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Math Help - Investigate the sequence?

  1. #1
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    Investigate the sequence?

    I have difficulty to investigate the following sequence to be convergent or divergent and find limit?

    Investigate the sequence?-gf.jpg
    Last edited by CaptainBlack; January 23rd 2011 at 03:40 AM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by khyratmath123 View Post
    I have difficulty to investigate the following sequence to be convergent or divergent and find limit?
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    To show convergence, show it is bounded and monotonically increasing. I believe you could show  a_n \leq 2 via induction.

    Once you know the limit exists (call it  L ), we can find it by noticing  L = \sqrt{2+L} .
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  3. #3
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    Are you sure it's not

    \displaystyle a_1 = \sqrt{2} and \displaystyle a_{n + 1} = \sqrt{2 + a_n}?

    If so, this is a relatively easy limit to evaluate...

    It should be clear that the sequence goes \displaystyle \left\{\sqrt{2}, \sqrt{2 + \sqrt{2}}, \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \dots \right\}.

    So you want to see if \displaystyle \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}} can be evaluated (has a limit...). Call this limit \displaystyle L.

    If \displaystyle L = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}} then

    \displaystyle L^2 = 2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots} }}}}

    \displaystyle L^2 = 2 + L

    \displaystyle L^2 - L - 2 = 0

    \displaystyle (L-2)(L+1) = 0

    \displaystyle L - 2 = 0 or \displaystyle L+1 = 0

    \displaystyle L = 2 or \displaystyle L = -1.


    It should be clear that \displaystyle L = -1 is an extraneous solution which came from the original squaring of the equation, so the limit is \displaystyle L = 2.
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  4. #4
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    thanks sir
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  5. #5
    MHF Contributor chisigma's Avatar
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    The difference equation can be written as...

    \displaystyle \Delta_{n}= a_{n+1}-a_{n} = \sqrt{2+a_{n}} -a_{n} = f(a_{n}) (1)

    The function f(*) is illustrated here...



    There is only one 'attractive fixed point' in x_{0}=2 but that's only a necessary condition of convergence. In this case however is |f(x)|\le |x_{0}-x| [red line...] so that any a_{0}>-2 will generate a sequence monotonically convergent at x_{0}...

    Kind regards

    \chi \sigma
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