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Prove the upper bound

  1. Feb 15, 2005 #1
    Hey guys,

    I have a sequence, [tex]\sqrt{2}[/tex], [tex]\sqrt{2 \sqrt{2}}[/tex], [tex]\sqrt{2 \sqrt{2 \sqrt{2}}}[/tex], ...

    Basically, the sequence is defined as x1 = root 2
    x(n+1) = root (2 * xn).

    I need to show that this sequence converges and find the limit.

    I proved by induction that this sequence increases. Since it increases, its bounded below by root 2. I need to show that it is bounded above by 2. Then I can use the Monotone Convergence Theorem to show that this sequence converges.

    Any ideas?
     
    Last edited: Feb 16, 2005
  2. jcsd
  3. Feb 15, 2005 #2
    Suppose the sequence is given in terms of n by

    [tex]a_n = 2^{ \frac{2^n - 1}{2^n}}[/tex],

    where the first term is given by [itex]a_1[/itex].

    Show that it satisfies the recurrence relation.

    --J
     
  4. Feb 15, 2005 #3
    Thanks I got it from that. But can someone tell me how to do the root thing? Is the code LaTeX code or what is it?

    Also, how can I prove the limit of that sequence = 2?

    Is there some theorem that says that the limit of an increasing bounded sequence is equal to the sup of that sequence?
     
  5. Feb 15, 2005 #4
    Oh nevermind about the limit part. It was proved in the book with the Monotone Convergence Theorem. I just didn't see it right away.
     
  6. Feb 15, 2005 #5
    \sqrt{2 a_n} gives [itex]\sqrt{2a_n}[/itex].

    --J
     
  7. Feb 16, 2005 #6
    Okay I guess I am just stupid then. Where exactly do you type \sqrt{2 a_n} ? I thought it was to surround it by CODE tags but that didn't do it and I tried typing it just by itself, with and without the \. What do I do then?
     
  8. Feb 16, 2005 #7
    [ tex ] \sqrt{2a_n}[ /tex ]
    [tex]\sqrt{2a_n}[/tex]

    and

    [ itex ] \sqrt{2a_n}[ /itex ]
    [itex] \sqrt{2a_n}[/itex]

    Remove the spaces from the tags to get them to work.

    --J
     
  9. Feb 16, 2005 #8
    Okay got it. Thanks for all your help.
     
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