Let $(x_k)_{k \in \mathbb{N}} \subset \mathbb{R}^n$ be a sequence, suppose that there exists $c \in [0,1)$ and $n_0 \in \mathbb{N}$ such that $ \vert \vert x_{k+1} - a \vert \vert \leq c \vert \vert x_k-a \vert \vert$, for every $k>k_0$. Show that $\lim x_k = a$.
I'm trying to show that $(x_k)$ is a Cauchy sequence, then it converges for some $b \in \mathbb{R}^n$. So, it is easy showing that $b=a$.
However, I couldn't show that it is a Cauchy sequence. I've tried to use the triangle inequality, but it didn't work. I don't know if it'll be usefull, but I got the inequality bellow:
$$\vert \vert x_{n+m} - x_n \vert \vert \leq (c^m +1) \vert \vert x_m - a \vert \vert, $$
for $m,n >k_0$, of course.
I'd appreciate if you could help me :)
Cauchy sequences are not the way I'd go.
Let $y_n = x_n - a$. Note that it suffices to prove $y_n \to 0$, as this implies $x_n = y_n + a \to 0 + a = a$.
Our relation states that $\|y_{n+1}\| \le c\|y_n\|$. By induction, we then get $$0 \le \|y_n\| \le c^n\|y_0\|.$$ But $c^n \|y_0\| \to 0 \|y_0\| = 0$, since $c \in [0, 1)$. Thus, by squeeze theorem, $y_n \to 0$, and hence $x_n \to a$.