contractive sequence

**particlejohn** · July 11th 2008, 08:41 AM

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

We claim that if $n \in \bold{N}$ , then $|x_{k}-x_{k+1}| > c^{k-1}|x_{1}-x_{2}|$ . For $k = 1$ , $|x_{1}-x_{2}| > |x_{1}-x_{2}|$ , which is false. So it cannot be a contractive sequence?

**particlejohn** · July 11th 2008, 09:18 AM

although this seems too simple.

**ThePerfectHacker** · July 11th 2008, 09:40 AM

Originally Posted by particlejohn

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

Note that $x_1 \not = x_2$ . Because if this was not the case then $0 = |x_2-x_1| > c|x_1 - x_0|$ which is impossible. But as you showed $|x_{n+1} - x_n| > c^{n-1}|x_1-x_2| \geq |x_1-x_2|$ . Let $0<\epsilon < |x_1-x_2|$ . Then $|x_{n+1} - x_n| > \epsilon$ for $n\geq 2$ . Therfore the sequence $\{x_n\}$ is not Cauchy and therefore not convergent.

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