How to prove that this series is a metric: $d(x,y):=\sum_{i=0}^\infty \frac{|x_i -y_i|}{2^i (1+|x_i-y_i|)}$ I have to prove that this is a metric: $$d(x,y):=\sum_{i=0}^\infty \frac{|x_i -y_i|}{2^i (1+|x_i-y_i|)}$$ 
The only thing that I can't prove is the triangle inequality, it's really hard. I know that $|x_i -y_i|\le |x_i -z_i|+|z_i -y_i|$, and I want to get: $$\frac{1}{2^i (1+|x_i-y_i|)} \le \frac{1}{2^i (1+|x_i-z_i|)}+\frac{1}{2^i (1+|z_i-y_i|)}$$ (I think...) The closest thing I've got is: $2^i (1+|x_i-y_i|)\le 2^i (1+|x_i-z_i|)+2^i (1+|z_i-y_i|)$ but... that's not quite what i want. I don't know how to get this...
 A: Do you know that
$$d(a,b) = \frac{|a-b|}{1 + |a-b|}$$
is a metric on $\mathbb{C}$? First prove that this is a metric. The triangle inequality which you prove here will immediately imply what you want.
To prove triangle inequality in this case, you may use that the function $f(t) =  \frac{t}{1+t}$ is monotonic.

EDIT: As requested in the comments, I will complete the proof as follows:
$$ |a-b| + |b-c| \geq |a-c| \Rightarrow \frac{|a-b|+ |b-c|}{1 + |a-b| + |b-c|} \geq \frac{|a-c|}{1 + |a-c|} = d(a,c)$$
(using the fact that the function $f$ is increasing)
Now 
$$d(a,b) + d(b,c) = \frac{|a-b|}{1 + |a-b|} + \frac{|b-c|}{1+|b-c|} \geq \frac{|a-b|+ |b-c|}{1 + |a-b| + |b-c|}$$
Combining above two, we get the triangle inequality.

Generalities: In fact there is nothing special about the metric on $|\cdot|$ on $\mathbb{C}$. If $(X,d)$ is any metric space, then $d_{1}$ is an equivalent metric on $X$ where
$$d_{1}(x,y) = \frac{d(x,y)}{1 + d(x,y)}$$
Note that $d_{1}(x,y) < 1 \;\forall x,y \in X$.
And then $d_{2}$ is a metric on $X^{\mathbb{N}}$ (the space of all $X-$ valued sequences) where $d_{2}$ is given as
$$ d_{2}\left((x_{i}),(y_{i})\right) = \sum\limits_{i=1}^{\infty}\frac{1}{2^{i}}d_{1}(x_{i},y_{i}) $$
