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Prove n^(1/n) tends to 1 as n tends to infinity

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Need to prove n^(1/n) tend to 1 as n tends to infinty

    2. Relevant equations

    3. The attempt at a solution

    Have tried comparing to n^(1/n)=(1+h) and using binomial series but no joy..please help
    Last edited: Nov 21, 2008
  2. jcsd
  3. Nov 21, 2008 #2


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    The sequence [itex]n^{1/n}[/itex], as n goes to infinity, converges to a if the function [itex]x^{1/x}[/itex] converges to a as x goes to infinity. If we set [itex]y= x^{1/x}[/itex] then ln(y)= (ln x)/x which is of the "infinity/infinity" form so we can use L'Hopital's rule.
  4. Nov 21, 2008 #3
    One direct method that comes to my mind is to show that for any [itex]\varepsilon>0[/itex] there exist N such that
    n^{1/n}\leq 1+\varepsilon
    (It is easy to see that
    n^{1/n}\geq 1
    for all n>N.
    The first equation is equivalent to [itex]n\leq(1+\varepsilon)^n=1+n\varepsilon+\dots[/itex]

    Do you see how to choose N?
  5. Nov 21, 2008 #4
    wow..briliantly simple using l'hopital...though i have not yet come across this theorem at uni...still will impress the tutors. thanks
  6. Nov 21, 2008 #5
    I cant see how to choose N....could you give me another hint?
  7. Nov 21, 2008 #6
    I should have included the next term in the binomial expansion:smile:

    n\leq 1+n\varepsilon+\frac{n(n-1)}{2}\varepsilon^2\Leftrightarrow \dots
    You just have to solve this for n>... and take the next larger integer for N.
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