Let $S$ be the set of all real sequences $x=\{x_n\}$, $d: S\times S \rightarrow \mathbb R$ be defined by:
$$d(x,y)=\sum_{n=1}^{\infty} \frac{|x_n-y_n|}{2^{n}[1+|x_n-y_n|]}.$$
Show that $(S,d)$ is a complete metric space.
I want to say we can use the Cantor Intersection property or an isometry here to prove completeness. However, I am having trouble getting started. Any suggestions?
Hint: Pick a Cauchy sequence $\{x_n\}$ in $S$. Pick $\varepsilon>0,k\in\mathbb{N}\setminus\{0\}$.
$$\begin{align} \exists N:m,n\ge N&\implies d(x_{m},x_n)<{\varepsilon\over2^k(\varepsilon+1)}\\ &\implies\sum_{j=1}^\infty{|x_{m,j}-x_{n,j}|\over2^{j}[1+|x_{m,j}-x_{n,j}|]}<{\varepsilon\over2^k(\varepsilon+1)}\\ &\implies {|x_{m,k}-x_{n,k}|\over2^{k}[1+|x_{m,k}-x_{n,k}|]}<{\varepsilon\over2^k(\varepsilon+1)}\\ &\implies |x_{m,k}-x_{n,k}|<\varepsilon \end{align} $$ Therefore $y_n=x_{n,k}$ is a Cauchy sequence in $\mathbb{R}$ converging to the limit $x^*_{k}$. Prove that $x^*$ is the limit of $\{x_n\}$ and you're done.
Note: If $x_n\in S$, $x_n$ is a sequence and by $x_{n,k}$ we mean its $k$-th term.
