I have troubles simplifying this quotient of factorials

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Homework Help Overview

The discussion revolves around finding the limit of the expression (n!)² / (2n)! as n approaches infinity, as posed in an exercise from Mary L. Boas' book Mathematical Methods in the Physical Sciences. The participants explore the growth rates of factorials and their implications for the limit.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the instinctive belief that (2n)! grows faster than (n!)², leading to a limit of zero, but express uncertainty about how to formally prove this. Some suggest using Stirling's approximation or writing out terms for specific values of n to analyze the behavior of the sequence.

Discussion Status

There is an ongoing exploration of different approaches, including the potential use of comparison tests and the squeeze theorem. Some participants express confusion about the application of these concepts to sequences rather than series, while others encourage continued investigation into upper bounds and comparisons.

Contextual Notes

Participants note the lack of formal equations provided in the original problem statement and discuss the constraints of their current understanding of comparison tests in the context of limits involving factorials.

pylauzier
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Homework Statement



I'm trying to self-study Mary L. Boas' book Mathematical Methods in the Physical Sciences. One of the exercices asks the reader to find the limit of n -> ∞ (n!)2 / (2n)!

Homework Equations



None


The Attempt at a Solution



Instinctively I know that (2n)! grows faster than (n!)^2, so I know the answer is zero. However I have absolutely no idea how to prove it. Can the (2n)! term be rewritten to somehow cancel out the (n!)^2 term on the numerator?
 
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pylauzier said:

Homework Statement



I'm trying to self-study Mary L. Boas' book Mathematical Methods in the Physical Sciences. One of the exercices asks the reader to find the limit of n -> ∞ (n!)2 / (2n)!

Homework Equations



None


The Attempt at a Solution



Instinctively I know that (2n)! grows faster than (n!)^2, so I know the answer is zero. However I have absolutely no idea how to prove it. Can the (2n)! term be rewritten to somehow cancel out the (n!)^2 term on the numerator?

If you want to be fancy you could use Stirlings approximation. If not, just write it out for some n. Like (1/1)*(1/2)*(2/3)*(2/4)*(3/5)*(3/6) for n=3. Does that suggest a comparison test?
 
Dick said:
If you want to be fancy you could use Stirlings approximation. If not, just write it out for some n. Like (1/1)*(1/2)*(2/3)*(2/4)*(3/5)*(3/6) for n=3. Does that suggest a comparison test?

edit: Wait, I messed up.
 
Dick said:
If you want to be fancy you could use Stirlings approximation. If not, just write it out for some n. Like (1/1)*(1/2)*(2/3)*(2/4)*(3/5)*(3/6) for n=3. Does that suggest a comparison test?

I know how to compare sums, but I've never learned to compare sequences. What am I looking for? For example if I simplify the product given by n=3, I can see that the denominator in my sequence is bigger than the denominator of 1/n. Since lim n-> infinity of 1/n tends to zero, does that prove that lim -> infinity (n!)^2 / (2n)! will also tend towards zero?

Edit: I looked up stirling's approximation. Would I simply have to replace the n terms in the formula with 2n if I want to use it to approximate (2n)! ? Sorry for another dumb question.
 
Last edited:
pylauzier said:
I know how to compare sums, but I've never learned to compare sequences. What am I looking for? For example if I simplify the product given by n=3, I can see that the denominator in my sequence is bigger than the denominator of 1/n. Since lim n-> infinity of 1/n tends to zero, does that prove that lim -> infinity (n!)^2 / (2n)! will also tend towards zero?

Edit: I looked up stirling's approximation. Would I simply have to replace the n terms in the formula with 2n if I want to use it to approximate (2n)! ? Sorry for another dumb question.

All of the terms in your series are <=1. About half of them are equal to 1/2. Keep thinking about it. Don't use Stirling's if you had to look it up. It's overkill. A comparison would be much nicer. I'm thinking about a geometric sequence.
 
Dick said:
All of the terms in your series are <=1. About half of them are equal to 1/2. Keep thinking about it. Don't use Stirling's if you had to look it up. It's overkill. A comparison would be much nicer.

I'm not sure I understand what I'm supposed to compare. I thought the comparison test was for series (sums of terms), not simple limits. If I'm not asking for too much, would you care pointing me towards the right direction? I even searched in my old calc II book and there is no mention of a comparison test for sequences (factorials). :(
 
pylauzier said:
I'm not sure I understand what I'm supposed to compare. I thought the comparison test was for series (sums of terms), not simple limits. If I'm not asking for too much, would you care pointing me towards the right direction? I even searched in my old calc II book and there is no mention of a comparison test for sequences (factorials). :(

Ok, then call it a "squeeze theorem". You can probably look that up. Your sequence is >=0. I'm not giving up on you yet. I'm just trying to get you to explain why your "instinctive" answer is right. What's good upper bound?
 
Ok, I think I understand now. The nth term will be smaller than (1/2^n), since all the terms are smaller than one and because there is n 1/2 terms.

So lim n -> ∞ (n!)2 / (2n)! =< lim n -> ∞ (1/2^n)

Since lim n -> ∞ (1/2^n) = 0, it follows that

lim n -> ∞ (n!)2 / (2n)! =< 0
 
pylauzier said:
Ok, I think I understand now. The nth term will be smaller than (1/2^n), since all the terms are smaller than one and because there is n 1/2 terms.

So lim n -> ∞ (n!)2 / (2n)! =< lim n -> ∞ (1/2^n)

Since lim n -> ∞ (1/2^n) = 0, it follows that

lim n -> ∞ (n!)2 / (2n)! =< 0

Sure, that's it!
 
  • #10
Dick said:
Sure, that's it!

Thanks a lot for your help!
 

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