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I have troubles simplifying this quotient of factorials

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data

    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)!

    2. Relevant equations

    None


    3. 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?
     
  2. jcsd
  3. Mar 7, 2012 #2

    Dick

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    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?
     
  4. Mar 7, 2012 #3
    edit: Wait, I messed up.
     
  5. Mar 7, 2012 #4
    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: Mar 7, 2012
  6. Mar 7, 2012 #5

    Dick

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    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.
     
  7. Mar 7, 2012 #6
    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). :(
     
  8. Mar 7, 2012 #7

    Dick

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    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?
     
  9. Mar 7, 2012 #8
    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
     
  10. Mar 7, 2012 #9

    Dick

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    Sure, that's it!
     
  11. Mar 7, 2012 #10
    Thanks a lot for your help!
     
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