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Convergence n^(1/n)

  1. Jun 1, 2010 #1
    Hello, I was wondering how to proof [tex]a_n = n^{1/n} \to 1.[/tex]

    Doing it straight from the definition got me nowhere. But I was thinking. It is obvious that [tex]\liminf a_n \geq 1[/tex] (since otherwise for big n you could get n^(1/n) < 1 <=> n < 1). And I also have already a proof of [tex]\limsup x_n^{1/n} \leq \limsup \frac{x_{n+1}}{x_n}[/tex] (which is a general result for any row x_n).

    But proving [tex]\limsup \frac{a_{n+1}}{a_n} \leq 1[/tex] seemed to be harder than I thought.

    So I'm completely stuck. Any ideas?

    Thank you,
    mr. vodka
  2. jcsd
  3. Jun 1, 2010 #2
    I don't think you need to use lim sup/inf here.


    So you just need to show that ln(n)/n-->0. Try using l'Hopital's rule.
  4. Jun 1, 2010 #3
    Hm, I'd like to proof it without the use of the exponential function. It's namely introduced in my Analysis course before the exp function, and it's actually used in a proof about power series, which is later used to introduce the e-function. Thanks for your help!
  5. Jun 1, 2010 #4


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    Science Advisor

    If you don't want to use exp function, how about using ln (natural log)?
    ln(an) = ln(n)/n -> 0.
  6. Jun 1, 2010 #5
    Well, we defined that as the inverse of e. Any possibilities without e (or ln)? I appreciate the help though! I find the ln/e proofs very elegant, but I hope you understand I'm going to choose for logical consistency in my course :)
  7. Jun 1, 2010 #6


    Staff: Mentor

    I'm not sure that there's a way around that doesn't use exp or ln.
  8. Jun 1, 2010 #7
    Oh... But it was left as an exercise for us in our course, so there must be. Hmmm, maybe I should contact one of the math assistants for this one then. Thank you guys for your time.
  9. Jun 2, 2010 #8
    For those that are interested: http://myyn.org/m/article/limit-of-nth-root-of-n/ [Broken]
    Last edited by a moderator: May 4, 2017
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