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Limit of the nth root of (n ln n)

  1. Oct 27, 2009 #1

    txy

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    1. The problem statement, all variables and given/known data

    Find [tex]lim_{n \rightarrow \infty} \sqrt[n]{n ln(n)}[/tex].
    Do not use L'Hospital's Rule or Taylor Series.

    2. Relevant equations


    3. The attempt at a solution

    I suspect I need to set up some inequality for this and then apply Squeeze Theorem. But I can't find any inequality that can help me do this. Any help would be appreciated.
     
  2. jcsd
  3. Oct 27, 2009 #2

    Dick

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    Start by taking the log of the expression and trying find the limit of that.
     
  4. Oct 27, 2009 #3
    You should find something like:

    [tex] n \ln(L) = ln( n \cdot ln (n) ) = ln(n) + \ln(ln(n)) [/tex]
     
  5. Oct 27, 2009 #4

    lurflurf

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    does
    n<n log n<n^(1+epsilon)
    help?
     
  6. Oct 28, 2009 #5

    txy

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    That's great. In either case, I need to use the fact that ln n < n for all n > 1. How do I prove this?
     
  7. Oct 28, 2009 #6

    Dick

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    If you need to prove it I think the easiest way is to define f(x)=x-ln(x). Notice f(1)=1. Now can you show f'(x)>0 for x>1? That would show f(x) is positive for x>1.
     
  8. Oct 28, 2009 #7

    txy

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    But this question that I posed is in the Limits chapter of my textbook, which comes before the chapter on Differentiation. I wonder if there's another way to prove that inequality. Perhaps I could change it to the form x < e^x , but I still can't make any progress.
     
  9. Oct 28, 2009 #8

    Dick

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    Ok, I think you might be fussing a little to much over this proof, but how about using the definition of e^x. e^x=limit (1+x/n)^n as n->infinity? Can you use that, together with the binomial theorem to prove e^x>x for x>0?
     
  10. Oct 29, 2009 #9

    txy

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    Yes now I can prove it. Thanks!
     
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