1. ## 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?

2. although this seems too simple.

3. Originally Posted by particlejohn
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.