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


by looserlama
Tags: convergence, hard, limit
looserlama
looserlama is offline
#19
Mar14-12, 04:10 PM
P: 30
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?
emailanmol
emailanmol is offline
#20
Mar14-12, 04:38 PM
P: 297
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?
emailanmol
emailanmol is offline
#21
Mar14-12, 04:49 PM
P: 297
Also

(2n-1)(2n-3)...(3)(1) ≤ 2^n*(n-1)!

Is not true.
The LHS has n terms giving rise to n^n and RHS has only n-1 terms.

We know that when limit temds to infinity only the highest power factor matters.
So LHS will be greater
looserlama
looserlama is offline
#22
Mar14-12, 08:53 PM
P: 30
Quote Quote by emailanmol View Post
Also

(2n-1)(2n-3)...(3)(1) ≤ 2^n*(n-1)!

Is not true.
The LHS has n terms giving rise to n^n and RHS has only n-1 terms.

We know that when limit temds to infinity only the highest power factor matters.
So LHS will be greater
I'm pretty sure that's wrong, just from me computing 2n(n-1)! - [itex]\frac{n!}{n!!}[/itex] = 2n(n-1)! - (2n-1)(2n-3)...(3)(1) with vary large numbers. For all of them it was always positive and it was increasing as n got larger.

Also, if we try dividing (2n-1)(2n-3)...(3)(1) by 2n, so dividing each term by 2, gives (n-1/2)(n-3/2)...(3/2)(1/2).

So (2n-1)(2n-3)...(3)(1)/2n(n!) = (n-1/2)(n-3/2)...(3/2)(1/2)/n!

Here the top one is obviously smaller than the bottom, but how do we prove that it goes to zero with this?
emailanmol
emailanmol is offline
#23
Mar15-12, 06:36 PM
P: 297
(2n-1)(2n-3)...(3)(1) ≤ 2^n*(n-1)!

Is not true for n -> infinity.
If u have doubts regarding this you can always plug in the expression into wolfram alpha.


For your second part , each factor of numerator is less than the corresponding denominator.

The maximum value for each expression could be 0.999999....9999 (i.e just less than 1)
That gives u something like (0.999....999)^t where t tends to infinity.(How?)
And we know that will be 0.(Why?)




(Remember you cannot compare coefficients of highest factor in this question to get value of limits as the highest power on both numerator and denominator is n^n.
That rule is valid only for n^a where n is a fixed constant)
That rule only applies while


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