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Homework Help: How would I find what a recursive sequence converges to a certain point?

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Given the sequence {a(n)}, which is refined recursively by

    a(1)=1, a(n+1)=√(1+a(n))

    Prove that the sequence converges and find it's limit.

    2. Relevant equations

    3. The attempt at a solution

    I proved that it converges with two pretty easy induction cases.

    I've also found what it converges to, the golden ratio w/ a calculator. How would I go about this not using a calculator? Usually one would just consider the infinite case but here I'm not sure how to do that.
    Last edited: Dec 16, 2008
  2. jcsd
  3. Dec 16, 2008 #2


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    Hi Quantumpencil! :smile:

    (have a square-root: √ :wink:)
    Just put n = n+1: a(n)=√(1+a(n)). :wink:
  4. Dec 16, 2008 #3
    Right, I got that; but don't you then end up with an infinite mess that's hard to evaluate?

    a(n)=[tex]\sqrt{1+(a(n)}[/tex]; still forces you to substitute in a lot of a(n)'s; Is there no way to just take the limit and get phi?
  5. Dec 16, 2008 #4


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    uhh? :confused: a(n) is fixed

    if you prefer, just use a …

    a = √(1 + a). :smile:
    (and what happened to that √ that i gave you? :wink:)​
  6. Dec 16, 2008 #5
    What I'm saying is since each of the subsequent terms depends on the one before it... that sort of seems like magic to me. when you go to infinity there negligible difference between the n+1th and nth term in the sequence, but still...

    Even with a(n)=[tex]\sqrt{1+a(n)}[/tex], what would I take a(n) to obtain the Golden ratio?

    Sorry if I'm not making sense; I'm confused as to how making this equivalence helps us find the limit; we still don't have an (n) to take the limit with respect to...

    Oh, and I noticed the sigma and clicked on it. Pretty cool stuff ;p
  7. Jan 1, 2009 #6
    Assume that the series converges.
    Let [tex]lim_{n\rightarrow\infty}a(n)=l[/tex].
    [tex]l = \sqrt{1+l}[/tex]
    [tex]l=(1+\sqrt{5})/2[/tex] or [tex](1-\sqrt{5})/2[/tex]
  8. Jan 1, 2009 #7


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    I have to disagree with tiny-tim here, at least with the way he said it. Saying "put n= n+1" is exactly the same as saying 0= 1! What he meant, I'm sure, is that the sequence {a(n)} necessarily has the same limit as the sequence {a(n+1)}, the same sequence, "shifted" by 1. That leads to what evail is saying:
    [tex]\lim_{n\rightarrow\infty} a(n)= \sqrt{1+ \lim_{x\rightarrow\infty} a(n+1)}[/tex]
    so if the limit is "[itex]l[/itex]", [itex]l= \sqrt{1+ l}[/itex] which has the solutions evail shows.

    Of course, the limit is one of [itex](1+ \sqrt{5})/2[/itex] or [itex](1- \sqrt{5})/2[/itex]. Do you see how to determine which one? a(1)= 1>0 and it is easy to see that this sequence is increasing. The limit can't be negative!
  9. Jan 2, 2009 #8
    Thanks HallsofIvy for completing my solution! i forgot that a(n+1)>a(n) for all n>1. hahaha.. so i had to reject the negative part :)
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