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Thread: contractive sequence

  1. #1
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    contractive sequence

    If $\displaystyle x_n $ is a sequence and there is a $\displaystyle c \geq 1 $ such that $\displaystyle |x_{k+1}-x_{k}| > c|x_{k}-x_{k-1}| $ for all $\displaystyle k > 1 $, then can $\displaystyle x_n $ converge?

    We claim that if $\displaystyle n \in \bold{N} $, then $\displaystyle |x_{k}-x_{k+1}| > c^{k-1}|x_{1}-x_{2}| $. For $\displaystyle k = 1 $, $\displaystyle |x_{1}-x_{2}| > |x_{1}-x_{2}| $, which is false. So it cannot be a contractive sequence?
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  2. #2
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    although this seems too simple.
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  3. #3
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    Quote Originally Posted by particlejohn View Post
    If $\displaystyle x_n $ is a sequence and there is a $\displaystyle c \geq 1 $ such that $\displaystyle |x_{k+1}-x_{k}| > c|x_{k}-x_{k-1}| $ for all $\displaystyle k > 1 $, then can $\displaystyle x_n $ converge?
    Note that $\displaystyle x_1 \not = x_2$. Because if this was not the case then $\displaystyle 0 = |x_2-x_1| > c|x_1 - x_0| $ which is impossible. But as you showed $\displaystyle |x_{n+1} - x_n| > c^{n-1}|x_1-x_2| \geq |x_1-x_2|$. Let $\displaystyle 0<\epsilon < |x_1-x_2|$. Then $\displaystyle |x_{n+1} - x_n| > \epsilon$ for $\displaystyle n\geq 2$. Therfore the sequence $\displaystyle \{x_n\}$ is not Cauchy and therefore not convergent.
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