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Series question

  1. Mar 16, 2008 #1


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    1. The problem statement, all variables and given/known data
    find [tex]lim_{n \rightarrow \infty} n^n q^{2^n}[/tex] where |q|<1

    2. Relevant equations

    3. The attempt at a solution

    I know that [tex]lim_{n \rightarrow \infty} n^{\alpha} q^{n} = 0[/tex] So it looks as if the limit in question should also be equal to 0. But how can I prove this?
    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 17, 2008 #2


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    I just want to check, but there isn't supposed to be a summation in there, is there? If not, then you mean you're looking for the limit of a sequence, no?

    What about looking at the limit of the ratio of a_n+1 / a_n ? The factors that don't go to one as n goes to infinity are (n+1) · q^(2^n). With |q|<1, the second term goes to zero much faster than n+1 grows. The ratio of consecutive terms goes to zero, so the sequence should go to zero. That sound OK?
  4. Mar 17, 2008 #3

    Gib Z

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    Theres always the 'common sense' response: Since it asks to find [tex]\lim_{n\to \infty} n^n q^{(2^n)}[/tex] where |q| < 1, the limit must have the same value for all |q|<1. When q=0, the sequence is equal to 0 for all terms =]

    EDIT: Alternative reasoning: [tex]\lim_{n\to \infty} n^n q^{(2^n)} = \lim_{n\to \infty} e^{\ln \left( n^n q^{(2^n) \right)} = \lim_{n\to \infty} e^{( n \ln n + 2^n \ln q)}[/tex]. Since |q|< 1, ln q is some negative constant. We know the natural log grows slower than any power of x, so the first term in the log can not be larger than say, n^2. 2^n is of a higher order, so exponent part goes to negative infinity, and e raised to that is 0. Reword my logic in more rigorous terms please, this is shameful :(
    Last edited: Mar 17, 2008
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