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Math Help - cauchy convergence

  1. #1
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    cauchy convergence

    Hi I am having a bit of trouble with a couple of questions in regards to convergences.
    Let the sequence \{a_{n}\}^\infty _{1}

    let a_{1} =1, a_{2} = 2, a_{n+2} = \frac{a_{n+1} + a_{n}}{2}

    Prove that |a_{n+2} - a_{n+1}| = \frac {1}{2^n}

    Hence prove that \{a_{n}\}^\infty_{1}

    From my attempts at part 1 I have been able to show that for n=0 and n=1 that the function will be equal to \frac{1}{2^0} and \frac{1}{2^1} respectively. However I am having trouble actually proving for n terms that it is true. I am able to get to |\frac{a_{n+1} - a_{n-1}}{2}| but am completely stuck there on.

    For the second part I have very little idea as to how I can prove that the sequence converges using my result from part 1.
    Thanks you for any input in advance :d
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Use induction for part 1.

    |a_{m+2}-a_{m+1}|=|\frac{a_{m+1}+a_m}{2}-a_{m+1}|

    = |\frac{a_m-a_{m+1}}{2}|

    Thus, if |a_{m+1}-a_{m}|=\frac{1}{2^{m-1}},

    |a_{m+1}-a_{m}|=\frac{1}{2*2^{m-1}}=\frac{1}{2^m}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Setting \alpha_{n} = a_{n+1} - a_{n} the difference equation becomes...

    \displaystyle \alpha_{n+1}= -\frac{\alpha_{n}}{2} , \alpha_{0}=1 (1)

    ... and its solution is...

    \alpha_{n} = (-\frac{1}{2})^{n} (2)

    The sequence a_{n} is given by...

    \displaystyle a_{n}= a_{0} + \sum_{k=0}^{n} \alpha_{n} = 1+\frac{2}{3}\ \{1-(-\frac{1}{2})^{n}\} (3)

    ... and from (3) You derive that is...

    \displaystyle \lim_{n \rightarrow \infty} a_{n} = \frac{5}{3} (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 5th 2010 at 09:33 AM. Reason: added a0...
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