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A Hard Factorial Limit

  1. Jul 3, 2011 #1
    1. The problem statement, all variables and given/known data
    I need to evaluate the following limit:

    [itex]lim_{n\rightarrow\infty}\left(\frac{1}{n!}\right)^{\frac{1}{3n}}[/itex]

    Now, I think I have solved it, but my methods are not rigorous. I am looking to see if my assumptions are sound and if there is an easier way to solve this limit.

    2. Relevant equations
    The Squeeze Theorem


    3. The attempt at a solution
    My main approach to solving this limit is to use the squeeze theorem. I think it is safe to assume that the the expression must be greater than or equal to zero; this provides my lower bound. I want to try and find an upper bound with the following form:

    [itex]\left(\frac{1}{n}\right)^{\frac{1}{b}}[/itex]

    Where b is some fixed integer. By letting this be the upper bound, I end up with the following inequalities:

    [itex]0 \leq \left(\frac{1}{n!}\right)^{\frac{1}{3n}} \leq \left(\frac{1}{n}\right)^{\frac{1}{b}}[/itex]

    The problem is to find an integer b that makes the above inequalities true for some value of n greater than 1. This required a bit of guessing and checking since I have no idea how to find exact values to the following inequality (it is the one immediately above involving the expressions with exponenets except that it is rewritten):

    [itex]n!^{\frac{b}{3}} - n^{n} \geq 0[/itex]

    When I let b = 4, the inequalities become true at around n = 50. This should let me apply the squeeze theorem to conclude that the limit must equal zero, right?
     
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  3. Jul 3, 2011 #2

    hunt_mat

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    How about looking at [itex]n!\leqslant n^{n}[/itex]? What does that tell you?
     
  4. Jul 3, 2011 #3
    I did use a similar inequality to determine that b cannot be equal to 1, 2, and 3, but it doesn't help in determining if there exists an n greater than 1 such that every n after makes [itex]n!^{\frac{4}{3}}-n^{n} \geq 0[/itex] a true statement.

    Now, if I used the inequality [itex]n! \leq n^{n}[/itex] to help with finding an upper bound you end with the following:

    [itex]\left(\frac{1}{n!}\right)^{\frac{1}{3n}} \geq \left(\frac{1}{n^{n}}\right)^{\frac{1}{3n}}[/itex]

    Before simplifying, it is clear that [itex]n! \leq n^{n}[/itex] is not a very helpful fact for finding an upper bound since the expression which is supposed to be an upper bound becomes a lower bound. In fact, after simplifying you find out that the right hand side of the inequality turns into the form of upper bound I was originally looking for when b is equal to 3. It shows that b cannot be equal to 3.

    This is what I see, am I missing something?
     
  5. Jul 3, 2011 #4

    SammyS

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    What about comparing [itex]\displaystyle \left(\frac{1}{n!}\right)^{1/(3n)} \text{ and } \left(\frac{1}{n}\right)^{1/(3n)}\ ?[/itex]

    or [itex]\displaystyle \left(\frac{1}{n!}\right)^{1/(3n)} \text{ and } \left(\frac{1}{3n}\right)^{1/(3n)}\ ?[/itex]
     
  6. Jul 3, 2011 #5
    If you compare the first two, the second expression approaches 1 as n increases without bound. If you compare the next two, the second expression approaches 1 as n increases without bound. If you examine the expression in the original limit, you can see that it does not approach one, it dips down below 1 and slowly approaches what looks like zero. Both of the suggested upper bounds are not useful for the squeeze theorem.
     
    Last edited: Jul 3, 2011
  7. Jul 3, 2011 #6

    SammyS

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    Yup! Silly mistake on my part.

    b = 4 does work, with [itex](n!)^{\frac{b}{3}} - n^{n} \geq 0[/itex], for n > 42 .

    However, b = 6, gives [itex]\displaystyle (n!)^{\frac{b}{3}} - n^{n} \geq 0[/itex], for n > 2 , and it's pretty easy to show this.

    Also, [itex]\displaystyle \lim_{n\to\infty}\left(\frac{1}{n}\right)^{1/6}=0[/itex]
     
  8. Jul 3, 2011 #7
    How do you go about solving for n when a specific b is input into the inequality?
     
  9. Jul 3, 2011 #8

    SammyS

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    Numerically. But [itex]\displaystyle (n!)^{2} - n^{n} = 0[/itex] is true for n = 1, 2.

    For b = 4, I used WolframAlpha. Then actually plugged in n=41, which gave a negative answer & n = 42 which gave a positive.
     
  10. Jul 4, 2011 #9
    Ah, thank you so much SammyS! And thank you everyone else! Letting b equal 6 is definitely much better than letting b equal 4.
     
  11. Jul 4, 2011 #10

    SammyS

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    You're welcome! & I apologize again for my mistaken first suggestion.
     
  12. Jul 4, 2011 #11
    Rewrite it as:
    [tex]
    \left(\frac{1}{n!}\right)^{\frac{1}{3 n}} = \exp{\left[-\frac{\ln{(n!)}}{3 n}\right]}
    [/tex]
    and use Stirling's asymptotic formula:
    [tex]
    \ln{(n!)} \sim \left(n + \frac{1}{2}\right) \, \ln{n} - n, \ n \rightarrow \infty
    [/tex]
     
  13. Jul 4, 2011 #12
    This method is much simpler! I love it! Thank you so much, Dickfore!
     
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