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Limit demonstration of sequence

  1. Apr 25, 2012 #1
    Hi guys

    I know this limit goes to infinity

    [tex]lim (3^n-n)[/tex]
    But how do I demonstrate it?

    Actually I know also that this type of limits goes to infinity

    [tex]lim \frac{a^n}{n^k},\forall a,k \in \mathbb{N},a>1[/tex]
    But I don't know how to prove it

    May you kindly help me?

    Many thanks
     
  2. jcsd
  3. Apr 25, 2012 #2


    One glamorous way to prove it:

    1) this is a positive sequence;

    2) the series [itex]\displaystyle{\sum_{n=1}^\infty \frac{n^k}{a^n}}\,\,[/itex] converges (D'alembert or Cauchy n-root test, say), thus [itex]\,\,\displaystyle{\frac{n^k}{a^n}\to_{n\to\infty} 0}[/itex] , so...

    DonAntonio
     
  4. Apr 25, 2012 #3
    Very interesting :) Really

    Is there any other proof, more common and less glamorous (without going into series)?
     
  5. Apr 25, 2012 #4


    Another way: define [itex]f(x):=\frac{a^x}{x^k}\,\,[/itex], show this function has a min. at some point, after which the derivative is always

    positive and thus the function's ascending, and since the function's always positive...

    DonAntonio
     
  6. Apr 26, 2012 #5
    But functions whose derivative is always positive after a certain minimum, don't mean they tend to infinity...

    For example:

    [tex]f(x)=-e^{-x}+1, x\geq 0[/tex]
    [tex]f(x)=-x, x<0[/tex]

    has a minimum at x=0 and after that is always ascending, though it tends to x=1
     
  7. Apr 26, 2012 #6
    ....
     
  8. Apr 26, 2012 #7
    Maybe with the second derivative which gives the concavity...
     
  9. Apr 26, 2012 #8
    Thank you DonAntonio

    I really apologise but I can't see the big picture...

    Facts:

    1. The function has a minimum at certain point
    2. After that minimum the derivative is always positive
    3. The function is always positive, i.e. f(x)>0, for all x in R

    How do I conclude that it tends to infinity?

    Sorry to bother...

    Thank you very much
     
  10. Apr 26, 2012 #9


    Well, I intended for you to complete the picture: since the function I used has no oblicuous and/or horizontal asymptotes, and since it is

    always positive, if after some definite point is derivative is positive then the function must an ascending after that point, so

    it MUST tend to [tex]\infty[/tex]

    DonAntonio
     
  11. Apr 26, 2012 #10
    Thank you very much DonAntonio

    I confess I was not reaching the part of the asymptotes...

    Thank you so very much indeed

    Greetings from Lisbon

    João
     
  12. Apr 26, 2012 #11


    Foi un prazer para mim.

    DonAntonio
     
  13. Apr 26, 2012 #12
    Muito obrigado mesmo pela atenção

    Melhores cumprimentos
     
  14. Apr 27, 2012 #13

    Office_Shredder

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    Gold Member

    The way I like to think about these kinds of ratios (an/nk) is by using this:
    [tex] \lim_{n\to \infty} \left( \frac{n+1}{n} \right)^k = 1[/tex]
    So when n is a large enough number, an/nk might be very small, we don't know. But each time I increase n by one after this point, the numerator increases by a factor of a, and the denominator increases by a factor which is very close to 1 (say, less than a1/2). From that point on the value of an/nk grows by a factor of at least a1/2 each time you increase n by one
     
  15. Apr 27, 2012 #14
    Very interesting thought, thank you very much...
     
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