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Cauchy Sequence

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the sequence of real numbers defined by [itex]x_{n + 1} = x_n + \frac{1}{x_n^2}, \, x_1 = 1[/itex] is not a Cauchy sequence.

    2. Relevant equations

    A sequence [itex]\{ p_n \}[/itex] is Cauchy if and only if, for all [itex]\varepsilon > 0[/itex], there exists an [itex]N > 0[/itex] such that [itex]d(p_n, p_m) < \varepsilon[/itex] for all [itex]m, n > N[/itex].

    3. The attempt at a solution

    We can assume that [itex]d[/itex] is the usual metric on [itex]\mathbb{R}[/itex]. I don't even see where to begin. I see that the sequence is monotonically increasing, so that
    [itex]1 = \frac{1}{x_1} > \frac{1}{x_2} > \frac{1}{x_3} > \dotsb.[/itex]
    So
    [itex]1 = \frac{1}{x_1^2} > \frac{1}{x_2^2} > \frac{1}{x_3^2} > \dotsb.[/itex]
    To me it looks like the sequence is in fact Cauchy. Please help!
     
  2. jcsd
  3. Feb 9, 2012 #2
    Well if you're trying to show that it is not Cauchy, state what it means for a sequence to not be Cauchy. That is where I would start :)
     
  4. Feb 9, 2012 #3
    The sequence isn't Cauchy if there exists an [itex]\varepsilon > 0[/itex] such that for all [itex]N > 0[/itex] there exists [itex]m, n > 0[/itex] such that [itex]d(x_m, x_n) \geq \varepsilon[/itex].

    ...

    still stuck

    ...
     
  5. Feb 9, 2012 #4

    LCKurtz

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    Do you have the theorem that a sequence is Cauchy if and only if it is convergent? And if so, what happens if you suppose ##\lim_{x\rightarrow \infty} = L##?
     
  6. Feb 9, 2012 #5
    Yes, I can assume the sequence is Cauchy if and only if it is convergent. The definition of limits that we are using states that [itex]\lim_{n \to \infty} x_n = L[/itex] if and only if
    [itex]\forall \varepsilon > 0 \, \exists N > 0 \, s.t. \, \forall n > N \quad d(L, x_n) < \varepsilon[/itex].

    Now can I somehow use the fact that [itex]x_n[/itex] is increasing to say that [itex]d(L, x_n)[/itex] is always increasing? And hence it is greater than or equal to [itex]\varepsilon[/itex] for some [itex]n[/itex]?
     
  7. Feb 9, 2012 #6

    LCKurtz

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    It's easier than that. If the sequence has a limit, what happens if you take the limit of both sides of your recursion?
     
  8. Feb 9, 2012 #7
    The limit of both sides of the recurrence should then equal the same thing, namely [itex]L[/itex]. Then I would have that

    [itex]\lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_n + \frac{1}{x_n^2} = L + \lim_{n \to \infty} \frac{1}{x_n^2} = L[/itex].

    Doesn't this simply prove that [itex]\lim_{n \to \infty} \frac{1}{x_n^2} = 0[/itex]?
     
  9. Feb 9, 2012 #8
    But then doesn't this imply [itex]\lim_{n \to \infty} x_n^2 = \infty[/itex]? completing the problem?
     
  10. Feb 9, 2012 #9

    Matterwave

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    Can that second limit be 0 while the first limit exists? Is there a number L such that 1/L^2=0?
     
  11. Feb 9, 2012 #10
    Exactly. Thank you LCKurtz for your patience! :)
     
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