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Thread: Convergence of Sequences

  1. #1
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    Convergence of Sequences

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

    Let $\displaystyle (x_n)$ be a sequence of complex numbers. If there is a constant $\displaystyle c<1$ such that $\displaystyle |x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$ for all $\displaystyle n\geq 2$, prove that $\displaystyle (x_n)$ 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; Mar 5th 2011 at 06:19 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 $\displaystyle (x_n)$ be a sequence of complex numbers. If there is a constant $\displaystyle c<1$ such that $\displaystyle |x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$ for all $\displaystyle n\geq 2$, prove that $\displaystyle (x_n)$ 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 $\displaystyle 0<c<1$ , and prove now that $\displaystyle |x_{n+1}-x_1|<c^{n-1}|x_2-x_1|$

    2) Now prove that $\displaystyle \forall k\in\mathbb{N}\,,\,\,|x_{n+k}-x_n}<p(c)|x_2-x_1|\,,\,\,with\,\, p(x)$ 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 $\displaystyle (x_n)$ be a sequence of complex numbers. If there is a constant $\displaystyle c<1$ such that $\displaystyle |x_{n+1}-x_n|\leq c|x_n-x_{n-1}|$ for all $\displaystyle n\geq 2$, prove that $\displaystyle (x_n)$ 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 $\displaystyle \Delta_{n} = x_{n+1} - x_{n}$ is ...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle |x_{n+1}-x_2|\geq 0$?
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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 $\displaystyle |x_{n+1}-x_2|\geq 0$?
    Do you follow this?
    $\displaystyle |x_{n+3}-x_{n+2}|\le c|x_{n+2}-x_{n+1}|\le c^2|x_{n+1}-x_{n}|\le c^3|x_{n}-x_{n-1}|$
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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 $\displaystyle |x_{n+2}-x_{n+1}|<c|x_{n+1}-x_n|$ 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 $\displaystyle |x_{n+2}-x_{n+1}|<c|x_{n+1}-x_n|$ 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 $\displaystyle |x_3-x_2|\le c|x_2-x_1|$. RIGHT?

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

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

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