Suppose $X = \{\{x_n\}_{n \in \mathbb{N}} : x_n \in \mathbb{R}$ $\forall n\geq 1\}$. Given that $d: X \times X \rightarrow \mathbb{R}$ such that $d(\{x_n\}_{n \in \mathbb{N}}, \{y_n\}_{n \in \mathbb{N}}) = \sum_{n = 1}^{\infty} 2^{-n}\frac{|x_n - y_n|}{1 + |x_n - y_n|}$ is a metric, show that $(X, d)$ is a complete metric space.
My attempt:
Consider $\{x^j\}_{j\geq 1}$ Cauchy in $d$. Fix $n \geq 1$. Then for all $0 < \epsilon < \frac{1}{2}$, there is an $j_{\epsilon} \in \mathbb{N}$ so that $d(x^{(j)}, x^{(k)}) < \frac{\epsilon}{2^n}$ for all $j, k \geq j_{\epsilon}$. So for $j, k \geq j_{\epsilon}$ we have
$\sum_{m = 1}^{\infty} 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$
$\implies 2^{-m}\frac{|x_m^{(j)} - x_m^{(k)}|}{1 + |x_m^{(j)} - x_m^{(k)}|} < \frac{\epsilon}{2^n}$
$\implies |x_m^{(j)} - x_m^{(k)}| < \frac{\epsilon}{1-\epsilon} \leq 2\epsilon$
So $\{x^{(j)}\}_{j\geq 1}$ is cauchy $\implies x_n^{(j)} \rightarrow x_n$ as $x \rightarrow \infty$.
Is this correct so far? I have come this far but I'm not sure how to complete the proof. Any assistance is much appreciated.