# contractive sequence

• July 11th 2008, 08:41 AM
particlejohn
contractive sequence
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?
• July 11th 2008, 09:18 AM
particlejohn
although this seems too simple.
• July 11th 2008, 09:40 AM
ThePerfectHacker
Quote:

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.