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Recursive sequence convergence

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]s_{1} = 1[/tex] and [tex]s_{n+1} = \sqrt{s_{n} + 1}[/tex] Assume this converges to [tex]\frac{1}{2}(1+\sqrt{5})[/tex] and prove it.


    2. Relevant equations



    3. The attempt at a solution

    I'm not really sure where to begin. I tried to set it up with [tex]|s_{n} - s| < \epsilon[/tex] but I'm not sure how to handle the recursiveness of the sequence.

    Any tips would be appreciated.
     
  2. jcsd
  3. Feb 12, 2009 #2
    OK new attempt. Using algebra I found:
    [tex]|\frac{2+s_{n}(1-\sqrt{5})}{2s_{n}}|< |\frac{2+M(1-\sqrt{5})}{2m}|<\epsilon[/tex]

    where m = inf([tex]s_{n}[/tex]) and M = sup([tex]s_{n}[/tex])

    Am I doing it right?
     
  4. Feb 12, 2009 #3

    Dick

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    I'm not sure how you found that, since you didn't show it, but one part of the question is probably easier than you are making it out to be. If s_n has a limit L, then s_n->L and s_n+1->L. So L (if it exists) must satisfy L=sqrt(L+1). Can you solve that for L? What are the possibilities? Showing the limit actually exists is a little harder. Any ideas on how to go about it?
     
  5. Feb 13, 2009 #4

    HallsofIvy

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    I'm not sure what you mean by "Assume this converges to [tex]\frac{1}{2}(1+ \sqrt{5})[/tex] and prove it".

    Here's the proof reading that literally:
    "If {sn} converges to [tex]\frac{1}{2}(1+ \sqrt{5})[/tex], then it converges to [tex]\frac{1}{2}(1+ \sqrt{5})[/tex]"!

    If you mean "assume that it converges" and then prove that the limit is [tex]\frac{1}{2}(1+ \sqrt{5})[/tex], take the limit on both sides of [tex]s_{n+1}= \sqrt{s_n+ 1}[/tex]
    Since the "s" terms on both sides give the same limit, if we call that limit L, we get [tex]L= \sqrt{L+ 1}[/itex] as Dick says.
     
    Last edited: Feb 14, 2009
  6. Feb 13, 2009 #5

    statdad

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    If you need to show the sequence converges, you can do it by showing

    1. The sequence is non-decreasing: [tex] s_1 \le s_2 \le \cdots \le s_n \cdots [/tex] (it's actually strictly increasing, but that's irrelevant)
    2. The sequence is bounded above - there is a number [tex] W [/tex] such that [tex] s_n \le W [/tex] for all values of [tex] n [/tex].

    Once you know it converges, you can proceed as the others have stated.

    If you are free to ASSUME it converges then you can skip my two steps and do as they correctly suggest.
     
  7. Feb 13, 2009 #6
    Blah, I did a terrible job of wording the problem. I think I may have gotten it, time will tell. Anyway, thank you for the help regardless! Next time I'll try better to word my questions. :)
     
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