# Pointwise convergence and convergence in metric

Let $s:=\mathcal{F}(\mathbb{N})=\{x = (x_j)_{j\in \mathbb{N}}|x+j\in \mathbb{C} \ \forall j\in \mathbb{N}\}$ be the space of all scalar sequences. And $$d(x,y):=\sum_{j=1}^\infty 2^{-j}\frac{|x_j-y_j|}{1+|x_j-y_j|},$$ provided it is a metric.

Now I want to prove $x\rightarrow y$ in $d(\cdot,\cdot)$ is equivalent to $x\rightarrow y$ componentwise, namely $x_j\rightarrow y_j$ for all $j\in\mathbb{N}$.

'$\Rightarrow$' is easy by formulating the contraposition.

How to prove "$\Leftarrow$"? Because pointwise convergence does not imply convergence in metric, I am stuck here.

Hint 1: Notice that $$x\mapsto \frac{x}{1+x}$$ is $0$ for $x=0$, it is positive for positive value of $x$ and moreover it is increasing on $[0,+\infty)$. Now use the fact that, for any $j\in\mathbb N$, $$2^{-j}\frac{|x_j-y_j|}{1+|x_j-y_j|}\leq d(x,y).$$
Hint 2: Notice that the sum $$\sum_{j=1}^{+\infty}2^{-j}$$ is convergent. Thereefore, for any $\varepsilon>0$ there is $j_0(\varepsilon)$ such that $$\sum_{j=j_0(\varepsilon)}^{+\infty}2^{-j}<\varepsilon.$$ Morever $$x\mapsto \frac{x}{1+x}<1,\;\forall x\in\mathbb R_{\geq 0}.$$ Now you are left with finitely may terms of the series to control.