What is the proof, that $\sum_{n=0}^{\infty} \frac{1}{2^n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$ is a metric? What is the proof, that $d(x,y) = \sum_{n=0}^{\infty} \frac{1}{2^n}\frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}$ is a metric? Where $x=(x_1,x_2,\cdots)$ , $y=(y_1,y_2,\cdots)$ and $d_n$ is a metric for $X_n$.
How does one prove the triangle inequality?
https://en.wikipedia.org/wiki/Metric_space#Product_metric_spaces

Since $d_i$ is metric, inequality $$d_i(x_i,z_i)\leq d_i(x_i,y_i)+d_i(y_i,z_i)$$ holds in i-th metric space from the product space. 
Also, metrics $d(x,y)$ and  $\frac{d(x,y)}{1+d(x,y)}$ are equivalent. 
(If $d$ is a metric, then $d/(1+d)$ is also a metric). 
Therefore, we have $$\frac{d_i(x_i,z_i)}{1+d_i(x_i,z_i)}\leq\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}+\frac{d_i(y_i,z_i)}{1+d_i(y_i,z_i)},$$
for every index $i$ and
$$\frac{1}{2^i}\frac{d_i(x_i,z_i)}{1+d_i(x_i,z_i)}\leq\frac{1}{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}+\frac{1}{2^i}\frac{d_i(y_i,z_i)}{1+d_i(y_i,z_i)},$$
for every $i$. By summing, we get $$d(x,z)\leq d(x,y)+d(y,z).$$
