Prove that $d$ is a metric on $X$. Let $X$ be a set of real sequences and $d(x,y)=\sum_{k=1}^{\infty}\dfrac{1}{2^k}\cdot \dfrac{\vert a_k-b_k \vert}{1+\vert a_k-b_k\vert}$ with $x=(a_k)_{k \in \mathbb{N}}$, $y=(b_k)_{k \in \mathbb{N}} \in X$.
Prove that $d$ is a metric on $X$.
I can easily checking $d(x,y) \ge 0$, $d(x,y)=0 \Leftrightarrow x=y$ and $d(x,y)=d(y,x)$.
I stuck at checking the convergence of this sequence and the condition $d(x,y)\le d(x,z)+d(z,y), \forall x,y,z \in X$. 
 A: Consider a function
$$
f(x) = \frac{x}{1+x}
$$
for nonnegative $x$.
We need to prove that it is subadditive
$$
f(a+b) \le f(a) + f(b).
$$
It is a concave function (check it, using the second derivative).
Note also that
$$
f(tx) = f(tx + (1-t)0) \ge tf(x) + (1-t)f(0) = tf(x).
$$
That's why we have
$$
f(a+b) = \frac{a}{a+b} f(a+b) + \frac{b}{a+b}f(a+b) \le
$$
$$
\le f\big(\frac{a}{a+b}\cdot (a+b)\big) + f\big(\frac{b}{a+b}\cdot (a+b)\big) =
$$
$$
= f(a) + f(b).
$$
A: 
Proposition:
  If $(X , d)$ is a metric space, then $d^*(x , y) = \frac{d(x , y)}{1 + d(x , y)}$ is a metric on $X$.

Assume and try to prove last result later. Now we just use it to solve your problem. Let assume that $d$ a metric the set $X$ of your context, and let $x = {(x_n)}_{n \in \mathbb{N}}$, $y = {(y_n)}_{n \in \mathbb{N}}$ and $z = {(z_n)}_{n \in \mathbb{N}}$ be three ''points'' in $X$, and let
$$
a_n = \frac{d(x_n , z_n)}{2^n} \qquad \mbox{and} \qquad b_n = \frac{d(x_n , y_n) + d(y_n , z_n)}{2^n}
$$
be two sequences on $[0 , \infty)$. Clearly, $a_n \leq b_n$ for all $n \in \mathbb{N}$, therefore
$$
\sum_{n = 1}^{\infty} \frac{d(x_n , z_n)}{2^n} \leq \sum_{n = 1}^{\infty} \frac{d(x_n , y_n) + d(y_n , z_n)}{2^n}\tag{1}\mbox{.}
$$
You are done if you know how to use last proposition (this last $d$ is the $d^*$ in the proposition. Then the both series in $(1)$ converge, as $d$ takes a value less than $1$ when we apply it two points in $X$), and if you know that $d_1$ is a metric (the metric on $\mathbb{R}$ induced by the norm $|\cdot|$).
