the first inequality holds. The second one is true because $(2n)! \leq 2^{n}(n!)^{2}$. It's just an algebra exercise but eventually it boils down to
$(2n-1)(2n-3)(2n-5)\cdot\cdot\cdot(7)(5)(3)(2) \leq (2n)(2n-2)(2n-3)\cdot\cdot\cdot(6)(4)(2)$. Both sides have n terms so you just need to match them up in pairs and compare them. The limit of the function all the way on the right approaches 0 as does the left.

Edit: I typoed in the far right equation. Everything is good now.

Last edited: Mar 13, 2012
13. Mar 13, 2012

Dick

I think if you use Stirling to check the limit of the expression on the right it goes to infinity. That's kind of a nonstarter.

14. Mar 13, 2012

scurty

Sorry, I made a typo in the equation on the right. How about now? It converges to 0 and is greater than the original equation for all n.

15. Mar 13, 2012

Dick

No, I'm still not buying it. 20! is much greater than (2^10)*(10!)^2. As I said the limit is 'delicate'.

16. Mar 13, 2012

scurty

You're right, sorry. When I got to this step,

$(2n)! = (2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdot\cdot\cdot(4)(3)(2)(1)\\$

I factored our a single 2 and made it

$(2n)! = 2(n)(2n-1)(n-1)(2n-3)(n-2)\cdot\cdot\cdot(2)(3)(1)(1)\\$

and then factored out a n! when it should have been

$(2n)! = 2^{n}(n)(2n-1)(n-1)(2n-3)(n-2)\cdot\cdot\cdot(2)(3)(1)(1)\\$.

Basically I factored out all the 2s from the even terms and wrote it as a single 2 instead of $2^{n}$.

Well, back to the drawing board... I never see mistakes in my own work until someone tells me I'm wrong and I go back carefully and then notice it. :/

17. Mar 13, 2012

Dick

That's ok. Checking with numbers often helps you to find gaffes like this. It's pretty easy if you've got a computer. :) I do it a lot. As should be obvious from my last post.

18. Mar 14, 2012

SammyS

Staff Emeritus
We haven't heard from OP, looserlama, since post #3.

19. Mar 14, 2012

looserlama

Yea I've been working on other stuff but I've been following the conversation.

So this is where I'm at now:

We've never seen stirling's approximation before so I don't think we're supposed to use it.
So I've just been focusing on the Squeeze theorem way:

It's easy to show (2n)!/4n((2n)!)2≤ (2n)!/4n(n!)2

But finding an upper bounding sequence that goes to zero is the hard part.

What I'm thinking now is, (2n)! = 2n(n!) [(2n-1)(2n-3)...(3)(1)]
The second part (stuff in square brackets) is ≤ 2n(n!)

So we have (2n)! ≤ 2n(n!)2n(n!) = 4n(n!)2 but if we try to makes this an upper bounding sequence (ie, (2n)!/4n(n!)2 ≤ 4n(n!)2/4n(n!)2= 1) then it doesn't converge to 0.

So I wanted to show (2n-1)(2n-3)...(3)(1) ≤ 2n(n-1)!, I know this is true but I don't know how to prove it.

Cause if we have that, then (2n)!/4n(n!)2 ≤ 4n(n!)(n-1)!/4n(n!)2= 1/n which goes to zero.

I just don't know how to prove that?

20. Mar 14, 2012

emailanmol

Its simple.
Expand
(2n)!
As (2^n)*(n!)*{(2n-1)(2n-3)(2n-5).......5*3*1)}

Now divide this by 4^n*(n!)^2

What do you get??

Its (2n-1)*(2n-3)*(2n-5)......3*1/((2^n)(n!))

There are n terms in numerator .Divide each by 2 to get rid of 2^n

What do you see?