1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sequences, convergance and limits.
Loading...