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Calculate the limit of $a_{n} = \frac{n^n}{(n-1)^n}$ when $n\rightarrow\infty$

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the limit of [tex]a_{n} = \frac{n^n}{(n-1)^n}[/tex] when [tex]n\rightarrow \infty[/tex], where n is an integer.

    2. Relevant equations

    3. The attempt at a solution
    [tex]a_{n} = \frac{n^n}{(n-1)^n} = \left(\frac{n}{n-1}\right)^{n} = e^{\ln\frac{n}{n-1}\right)^{n}} = e^{n\ln\frac{n}{n-1}} = e^{n\left(\ln(n)-\ln(n-1)\right)} = e^{-n\left(\ln(n-1)-\ln(n)\right)} = e^{-n\left(\ln\frac{n-1}{n}\right)} = e^{-n\left(\ln\left(1-\frac{1}{n}\right)\right)} = [/tex]que?
  2. jcsd
  3. Feb 26, 2009 #2


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    Science Advisor
    Homework Helper

    What is the limit of
    [tex]n \ln(1 - 1/n)[/tex]
    as [itex]n \to \infty[/itex]?
    When you're not sure, try expanding the logarithm in a series around n = infinity.
  4. Feb 26, 2009 #3


    Staff: Mentor

    For this limit--
    [tex]\lim_{n \rightarrow \infty} n ln(1 - 1/n)[/tex]
    a simpler approach might be to use L'Hopital's Rule, with the limit rewritten as a quotient:
    [tex]\lim_{n \rightarrow \infty} \frac{ln(1 - 1/n}{1/n})[/tex]
  5. Feb 26, 2009 #4
    While both suggestions do give me the end result I want, we have still not reached MacLaurin expansions, nor L'Hopitals rule. Much thanks for your help, however.
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