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Nested radicals and its convergence

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data

    This is supposed to be really easy, but I don't think my answer is good

    Consider this

    [tex]\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}[/tex]

    I was hinted that [tex]a_{n + 1} = \sqrt{1 + a_n}[/tex] for all n ≥ 0 and I am supposed to show that the sequence convergees



    3. The attempt at a solution

    Am I suppose to use [tex]a_{n +1}[/tex] converges or [tex]a_n[/tex]?

    Since the nested radicals go on to infinity, wouldn't it be better to write it as

    [tex]a_n = \sqrt{1 + a_n}[/tex]

    So that

    [tex]a^2 _n = 1 + a_n[/tex]

    We get a quadratic and solve (on Maple) we get

    [tex]\frac{1}{2}(\sqrt(5) + 1)[/tex]

    I rejected negative root because there is no way a negative root can occur in this sequence (we are just adding positive numbers and rooting it (I hope that's a word))
     
  2. jcsd
  3. Nov 28, 2011 #2

    HallsofIvy

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    Staff Emeritus
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    Those are just different numberings for the same sequence if one converge the other converges to the same thing.

    No, that's not true for any finite value of n. What is true is that if [itex]\lim_{n\to\infty} a_n= a[/itex], then [itex]\lim_{n\to\infty}\sqrt{1+ a_n}= \sqrt{1+ \lim_{n\to\infty} a_n}= \sqrt{1+ a}[/itex].

    Okay, except that that it should be the value of the limit, a, not [itex]a_n[/itex]

    Aw, c'mon! You use Maple to solve a quadratic equation? (Yes, that is the correct limit.)
     
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