Be $\{\ \overrightarrow{x_n} \}$ a sequence in $\mathbb R^k$. Suppose that $\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|\leq r\|\overrightarrow{x_n}-\overrightarrow{x}_{n-1}\|$ with $0\leq r<1$.
Show that the sequence $\{ \overrightarrow{x_n}\}$ is convergent.
Hint: prove that sequence $\{\ \overrightarrow{x_n} \}$ is a Cauchy sequence, using: $$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|\leq \|\overrightarrow{x_m}-\overrightarrow{x}_{m-1}\|+\|\overrightarrow{x}_{m-1}-\overrightarrow{x}_{m-2}\|+\cdots+\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|$$ And hypothesis of sequence (with $m>n$).
My solution:
$$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|\leq \|\overrightarrow{x_m}-\overrightarrow{x}_{m-1}\|+\|\overrightarrow{x}_{m-1}-\overrightarrow{x}_{m-2}\|+\cdots+\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|$$
using hypotesis of recurrence
$$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|\leq r\|\overrightarrow{x}_{m-1}-\overrightarrow{x}_{m-2}\|+r\|\overrightarrow{x}_{m-2}-\overrightarrow{x}_{m-3}\|+\cdots+r\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|$$
$$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|\leq r+r^2\|\overrightarrow{x}_{m-2}-\overrightarrow{x}_{m-3}\|+r^2\|\overrightarrow{x}_{m-3}-\overrightarrow{x}_{m-4}\|+\cdots+r^2\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|$$
$$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|\leq r(1+r+r^2+...+r^n)\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|$$
Where $1+r+r^2+...+r^n\leq\frac{1}{1-r}$
So $$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|\leq r(\frac{1}{1-r})\|\overrightarrow{x}_{n+1}-\overrightarrow{x_n}\|$$
So
$$\|\overrightarrow{x_m}-\overrightarrow{x_n}\|<\epsilon$$