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Proving a monotonic sequence is unbounded

  1. Sep 24, 2007 #1
    I'm trying to prove that the sequence
    [tex]x_1,x_2,_\cdots[/tex]
    of real numbers, where
    [tex]x_1=1[/tex] and [tex]x_{n+1}=x_n+\frac{1}{x_n^2}[/tex] for each [tex]n=1,2, \cdots[/tex]

    is unbounded.

    (sorry for the ugly latex! i don't know if there's a way to format that better)

    I'm thinking of proving by contradiction, assuming it is bounded and then somehow getting it to imply that the sequence is not increasing, but I'm not sure how to go about it.

    Any hints?
     
  2. jcsd
  3. Sep 24, 2007 #2

    morphism

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    If it were bounded, it would converge.
     
  4. Sep 24, 2007 #3

    HallsofIvy

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    And if it were to converge to, say, x, that limit would satisfy
    [tex]x= x+ \frac{1}{x}[/tex]
    What values of x satisfy that?
     
  5. Sep 24, 2007 #4
    :confused: I don't understand why this is true.
     
  6. Sep 24, 2007 #5

    HallsofIvy

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    Start with [itex]x_{n+1}= x_n+ 1/x_n[/itex] and take the limit, as n goes to infinity ,of both sides. If the sequence [itex]{x_n}[/itex] converges to some number, x, then each "[itex]x_n[/itex]" or "[itex]x{n+1}[/itex]" term will go to x.
     
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