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Math Help - Terms in sequence

  1. #1
    Senior Member I-Think's Avatar
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    Terms in sequence

    Define a sequence, a_1=\frac{1}{2} and a_{n+1}=\frac{1}{2+a_n}
    Prove that |a_{n+1}-a_{n+2}|<\frac{|a_n-a_{n+1}|}{4}

    Proof by induction

    When n=1, it is true
    \frac{1}{60}<0.04

    Assume true for n=k
    |a_{k+1}-a_{k+2}|<\frac{|a_k-a_{k+1}|}{4}

    Test n=k+1
    |a_{k+2}-a_{k+3}|<\frac{|a_{k+1}-a_{k+2}|}{4}

    4|\frac{1}{2+a_{k+1}}-\frac{1}{2+a_{k+2}}|<|\frac{1}{2+a_k}-\frac{1}{2+a_{k+1}}|

    4|\frac{a_{k+1}-a_{k+2}}{(2+a_{k+1})(2+a_{k+2})}|<|\frac{a_{k}-a_{k+1}}{(2+a_{k})(2+a_{k+1})}|

    By hypothesis, this is true for the numerators, now testing the denominators
    if
    (2+a_{k+1})(2+a_{k+2})>(2+a_k)(2+a_{k+1}) it is always true
    Simplifying
    a_{k+2}>a_k

    However, this is not always true.

    Now I need help. I am unsure how to proceed, or any of the preceding steps are wrong and I need to find a new method t solve the problem.
    All help greatly appreciated
    Last edited by I-Think; October 13th 2010 at 10:29 AM.
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  2. #2
    MHF Contributor

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    To compare them, right both in terms of a_{n+1} only.

    Since a_{n+1}= \frac{1}{2+ \frac{1}{a_n}}, for all n, a_{n+2}= \frac{1}{2+ \frac{1}{a_{n+1}}} so a_{n+1}- a_{n+2}= \frac{(a_{n+1})(2+ \frac{1}{a_{n+1}})}- 1}{2+ \frac{1}{a_{n+1}} = \frac{2a_{n+1}}{2+ \frac{1}{a_{n+1}}}.

    a_{n+1}= \frac{1}{2+ \frac{1}{a_n}}= \frac{a_n}{2a_n+ 1}
    2a_na_{n+1}+ a_{n+1}= a_n
    2a_na_{n+1}- a_n= -a_{n+1}
    a_n(1- 2a_{n+1})= a_{n+1}
    a_n= \frac{a_{n+1}{1- 2a_{n+1}}}
    so that
    a_n- a_{n+1}= \frac{a_{n+1}}{1- 2a_{n+1}}- a_{n+1}= \frac{-2a_{n+1}^2}{1- 2a_{n+1}}

    Now compare those.
    Last edited by HallsofIvy; October 13th 2010 at 11:40 AM.
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  3. #3
    Senior Member I-Think's Avatar
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    Thanks
    Last edited by I-Think; October 13th 2010 at 12:19 PM. Reason: Fixed latex errors
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