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Prove lim na^n = 0 when 0 < a < 1

  1. Aug 13, 2010 #1
    The problem statement, all variables and given/known data
    Prove lim na^n = 0 when 0 < a < 1.

    The attempt at a solution
    Without danger, we change from the discrete n to the continuous x so that now we have to prove that lim xa^x = 0. Let e > 0. We have to find an N such that xa^x < e for all x > N. Now if xa^x < e is the same as 1 < eb^x/x, where b = 1/a. Using L'Hospital's rule, we have that lim eb^x/x = lim e(ln b)b^x = oo, so there is an N such that 1 < eb^x/x. QED

    Is this the simplest way to prove this limit. For some odd reason, I feel that there is a simpler solution. Any tips?
     
  2. jcsd
  3. Aug 13, 2010 #2
    Apply L'Hopital's rule in the limit [tex] \lim x.a^x [/tex] after writing [tex] x.a^x = \frac{x}{a^{-x}} [/tex].
     
  4. Aug 13, 2010 #3

    Hurkyl

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    My first inclination is not to use calculus at all!

    Let a < u < 1. Prove that, eventually, (n+1)/n < u/a. Then, an easy inductive proof shows that 0 < n a^n < C u^n for some constant C.

    I suppose technically I've used some amount of calculus in that I invoke the Archimedean property, the squeeze theorem, and knowledge that u^n converges to 0.


    Come to think of it, the inductive step isn't really going to be conceptually much different from taking a derivative or a logarithmic derivative. (Though the analogy might seem opaque if you're just learning this stuff)
     
  5. Aug 13, 2010 #4

    Dick

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    I would do it by showing log(n*a^n)/n goes to log(a) using l'Hopital. Hence log(n*a^n) approaches -infinity.
     
  6. Aug 14, 2010 #5
    Thank you for the suggestions. I'm a little rusty on this stuff.
     
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