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Homework Help: Arthur Mattuck, Introduction to Analysis, Problem 5-7

  1. Oct 5, 2007 #1


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    This question is from Arthur Mattuck's "Introduction to Analysis", chapter 5, problem 5-7.

    1. The problem statement, all variables and given/known data
    Define a sequence recursively by [tex]a_{n+1}=\sqrt{2a_{n}}[/tex], [tex]a_{0}>0[/tex].

    (a) Prove that for any choice of [tex]a_{0}>0[/tex], the sequence is monotone and bounded.

    2. Relevant equations


    3. The attempt at a solution

    I've rewritten the sequence as

    [tex]a_n = \sqrt{2{\sqrt{2\sqrt{2\sqrt{2...\sqrt{2a_0}}}}}[/tex],

    and taken the ratio of [tex]a_{n+1}[/tex] and [tex]a_{n}[/tex], which leads me to the expression

    [tex]\frac{a_{n+1}}{a_n} = 2^{(1/2)^{n+1}}\sqrt{a_0}[/tex].

    From computing the sequence for a few initial values of [tex]a_0[/tex], I've been able to determine that the sequence is decreasing if [tex]a_0>2[/tex] and increasing if [tex]0<a_0<2[/tex]. However, I'm not sure how to show this from this ratio, i.e. how does


    which indicates that the sequence is decreasing, turn into [tex]a_0>2[/tex]?

    The rest of the problem comes easy enough. I'm able to show that it is bounded and that the limit is 2 (in part b of the question).
  2. jcsd
  3. Oct 5, 2007 #2


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    Homework Helper

    You are making this too complicated. Proving a_(n+1)>a_n translates into sqrt(2*a_n)>a_n. For what range of x is it true that sqrt(2x)>x.
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