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Quick Analysis Problem

  1. Feb 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Let S_1 = 1 and for n >= 1 let S_n+1 = sqrt(S_n + 1)

    e.g. s_2 = sqrt(2), s_3 = sqrt(sqrt(2) + 1), etc.

    Prove lim(S_n) = 1/2(1 + sqrt(5)).

    2. Relevant equations

    3. The attempt at a solution

    Not really sure how to approach this. This is right after the "Limit Theorems" section in our book, so I would assume I have to use some limit theorems. The tricky part is figuring out which ones to use! :rolleyes: Thanks for any help.
  2. jcsd
  3. Feb 12, 2008 #2
    Alright, two things:

    First you must find the possible limits. If they exists they will solve

    lim(S_n+1) = lim(sqrt[S_n + 1]).

    recall that lim(S_n+1) = lim(S_n) = s. So you just solve a quadratic equation.

    Now, you don't know if they exist, but just remember one thing:

    A monotone bounded sequence is convergent.

    See if you can prove that the sequence is bounded and monotone.

    If you can show that, then you know it must converge, and if it converges then it converges to either one of the two equilibrium (should be fairly obvious to which).
  4. Feb 12, 2008 #3
    Thanks for your response. I neglected to say in the first post that we can assume S_n converges, which makes the problem a lot easier. So I know that lim(S_n+1) = lim(sqrt[S_n + 1]) = lim(S_n) = s. I'm not sure what you mean by solving a quadratic equation.

    Edit: OK, I have lim(S_n+1)lim(sqrt[S_n + 1]) = lim(S_n+1 * sqrt[s_n + 1]) = s^2. Not sure if that helps me or not.
    Last edited: Feb 12, 2008
  5. Feb 12, 2008 #4
    you have S_n+1 = sqrt(S_n + 1) or equivalently
    S_n+1^2 = S_n + 1

    Assuming that S_n converges to s, we have lim(S_n+1) = lim(S_n) = s
    So s^2 = s + 1

    Solve for s, you will get two solutions, one of which is the one you posted.
    Now, you gotta show that it converges to that one and not the toher... which means that it will be either increasing or decreasing monotonically to it (which should be its boundary). That's what you have to show.
  6. Feb 12, 2008 #5
    Ok, I figured it out. Thanks for your help!
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