Problem 12, section 1.5 from Introductory Functional Analysis - Kreyszig. The problem:
Let $s:= S(\mathbb{C})$ be the space of all complex sequences, with metric $d$ defined for each $x=(\xi_j)_j, y=(\eta_j)_j\in s$ as $$d(x,y)=\sum_{j=1}^{+\infty}\dfrac{1}{2^j}\dfrac{\lvert \xi_j-\eta_j\rvert}{1+\lvert \xi_j-\eta_j\rvert}$$
Prove that $(s,d)$ is complete, using that $x_n\to x(n\to\infty)\iff \xi_k^{(n)}\to\xi_k(n\to\infty)$ for each $k\in\mathbb{N}$, where $x_n=\left(\xi_k^{(n)}\right)_k$ and $x=(\xi_k)_k$.
My attempt:
Let $(x_n)_n$ be a Cauchy sequence of $s$. We need to prove it converges, and we're asked to do it showing for each $k\in\mathbb{N}$ ,$\left(\xi_k^{(n)}\right)_n$ converges. Find a limit seems to be not possible, so I am trying to check if $\left(\xi_k^{(n)}\right)_n$ is a Cauchy sequence, because it is a sequence of $\mathbb{C}$.
We know $$\forall\varepsilon>0,\exists m\in\mathbb{N}: d(x_p,x_q)=\sum_{j=1}^{+\infty}\dfrac{1}{2^j}\dfrac{\left\lvert \xi_j^{(p)}-\xi_j^{(q)}\right\rvert}{1+\left\lvert \xi_j^{(p)}-\xi_j^{(q)}\right\rvert}<\varepsilon, \forall p,q\geq m$$
We fix $k\in\mathbb{N}$. Since $\dfrac{\varepsilon}{2^k}>0$, $\exists m\in\mathbb{N}$ such that $$\dfrac{\left\lvert \xi_j^{(p)}-\xi_j^{(q)}\right\rvert}{1+\left\lvert \xi_j^{(p)}-\xi_j^{(q)}\right\rvert}<\varepsilon,\forall p,q\geq m$$
Is this last fact enough to see $\left(\xi_k^{(n)}\right)_n$ is a Cauchy sequence?
Yes, it is. Just note that $\frac {|a|}{1+|a|} <\epsilon$ iff $|a|<\frac {\epsilon} {1-\epsilon}$. (Without loss of genearlity assume that $0 <\epsilon <\frac 1 2 $ so that $\frac {\epsilon} {1-\epsilon} <2\epsilon$).