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Limit of factorial functions

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the folowing holds:
    [tex]\lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \infty[/tex]


    2. Relevant equations
    It can be shown that
    [tex]\frac{(n!)^2 4^n}{(2n)!} = \prod_{k=1}^n \frac{4k}{k+n}[/tex]


    3. The attempt at a solution
    If I can proove that
    [tex]\lim_{n\to\infty} \ln\left[\frac{(n!)^2 4^n}{(2n)!} \right] = \infty[/tex]
    then I am done since the logarithm is a strictly increasing function.

    By using the "Relevant equation" I have
    [tex]\ln\left[\frac{(n!)^2 4^n}{(2n)!} \right] = \ln\left[\prod_{k=1}^n \frac{4k}{k+n}\right] = \sum_{k=1}^n \ln\left(\frac{4}{1+\frac{n}{k}}\right)[/tex]

    But from here I don't get any further...
     
    Last edited: Mar 4, 2013
  2. jcsd
  3. Mar 4, 2013 #2

    Dick

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    Can't you just use Stirling's approximation on the factorials in the original problem?
     
  4. Mar 4, 2013 #3
    I approached it a little differently than you - here was my thinking:

    in the numerator distribute the 2's so that you get two products of only even numbers: ##\frac{(2*4*6*...*2n)^2}{2n!}## This would allow you to cancel out the even terms in the denominator and get ##\frac{2*4*6*...*2n}{1*3*5*...*2n-1}##. I then Identify this as ##\prod_{k=1}^n \frac{2k}{2k-1}##. From here you should be able to use logs to convert this product to a sum and use the integral test (or perhaps something else) to show that the sum must diverge.

    Good Luck!
     
  5. Mar 4, 2013 #4

    Ray Vickson

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    If you don't like to use Stirling's approximation, you can use a modification of it to get both lower and upper bounds: for any integer ##k \geq 1## we have
    [tex] \sqrt{2 \pi}k^{k + \frac{1}{2}}e^{-k} < k! < \sqrt{2 \pi}k^{k + \frac{1}{2}}e^{-k+ \frac{1}{12k}} [/tex]
    This works well even for small k; for example, for k = 1 it gives
    .9221370087 < 1 < 1.002274449 and for k = 2 it gives 1.919004351 < 2 < 2.000652047 .

    For a simple proof, see Feller, "Introduction to Probability Theory and Its Applications, Vol I, Wiley (1967).
     
    Last edited: Mar 4, 2013
  6. Mar 4, 2013 #5
    Thanks for the replies! The method hapefish proposed worked very vell. The sum diverges by the comparison test with the harmonic series. I also solved the problem using stirlings formula, thanks a lot guys! :)
     
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