Convergence with respect to $d_1(x,y)=\sum^\infty_{i=1}\frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|}$ is equivalent to pointwise convergence? This was part of an exercise given to the students in my class last week which I wasn't sure how to do. The question starts by studying the metric $d_1$ defined on $\mathbb{R}^\infty=\{x=(x_i)\mid i\in\mathbb{N}\}$ given by
$$d_1(x,y)=\sum^\infty_{i=1}\frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|},\qquad x,y\in\mathbb{R}^\infty.$$
The students are asked to show that this is indeed a metric and compare it with another metric defined on the same space, none of which is difficult. Finally the question asks to show that if $(x^{(n)})_{n\in\mathbb{N}}\subseteq\mathbb{R}^\infty$ is a sequence, then its convergence to a point $x\in\mathbb{R}^\infty$ with respect to $d_1$ is equivalent to its pointwise convergence to the same point. 
As I understand it, pointwise convergence should mean that for each $i\in\mathbb{N}$, and each real $\epsilon>0$ there exists $N=N(\epsilon,i)\in\mathbb{N}$ such that $|x^{(n)}_i-x_i|<\epsilon$ whenever $n>N$.

Now one direction is clear to me, but proving that pointwise convergence implies $d_1$-convergence is not. Can anyone help me out with this direction?

What I need to show is that given any real $\delta>0$ there exists $M=M(\delta)\in\mathbb{N}$ such that
$$d_1(x^{(n)},x)=\sum^\infty_{i=1}\frac{1}{2^i}\frac{|x^{(n)}_i-x_i|}{1+|x^{(n)}_i-x_i|}<\delta$$
whenever $n>M$, using the assumption of pointwise convergence which I believe to have interpreted correctly just above. The problem seems to be that whilst a suitable $M=M(\epsilon,i)$ exists for each $i$, it is by no means clear to me that this collection of $M$s is bounded above.
Any help would be greatly appreciated.
 A: So this is the answer I ended up giving my students. I'll use the notation introduced in the question itself.
We assume that $(x^{(n)})_{n\in\mathbb{n}}$ is a sequence converging pointwise to a point $x\in\mathbb{R}^\infty$. We also assume given a real $\delta>0$. Now, since for any $n$, the value $d_1(x^{(n)},x)$ is bounded above by the absolutely convergent series $\sum^\infty_{n=1}\frac{1}{2^n}=1$ we can find for the given sequence $(x^{(n)})_{n\in\mathbb{n}}$ an integer $K$ such that the inequality
$$\sup_{n\in\mathbb{N}}\left\{\sum^\infty_{i=K+1}\frac{1}{2^i}\frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}\right\}<\frac{\delta}{2}$$
is satisfied. Notably $K$ is independent of $n$.
Now for each $i=1,\dots,K$ we use the assumption of pointwise convergence to get integers $N_i$, $i=1,\dots,K$, such that 
$$|x_i^{(n)}-x_i|<\frac{\delta}{\frac{2K}{2^i}-\delta}.$$
Note that if necessary we are free to replace $K$ be any larger (but finite) integer so as to guarantee that $\frac{2K}{2^i}-\delta\neq 0$ for each $i$.
After a little rearranging we get from this that
$$\sum^K_{i=1}\frac{1}{2^i}\frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}<\sum^K_{i=1}\frac{\delta}{2K}=K\cdot \frac{\delta}{2K}=\frac{\delta}{2}$$
whenever $n>N:=\max_{i}n_i$, and we use this to get
$$d_1(x^{(n)},x)=\sum^\infty_{i=1}\frac{1}{2^i}\frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}=\sum^K_{i=1}\frac{1}{2^i}\frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}+\sum^\infty_{i=K+1}\frac{1}{2^i}\frac{|x_i^{(n)}-x_i|}{1+|x_i^{(n)}-x_i|}<\frac{\delta}{2}+\frac{\delta}{2}=\delta$$
which is enough to conclude that the sequence $(x^{(n)})_{n\in\mathbb{N}}$ converges to $x$ with respect to the metric $d_1$. This is exactly what we needed to show, so we are done.
