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Proving a sequence has a lower bound

  1. Dec 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the sequence ##\frac{1}{2}(x_n + \frac{2}{x_n})= x_{n+1}##, where ##x_1 =1##:

    Prove that ##x_n## is never less than ##\sqrt{2}##, then use this to prove that ##x_n - x_{n+1} \ge 0## and conclude ##\lim x_n = \sqrt{2}##.

    2. Relevant equations


    3. The attempt at a solution
    Graphing the equation ##\frac{1}{2}(x_n + \frac{2}{x_n})= y##, it's easy to see that that it is greater than ##\sqrt{2}## for all positive values of ##x##. I'm just not sure how to do this algebraically.

    Even if we take it that the lowest bound is ##\sqrt{2}##, we can do $$x_n - x_{n+1} = x_n - \frac{x_n}{2}-\frac{1}{x_n} = \frac{x_n}{2} - \frac{1}{x_n} \ge 0$$ and if the minimum ##x_n = \sqrt{2}##, we can see that it will always be ##\ge 0##. I'm not sure how to do this without hand-waiving the first part, and I also don't know how to get from this to finding the limit of the sequence.
     
  2. jcsd
  3. Dec 30, 2016 #2

    Delta²

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    Gold Member

    To find the limit is easy, if the limit exists then ##\lim{x_n}=\lim{x_{n+1}}=L## , use that and take the limit on both sides of the defining recurrence equation to make (using of the properties of limit for addition and division) a quadratic equation for L that is ##\frac{1}{2}(L+\frac{2}{L})=L##.

    To prove that all the terms (except the first) are bigger than ##\sqrt2## maybe induction on n could work.
     
  4. Dec 30, 2016 #3

    pasmith

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    Homework Helper

    Presumably this is for [itex]n \geq 2[/itex], since [itex]x_1 = 1 < \sqrt{2}[/itex].

    Show that [tex]x_{n+1}^2 = 2 + \frac{(x_n^2 - 2)^2}{4x_n^2}.[/tex]

    The condition [itex]x_n - x_{n+1} \geq 0[/itex] is saying that the sequence is monotonic decreasing. A sequence which is monotonic decreasing and bounded below converges.
     
  5. Dec 30, 2016 #4

    Ray Vickson

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    Look at the arithmetic-geometric inequality; see
    http://jwilson.coe.uga.edu/emt725/AMGM/AMGM.html
     
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