Let $(X,d)$ be a complete metric space, $A \subseteq X $ closed, $f:A \rightarrow A$ a contraction .
In the proof of Banach fixed-point theorem we want to show that $x_{n+1} = f(x_n)$ is a Cauchy-sequence. Why isn't it enough to show that for some $0\leq L < 1$ the following holds: with $x_0 \in A $ arbitrary:
$$d(x_n,x_{n+1})< L^n d(x_1,x_0)$$
Is there a non-cauchy sequence that fulfills that condition? I fell for that trap, but I cannot find an example.
It seems that it implies Cauchyness.
Let $m > n \in \mathbb{N}$.
\begin{align}d(x_m,x_n)&\le d(x_m,x_{m-1})+d(x_{m-1},x_{m-2}) + \ldots + d(x_{n+1},x_n)\\ &\le (L^{m-1} + L^{m-2} + \ldots L^n)\,d(x_1,x_0)\\ &= L^n(1 + L + \ldots + L^{m-n-1})\,d(x_1,x_0)\\ &= L^n\frac{1-L^{m-n}}{1-L}\,d(x_1,x_0)\\ &= \frac{L^n - L^m}{1-L}\,d(x_1,x_0)\\ &\le \frac{L^n}{1-L}\,d(x_1,x_0)\\ &\xrightarrow{m,n\to\infty}0 \end{align}