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Math Help - Convergence of Sequences

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    Convergence of Sequences

    This proof is in my notes. I would appreciate any guidance that anyone has to offer.

    Let (x_n) be a sequence of complex numbers. If there is a constant c<1 such that |x_{n+1}-x_n|\leq c|x_n-x_{n-1}| for all n\geq 2, prove that (x_n) is convergent.

    Judging by where this is in my notes I'm assuming I might be supposed to use the definition of a Cauchy sequence to prove this.
    Last edited by zebra2147; March 5th 2011 at 07:19 PM.
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    Quote Originally Posted by zebra2147 View Post
    This proof is in my notes. I would appreciate any guidance that anyone has to offer.

    Let (x_n) be a sequence of complex numbers. If there is a constant c<1 such that |x_{n+1}-x_n|\leq c|x_n-x_{n-1}| for all n\geq 2, prove that (x_n) is convergent.

    Judging by where this is in my notes I'm assuming I might be supposed to use the definition of a Cauchy sequence to prove this.

    Hints:

    1) I assume 0<c<1 , and prove now that |x_{n+1}-x_1|<c^{n-1}|x_2-x_1|

    2) Now prove that \forall k\in\mathbb{N}\,,\,\,|x_{n+k}-x_n}<p(c)|x_2-x_1|\,,\,\,with\,\, p(x) an integer polynomial

    with zero free coefficient.

    Tonio
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by zebra2147 View Post
    This proof is in my notes. I would appreciate any guidance that anyone has to offer.

    Let (x_n) be a sequence of complex numbers. If there is a constant c<1 such that |x_{n+1}-x_n|\leq c|x_n-x_{n-1}| for all n\geq 2, prove that (x_n) is convergent.

    Judging by where this is in my notes I'm assuming I might be supposed to use the definition of a Cauchy sequence to prove this.
    In You define \Delta_{n} = x_{n+1} - x_{n} is ...

    \displaystyle x_{n} = x_{0} + \sum_{k=0}^{n} \Delta_{k} (1)

    Now if there is a constant c<1 such that for n 'large onough' is |\Delta_{n}|\le c\ |\Delta_{n-1}| , then for the ratio test the series \displaystyle \sum_{n=0}^{\infty} \Delta_{n} converges...

    Kind regards

    \chi \sigma
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    I know I'm probably being stupid but how would you recommend proving that? Would you use the triangle inequality by starting with |x_{n+1}-x_2|\geq 0?
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    Quote Originally Posted by zebra2147 View Post
    I know I'm probably being stupid but how would you recommend proving that? Would you use the triangle inequality by starting with |x_{n+1}-x_2|\geq 0?
    Do you follow this?
    |x_{n+3}-x_{n+2}|\le c|x_{n+2}-x_{n+1}|\le c^2|x_{n+1}-x_{n}|\le c^3|x_{n}-x_{n-1}|
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    Super Member girdav's Avatar
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    The Tonio's 1) can be shown by induction.
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    Plato...
    I guess I don't really follow that... how do we know that |x_{n+2}-x_{n+1}|<c|x_{n+1}-x_n| and so on.... Isn't this assuming that the sequence is strictly increasing or decreasing? Or am I confused?
    I think if I understood this completely then a proof by induction would be relatively easy...
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  8. #8
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    Quote Originally Posted by zebra2147 View Post
    how do we know that |x_{n+2}-x_{n+1}|<c|x_{n+1}-x_n| and so on.... Isn't this assuming that the sequence is strictly increasing or decreasing? Or am I confused?
    Absolutely not.
    It is simply using the given.
    From the given, we know |x_3-x_2|\le c|x_2-x_1|. RIGHT?

    Step up one. |x_4-x_3|\le c|x_3-x_2|.
    In that line replace |x_3-x_2| to get |x_4-x_3|\le c^2|x_2-x_1|.

    Step up one again. |x_4-x_3|\le c^3|x_2-x_1|.

    Now prove by induction that
    |x_{n+1}-x_n|\le c^{n-1}|x_2-x_1|
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