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Need help with lim sqrt(n)/n! as n-> infinity

  1. Nov 23, 2004 #1
    Hi,

    could you help me a bit with this limit?

    [tex]
    \lim_{n \rightarrow \infty} \sqrt[n]{n!}
    [/tex]

    Sure it should be more than

    [tex]
    \lim_{n \rightarrow \infty} \sqrt[n]{n}
    [/tex]

    But, when I write it as

    [tex]

    \lim_{n \rightarrow \infty} \sqrt[n]{n!} = \lim_{n \rightarrow \infty} \sqrt[n]{n} . \lim_{n \rightarrow \infty} \sqrt[n]{n-1} . \lim_{n \rightarrow \infty} \sqrt[n]{n-2} ... \lim_{n \rightarrow \infty} \sqrt[n]{1}
    [/tex]

    each term goes to 1, so I thought the limit could be 1, but that would be strange...

    Thank you.
     
  2. jcsd
  3. Nov 23, 2004 #2
    Hint: Use the corollary that (1+h)^ n > 1+ hn. Also you know that limit of nth root n is 1.
     
  4. Nov 23, 2004 #3

    arildno

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    Regard the following trick (regard even values of n):
    [tex]n!=n^{n}\prod_{i=1}^{\frac{n}{2}}(1-\frac{i-1}{n})\prod_{j=1}^{\frac{n}{2}}(1-\frac{\frac{n}{2}+j-1}{n})[/tex]
    Now,
    [tex]\prod_{i=1}^{\frac{n}{2}}(1-\frac{i-1}{n})\geq\prod_{i=1}^{\frac{n}{2}}(1-\frac{1}{2})=(\frac{1}{2})^{\frac{n}{2}}[/tex]
    And:
    [tex]\prod_{j=1}^{\frac{n}{2}}(1-\frac{\frac{n}{2}+j-1}{n})\geq\frac{1}{n^{\frac{n}{2}}}[/tex]

    Hence,
    [tex]n!\geq{n}^{n}(\frac{1}{2})^{\frac{n}{2}}\frac{1}{n^{\frac{n}{2}}}=(\frac{n}{2})^{\frac{n}{2}}[/tex]

    This should help you..
     
  5. Nov 23, 2004 #4
    Thank you, but I can't see how could I use the inequality you suggested. Could you be more specific please?
     
  6. Nov 23, 2004 #5
    So the thing you say is that

    [tex]
    n! > \sqrt{ \frac{n^{n}}{2^{n}}}
    [/tex]

    I know this thing goes to infinity. And is THIS the reason, why also

    [tex]
    \sqrt[n]{n!}
    [/tex]

    goes to infinity? But there is n-th root, why doesn't it make a difference?
     
  7. Nov 23, 2004 #6
    Btw I can't imagine I would think up such a trick while writing a test
    :mad:
     
  8. Nov 23, 2004 #7

    BobG

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    You must be Canadian.
     
  9. Nov 24, 2004 #8
    Why should I be Canadian?
     
  10. Nov 24, 2004 #9

    arildno

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    Of course, we have:
    [tex](n!)^{\frac{1}{n}}>((\frac{n}{2})^{\frac{n}{2}})^{\frac{1}{n}}=\sqrt{\frac{n}{2}}[/tex]
     
  11. Nov 24, 2004 #10

    BobG

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    You speak Canadian (see bold).

    You open the lights and close lights, as well, right?
     
  12. Nov 24, 2004 #11
    What's strange on writing ?
     
  13. Nov 24, 2004 #12

    rpc

    User Avatar


    because Americans "take" a test
     
  14. Nov 24, 2004 #13
    Oh I see it now. It doesn't definitely mean I'm Canadian :) I'm just not that familiar with these phrases (although I know this one... :approve: )
     
  15. Nov 24, 2004 #14
    1st off: the reason why your method is wrong, is because you make the equality:
    [tex]
    \lim_{n \rightarrow \infty} \sqrt[n]{n!} = \lim_{n \rightarrow \infty} \sqrt[n]{n}\lim_{n \rightarrow \infty} \sqrt[n]{n-1} \ldots
    [/tex]

    But when you take the limit of [itex]n\rightarrow\infty[/itex] the number of limits doesn't change as n increases. So you start of with n limits, multiplied with each other. Each individual limit has a limit of 1. But then you don't take into account that the number of limits multiplied together also changes, when n increases.

    Anyways:
    [tex]
    \lim_{n \rightarrow \infty} \sqrt[n]{n!} =
    [/tex]
    [tex]
    \lim_{n \rightarrow \infty} e^{\frac{1}{n}\ln{n!}}
    [/tex]

    Note again, that you can't write this as:
    [tex]
    \lim_{n \rightarrow \infty} e^{\frac{1}{n}(\ln{n}+\ln{n-1}+\ln{n-2}+...)} =
    [/tex]
    and then take the limit of each logarithm seperately. Again, because the number of logarithms also changes with n.

    Instead, you could for instance use the Stirling inequality:
    [tex]
    \ln n! > n\ln{n} - n
    [/tex]

    [tex]
    \lim_{n \rightarrow \infty} e^{\frac{1}{n}\ln{n!}} >
    [/tex]
    [tex]
    \lim_{n \rightarrow \infty} e^{\frac{1}{n}(n\ln{n}-n)} = \infty
    [/tex]
     
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