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Sequences, convergance and limits.

  1. Nov 24, 2009 #1
    1. The problem statement, all variables and given/known data
    An = (n+3)^1/(n+3)
    Converge or diverges? Find the limit of the convergent sequences.

    2. Relevant equations

    3. The attempt at a solution

    I have the solution but I don't understand it. I'm looking to get it.

    It shows me taking the limit as x->infinity of x^1/x and determine that it is 1. (Thus the series converges). I am not sure at all how to find the limits though, I'm lost in my calculus class. Any guidance would be great.
  2. jcsd
  3. Nov 24, 2009 #2
    Use the fact that [tex] m^{1/m} = e^{\ln (m^{1/m})} [/tex] and then for example the L'Hospital rule to calculate the limit for a large m.
  4. Nov 25, 2009 #3
    So my f(x) would be my original

    and the g(x) would be my e^lnn(n+3)^(1/n+3)

    I put f'(x) over g'(x) and keep going untill it's not 0/0, whatever I do get beside 0/0 is my limit?

    I'm honestly stuck trying to differentiate my g(x) ;X Is this the only way to go about this?
  5. Nov 25, 2009 #4


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    Take the log of your expression. Then you should get something of the form f(n)/g(n). You won't get that till you take the log.
  6. Nov 25, 2009 #5
    log of n+3?

    I don't see how that gets me anywhere?
  7. Nov 25, 2009 #6


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    No, log((n+3)^(1/(n+3))).
  8. Nov 25, 2009 #7
    you want me to derive that then? Sorry I'm lost at the moment.
  9. Nov 25, 2009 #8


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    I want you to use the rules of logs and turn that into a quotient of infinity/infinity form, then use l'Hopital.
  10. Nov 27, 2009 #9
    (n+3)^[1/(n+3)] is just the same as saying x^(1/x).

    x^(1/x) = e^(1/x lnx)

    the limit as x -> infinity, lim x^(1/x) = lim e^(1/x lnx) = e^lim (lnx / x)

    lim (lnx / x) = lim (1/x) {Le Hopital's rule}
    = 0

    Thus, e^0 = 1

    and the series is convergent
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