Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of a series help

  1. Sep 8, 2009 #1
    I want to calculate the following:

    [tex]
    \displaystyle\lim_{k\to\infty}\frac{n_k}{d_k}
    [/tex]
    where,
    [tex]
    n_0 = 2
    [/tex]
    [tex]
    d_0 = 1
    [/tex]
    [tex]
    n_k = 2n_k_-_1 +d_k_-_1
    [/tex]
    [tex]
    d_k = n_k_-_1 + d_k_-_1
    [/tex]

    For the life of me I have no idea how to do this. By the way, the answer is supposed to be
    [tex]\frac{1 + \sqrt{5}}{2}[/tex]

    This is not a homework problem. I was doing an electrical engineering problem and to solve the problem this series was magically solved.

    Please any help is appreciated.
     
  2. jcsd
  3. Sep 8, 2009 #2

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

  4. Sep 8, 2009 #3
    I'm intrigued... as the answer it is supposed to be is the golden ratio.
     
  5. Sep 8, 2009 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Okay, you may proceed as follows:
    [tex]\frac{n_{k}}{d_{k}}=\frac{2n_{k-1}+d_{k-1}}{n_{k-1}+d_{k-1}}=1+\frac{1}{1+\frac{d_{k-1}}{n_{k-1}}}[/tex]

    Assuming that a limit exists as k (and therefore k-1) trundles off into infinity, call that limit "x".

    Thus, you get the equation,
    which ought to be easily solved.
    [tex]x=1+\frac{1}{1+\frac{1}{x}}[/tex]

    Note that this is simply another way of writing the continued fractions representation of the golden ratio.
     
    Last edited: Sep 8, 2009
  6. Sep 8, 2009 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Another way to proceed, would be to first solve for the one sequence, then for the seconde, and finally solve for the limiting ratio.

    We can start with rearranging the second equation:
    [tex]n_{k-1}=d_{k}-d_{k-1}\to{n_{k}=d_{k+1}-d_{k}(*)[/tex]

    Inserting these relations into the first, we get:
    [tex]d_{k+1}-d_{k}=2(d_{k}-d_{k-1})+d_{k-1}\to{d}_{k+1}-3d_{k}+d_{k-1}=0[/tex]
    Noting from (*) that we have [itex]d_{0}=1,d_{1}=3[/itex], you should be able to solve for the d-sequence.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit of a series help
  1. Help with a series (Replies: 12)

  2. Limit of Series (Replies: 6)

Loading...