Proof metric space with distance function 
Thats the first time i have to do such an proof but don't know how, never seen or done this before. Especially (iii).
Let $X$ be the Set of all complex sequences.
$$
d((a_n),(b_n)) := \sum^\infty_{i=0} \frac{1}{2^{i+1}}\frac{\left | a_i-b_i \right |}{1+\left | a_i-b_i \right|}, ((a_n),(b_n) \in X)
$$
Proof that $(X,d)$ is an metric space.
Definition of metric space says:


*

*$d((a_n),(b_n)) \geq 0 $ and $d((a_n),(b_n))=0 \Leftrightarrow (a_n)=(b_n)$

*$d((a_n),(b_n)) = d((b_n),(a_n))$

*$d((a_n),(c_n) \leq d((a_n),(b_n))+d((b_n),(c_n)) \ Triangle \ inequality $


Can someone help me please
Thanks 
Landau.

Add on Daniel Fisher's comment
When $f(t)= \frac{t}{1+t}>0$ is concave for $t>0$, then $$ f(|a-b|)
+f(|b-c|)\geq f(|a-b| + |b-c|)\geq f(|a-c|)$$ where $a,\ b,\ c$ are
points in a metric space $(X,d=|\ |)$. Hence $(X,\frac{d}{1+d})$ is
a metric space.
