Show that the metric space $(X,d)$ is complete Let $ X $ be the set of all complex sequences and let $ d: X \times X \rightarrow \mathbb{R} $ be defined by
$$
d\left(\left(x_{n}\right)_{n \in \mathbb{N}},\left(y_{n}\right)_{n \in \mathbb{N}}\right)=\sum \limits_{n=1}^{\infty} 2^{-n} \frac{\left|x_{n}-y_{n}\right|}{1+\left|x_{n}-y_{n}\right|}
$$
for all sequences $ \left(x_{n}\right)_{n \in \mathbb{N}},\left(y_{n}\right)_{n \in \mathbb{N}} \in X $.
Show that $(X,d)$ is complete.
Idea of the proof:
To show that (X,d) is complete, you must show that every Cauchy sequence $(x_n)_{n∈\mathbb{N}}$ from X is a convergent sequence. This means that for each Cauchy sequence $(x_n)_{n∈\mathbb{N}}$, you must find a value x∈X such that x is a limit of the sequence $(x_n)_{n∈\mathbb{N}}$.
Attempt: Let $(x_n)_{n∈\mathbb{N}}$ of X be Cauchy. After presupposition there is an $N∈\mathbb{N}$ for each $\epsilon<0$, so that is valid for all $n,m \ge N$: $d(x_{m},x_n)<{\varepsilon\over2^k(\varepsilon+1)}\\$ $$\begin{align}
&\implies\sum_{j=1}^\infty{|x_{m,j}-x_{n,j}|\over2^{j}[1+|x_{m,j}-x_{n,j}|]}<{\varepsilon\over2^k(\varepsilon+1)}\\
&\implies {|x_{m,k}-x_{n,k}|\over2^{k}[1+|x_{m,k}-x_{n,k}|]}<{\varepsilon\over2^k(\varepsilon+1)}\\
&\implies |x_{m,k}-x_{n,k}|<\varepsilon 
\end{align}
$$
Here I get stuck all the time, because I can't think of anything useful here.
I also already got the hint to use the following:
a sequence $(x_k)_{k∈\mathbb{N}}=((x^{(k)}_n)_{n∈\mathbb{N}})_{k∈\mathbb{N}}$ in X converges to $(x_n)_{n∈\mathbb{N}}$ if and only $x^k_n→x_n$ for k→∞ and every $n∈\mathbb{N}$.
But I don't know how to take advantage of it.
Does anyone have any other suggestions?
 A: Let $\{y_{n}\}_{n\in\Bbb{N}}=\{x_{n,1},x_{n,2},...\}_{n\in\Bbb{N}}$ be a Cauchy sequence in $X$.
Then for $\epsilon>0$ and $r\in\Bbb{N}$ we chose $\epsilon_{1}<\frac{\epsilon}{2^{r}(1+\epsilon)}$. Then there exists $n_{1}$ such that for all $m,n\geq n_{1}$ we have
$$d(y_{n},y_{m})=\sum_{k=1}^{\infty}\frac{1}{2^{k}}\frac{|x_{m,k}-x_{n,k}|}{1+|x_{m,k}-x_{n,k}|}<\epsilon_{1}$$ .
So for each $r$ we have
$\frac{1}{2^{r}} \frac{|x_{m,r}-x_{n,r}|}{1+|x_{m,r}-x_{n,r}|}<\epsilon_{1}\implies |x_{n,r}-x_{m,r}|<\frac{\epsilon_{1} 2^{r}}{1-2^{r}\epsilon_{1}}<\epsilon$ .
Hence the sequence $\{x_{n,r}\}_{n\in\Bbb{N}}$ is a Cauchy sequence in $\Bbb{C}$ for each $r$ and hence is convergent due to completeness of $\Bbb{C}$. Let $t_{r}$ be the limit for each $\{x_{n,r}\}$ . So we claim that $\{y_{n}\}\to\{t_{1},t_{2},...\}=t$ (say) .
We let again $\varepsilon>0$ be arbitrary .
Now $$d(y_{n},t)=\sum_{r=1}^{\infty}\frac{1}{2^{r}}\frac{|t_{r}-x_{n,r}|}{1+|t_{r}-x_{n,r}|}$$
We choose $m\in\Bbb{N}$ large such that $\sum_{r=m+1}^{\infty}\frac{1}{2^{r}}<\frac{\varepsilon}{2}$ .
Then $$\sum_{r=m+1}^{\infty}\frac{1}{2^{r}}\frac{|t_{r}-x_{n,r}|}{1+|t_{r}-x_{n,r}|}<\frac{\varepsilon}{2}$$.
As $\displaystyle\lim_{n\to\infty}x_{n,r}=t_{r}$ we get a $K\in\Bbb{N}$ such that $n\geq K\implies |t_{r}-x_{n,r}|<\frac{\epsilon}{2m}$ .
Then we have $\sum_{r=1}^{m}\frac{1}{2^{r}}\frac{|t_{r}-x_{n,r}|}{1+|t_{r}-x_{n,r}|}<m\cdot\frac{\varepsilon}{2m}=\frac{\varepsilon}{2}$.
Then we have by adding the above two inequalities that :-
$$d(y_{n},t)=\sum_{r=1}^{\infty}\frac{|t_{r}-x_{n,r}|}{1+|t_{r}-x_{n,r}|}<\varepsilon\,,\forall n\geq K$$ .
Thus $\{y_{n}\}$ converges to $t\in X$ .
