Sufficient condition that a sequence is a Cauchy sequence

Question

Is the following statement true:

Let $(X,d)$ be a metric space and $(x_n)_n \in X ^\mathbb{N}$. Then:

$\exists m \in \mathbb{N}:d(x_m,x_n)\to 0$ when $n \to \infty $ in $(\mathbb{R}, d_E) \implies (x_n)_n$ is a Cauchy sequence

My attempt:

Let $\epsilon >0$ be arbitrary. Choose $n_0$ such that $\forall n > n_0: d(x_m, x_n) < \epsilon/2$.

Now, for $p,q > n_0$, we have: $d(x_p,x_q) \leq d(x_m, x_p) + d(x_m,x_q) < \epsilon/2 + \epsilon/2 = \epsilon$

Hence, $(x_n)_n$ is Cauchy.

Now, my actual question:

In a certain proof, it is stated that for $0 < n < m$, we have:

$$d(x_m,x_n) \leq \frac{k^n}{1-k}d(x_1,x)$$ where $0 <k < 1$

Can I use the claim I made to conclude that $(x_n)_n$ is a Cauchy sequence, because the right hand side goes to $0$?

Answer 1

The LHS is not negative, so yes.