Why do we use this quotient $\frac{d_j(f,g)}{1+d_j(f,g)}$ in the metric of $C(\mathbb{R})$? I am trying to understand the metric we defined in class for $C(\mathbb{R})$ i.e. on the set of continuous functions $f: \mathbb{R} \to \mathbb{R}$. A little background: We developed some strong theory on the convergence of random variables for continuous functions defined on compacta and we define this metric in an effort to extend these results. We defined
$$d(f,g):= \sum_{j = 1}^{\infty}2^{-j} \frac{d_j(f,g)}{1+d_j(f,g)}$$
where
$$ \forall j \in \mathbb{N}: d_j(f,g):= \underset{-j \leq t \leq j}{\operatorname{sup}} |f(t)-g(t)| $$
Now I know so far that the infinite sum $\sum_{j = 1}^{\infty}2^{-j}$ converges to $1$ and thus what we do seems to me like giving each interval $J := [-j,j]$ a weight such that the total weights sum to $1$ and then evaluating the familiar supremum-norm on these weighted compacta. In that interpretation what happens around $0$ is most important to us since the interval $[-1,1]$ carries the highest weight. So far so good.
What I don't understand is the following.
Why does the sum contain (or what is the meaning of) this quotient:
(*) $$\frac{d_j(f,g)}{1+d_j(f,g)}$$
Why do we use this and not some other function of the $d_j(f,g)$ that guarantees convergence?
Edit:
I edited out the suggestion of
$$d(f,g):= \sum_{j = 1}^{\infty}2^{-j} d_j(f,g)$$
since my point was not that this would be a valid replacement. I was aware that it probably would not converge. Thanks to geometricK for the concrete example though. My question is more about the benefits of the function (*) over others that do give convergence.
 A: First notice that on any metric space $(S,\rho)$, the function
$$
d(x,y)=\frac{\rho(x,y)}{1+\rho(x,y)},\qquad x,y\in S
$$
is also a metric on $S$, and that the topologies defined by $(S,d)$ and $(S,\rho)$ are the same.
This is not the only way to obtained equivalent bounded metrics on $S$, A general procedure is to consider any bounded monotone nondecreasing continuous function $f:[0,\infty)\rightarrow[0,\infty)$  such that $f(t)=0$ iff $t=0$ and $f(t+s)\leq f(s)+f(t)$. Then
$$
d_f(x,y)=f(\rho(x,y))
$$
would give a bounded metric on $S$ that is equivalent (open sets are the same) to $\rho$.
Going back to your question, It is really not that important what (equivalent to the sup norm) metric $d_j$ on $C([-j,j])$ you take, however to keep the series
$$
\rho(f,g)=\sum^\infty_{j=1} 2^{-j}d_j(f,g)
$$
convergent (after all metrics take finite values), the terms $2^{-j}d_j(f,g)$ need to be controlled. In particular, one can always take another metric $d'_j$ in $C([-j,j])$ that is bounded, for example
$$
d'_j(f, g)=\min(1,d_j(f, g))
$$
is one such metric, and
$$
d''_j(f, g)=\frac{d_j(f, g)}{1+d(f, g)}
$$
is another such metric.
It is not difficult to show that if $d'_j$ and $d''_j$ are bounded (lets say all bounded by $1$ to make things simpler) equivalent metrics on $C([-j,j])$, then
\begin{align}
\rho'(f, g)&=\sum_j 2^{-j}d'_j(f, g)\\
\rho''(f, g)&=\sum_j 2^{-j}d''_j(f, g)
\end{align}
defined equivalent metrics on the space of continuous function on $\mathbb{R}$. the topology they generate, as you may be well aware, is that of uniform convergence in compact sets.
When studying regularity of functions (Li[schitz regularity for instance) the metric $d'_j(f, g) = \min(1,\|f-g\|_{u([-j,j]})$ me be more convenient as bounds become simpler.
A: My favourite metric in this setting is $$d(f,g)=\sup\{ d_j(f,g) \wedge 1/j: j\in\mathbb N\}$$ where $a\wedge b$ is the minimum of two real numbers. It is immediate that $d(f,g)<1/k$ if and only if $d_k(f,g)< 1/k$, and this is exactly what you want, e.g., to show that a sequence $f_n$ converges to $f$ if and only if $d_j(f_n,f)\to 0$ for all $j\in\mathbb N$.
A: $$d(f,g):= \sum_{j = 1}^{\infty}2^{-j} d_j(f,g)$$ is not convergent. We can take $\sum_j 2^{-j}\min \{{1, d_j(f,g)}\}$ and this is also frequently used.
