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Is this acceptable reasoning for the lim n!^2/(2n)! as n->∞

  1. Sep 20, 2016 #1
    1. The problem statement, all variables and given/known data
    (This isn't actually course work for me but the text book is used in course work at my school)

    Find the limit:
    [tex] \lim_{x\rightarrow ∞}\frac{(n!)^2}{(2n)!}[/tex]

    2. Relevant equations
    None other than what a factorial and limit are. 4! = (1)(2)(3)(4), for example.

    3. The attempt at a solution

    In examining (2n)!, it looks like:

    [tex](2n)! = (1)(2)(3)...(n)(n+1)...(2n-3)(2n-2)(2n-1)(2n)[/tex]
    which is
    [tex](n!)(n+1)...(2n-3)(2n-2)(2n-1)(2n)[/tex]
    So I can cancel one n! term, leaving:
    [tex]\frac{n!}{(n+1)(n+2)...(2n-2)(2n-1)(2n)}[/tex]
    Then expanding n! a little bit gives:
    [tex]\frac{(1)(2)...(n-2)(n-1)n}{(n+1)...(2n-2)(2n-1)(2n)}[/tex]
    And it's here I notice that n+1, the first term in what's left of my denominator, is larger than 1, the first term in my numerator, for all n>0, n∈ℕ. And of course, n+2>2 for all n>0, and so on, all the way to the very end, when 2n>n for all n>0.

    By this reasoning I argue that if n approaches infinity, the denominator will approach faster, and so the limit is zero.
    I know zero is the correct answer, but would the denominator simply being larger be a good enough reason to declare that the limit is zero?


    If not, here's my "Phase Two" plan.


    If I factor out 2n from the numerator I'd have (I think!):
    [tex]\frac{(1)(2)...(n-2)(n-1)n}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})n(2^n)}[/tex]
    Then I divide by n,
    [tex]\frac{(1)(2)...(n-2)(n-1)}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})(2^n)}[/tex]

    And now I have a high degree polynomial divided by a high degree polynomial TIMES an exponential function, and as was shown in another thread where I had questions regarding that specific topic, the exponential wins by a mile. So in this case, the limit has to be zero, because the denominator approaches infinity WAY faster.

    *I also just noticed that factoring out the 2n will also take out every single term in the numerator: 2n-4 = 2(n-2), so goodbye n-2 term; 2n-6 = 2(n-3), so goodbye n-3 term, and it should go that way all the way down. So I'd have 1 in the numerator and (n-(1/2))2n in the denominator, meaning this race to infinity is shamefully one sided (assuming my factoring of 2n isn't done incorrectly).


    Yes, no, maybe so? Any thoughts on a better way to do this?





    Thanks!
     
  2. jcsd
  3. Sep 20, 2016 #2
    Your first solution is fine but I would express the last part a bit cleaner.
    What you are doing is matching the terms one by one and note the one in the numerator is always smaller. But using this you clearly see that your expression is smaller than for example ##1/(n+1)## which has the limit zero as ##n\to \infty## so you bounded the limit above by zero while your expression is always positive.

    Also check out Stirling's formula for the behaviour of ##n!## at large ##n## and you see the same result.
     
  4. Sep 20, 2016 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Use Stirling's Formula;
    http://www.sosmath.com/calculus/sequence/stirling/stirling.html
    or
    https://en.wikipedia.org/wiki/Stirling's_approximation
     
  5. Sep 20, 2016 #4

    micromass

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    2016 Award

    You could use the ratio test for series (yes, I know you are dealing with a sequence here!).
     
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