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Try using a sterling approximation on the factorial

  1. Jul 24, 2009 #1
    Can somebody demonstrate :
    [tex]\[
    \frac{n}{{\sqrt[n]{{n!}}}} < \left( {1 + \frac{1}{n}} \right)^n
    \]
    [/tex]
    ITS not A HOMEWORK
     
  2. jcsd
  3. Jul 24, 2009 #2
    Re: Hi


    I suppose it wouldn't exactly be a proof but you could try using a sterling approximation on the factorial.
     
  4. Jul 24, 2009 #3
    Re: Hi

    There must be a solution. several days i try to solve this problem but i do not reach any results.
     
  5. Jul 24, 2009 #4

    berkeman

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    Staff: Mentor

    Re: Hi

    What application is the problem from?
     
  6. Jul 24, 2009 #5
    Re: Hi

    From math's corpus
     
  7. Jul 24, 2009 #6
    Re: Hi

    I think that I've figured out how to prove your inequality. I'm new to this board, but it seems that you are expected to show what you've already tried or where you got stuck, even if it's not homework. So, if you could give a brief summary of your work so far, that would help me to know what hints to give you. Also, your explanation that this problem came from "math's corpus" doesn't really mean anything in English. That's OK, but could you try again and be more specific? Thanks.
     
  8. Jul 24, 2009 #7

    berkeman

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    Re: Hi

    Good job Petek!
     
  9. Jul 24, 2009 #8

    CRGreathouse

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    Re: Hi

    There is a form of Stirling's approximation that gives a range in which the factorial falls. I suspect that would suffice to solve this problem.
     
  10. Jul 31, 2009 #9
    Re: Hi

    It's now been a week since the OP's last post (hope I didn't scare him/her off). Does anyone object to me posting my solution? I'd like others to check it for correctness.

    Petek
     
  11. Jul 31, 2009 #10

    CRGreathouse

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    Re: Hi

    Go ahead, that shouldn't be a problem.
     
  12. Jul 31, 2009 #11

    berkeman

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    Re: Hi

    I agree. Looks kind of like a puzzle...
     
  13. Jul 31, 2009 #12
    Re: Hi

    OK, here goes.



    Exercise: Show that [tex]\[\frac{n}{{\sqrt[n]{{n!}}}} < \left( {1 + \frac{1}{n}} \right)^n \][/tex] for all natural numbers n.

    Solution: We use without proof the following inequalities:

    [tex](1 + \frac{1}{k})^k \leq e \leq (1 + \frac{1}{k})^{k + 1} \right {(*)}[/tex]

    (where e is Euler's number) for all natural numbers k. A proof may be found in many calculus and real analysis books. (I couldn't find a good online reference for these inequalities. Does anyone know of one?)

    For k = 1, 2, ..., n - 1, multiply together the inequalities on the left side of (*):


    [tex] \prod_{k=1}^{n-1}(1 + \frac{1}{k})^{k} \leq e^{n -1} [/tex]


    The left side of this inequality equals [tex]\frac{n^n}{n!} [/tex] (To see this, rewrite [tex](1 + \frac{1}{k})[/tex] as

    [tex] \frac{k + 1}{k}[/tex], simplify the product to [tex]\[\frac{n^{n-1}}{(n-1)!}[/tex] and multiply by [tex]\frac{n}{n})[/tex].

    Therefore


    [tex]\frac{n^n}{n!} \leq e^{n-1}[/tex]


    or,


    [tex]\[\frac{n}{{\sqrt[n]{{n!}}}} \leq e^{\frac{n-1}{n}} = e^{1-\frac{1}{n}}[/tex]


    Now, in (*), raise the right inequality to the power [tex]1 - \frac{1}{n}:[/tex]


    [tex]e^{1-\frac{1}{n}} \leq (1 + \frac{1}{n})^{(n+1)(1-\frac{1}{n})} = (1 + \frac{1}{n})^{n-\frac{1}{n}}[/tex], which is strictly less than [tex](1 +

    \frac{1}{n})^{n}[/tex].

    Therefore, [tex]\[\frac{n}{{\sqrt[n]{{n!}}}} \leq e^{1-\frac{1}{n}} < (1 + \frac{1}{n})^{n}[/tex], as required.
     
  14. Jul 31, 2009 #13

    berkeman

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    Re: Hi

    Impressive!
     
  15. Aug 1, 2009 #14
    Re: Hi

    thanks for solving it is quite simlpe resolved, I know another solve more complex, it seems that yours is the easier
     
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