Property of a metric in the space of all the sequences of real numbers A few weeks ago I had this problem, the adjoint-teacher solved it on class, and I thought I understood, but now I'm rechecking and there are a few things that aren't clear for me.
So we defined this metric in the space of all the sequences of real numbers, $S$:
$$d(x,y):=\sum_{i=0}^\infty \frac{|x_i -y_i|}{2^i (1+|x_i-y_i|)}$$

If $\overline x^k=(x_i^k)$, $\overline x=(x_i)\in S$, prove that $$\lim_{k\to \infty}d(\overline x_i^k,\overline x)=0 \iff \lim_{k\to \infty}x_i^k=x_i \;\;\;\forall\;i\in\Bbb N$$

$(\Rightarrow)$ We have that if $k\rightarrow \infty$ then $d(x_i^k,x_i)\rightarrow 0$.
Is clear that: $$0\le \frac{|x_i^k -x_i|}{2^i (1+|x_i^k-x_i|)}\le d(x_i^k,x_i)$$
We know that $f(x)=\frac {x}{1+x}$ is an increasing function in $[0,\infty)$, (we did proved this) then it has a continuous inverse, namely $f^{-1}(x)=\frac{1}{1-x}$, then
$f^{-1}(f(|x_i^k-x_i|))\rightarrow f^{-1}(f(0))=0$ 
$\color{red}{\mathbf {(i)}}$ This last part is pretty clear, however didn't he forgot that what we have in the inequality is not $f(|x_i^k-x_i|)$ but $f(|x_i^k-x_i|)*\frac{1}{2^i}$? Did he overlooked that because ${1}/{2^i}$ is a constant respect to k?
($\Leftarrow$)Now we have that $x_i^k\rightarrow x_i$ if $k\rightarrow\infty\;\;\forall\;i\in\Bbb N$. Is clear that $f(|x_i^k-x_i|)<1$, using the same $f$ as before, then 
$$\sum_{i=0}^\infty \frac{|x_i^k -x_i|}{2^i (1+|x_i^k-x_i|)}\le \sum_{i=0}^\infty \frac{1}{2^i}=1$$
Let $\varepsilon>0$ then $\exists\; N\in \Bbb N$ such that $$\sum_{i=N}^\infty \frac{|x_i^k -x_i|}{2^i (1+|x_i^k-x_i|)}\le \sum_{i=N}^\infty \frac{1}{2^i}<\varepsilon$$
$\color{red}{\mathbf {(ii)}}$ I don't see why this inequality holds, or better, where did he got it? Is it because the last sum is absolute convergent (and I belive the first one is too abs. conv.)?
If $M<N$ the $\exists\; N_M$ such that $\forall\; r\in\Bbb N$ with $r>N_M$: $|x_M^r-x_M|<\varepsilon$
$\color{red}{\mathbf {(iii)}}$ What is he doing here? is he delimiting the rest of the elements? I don't understan what is the r doing.
Now we define $Ñ=\max \{N_1,\dots ,N_{N-1},N \}$, 
so if $P>Ñ$ then
$$\begin{align} \\
&\sum_{i=0}^\infty \frac{|x_i^k -x_i|}{2^i (1+|x_i^k-x_i|)}\\ 
& =\sum_{i=0}^{N-1} \frac{|x_i^k -x_i|}{2^i (1+|x_i^k-x_i|)}+\sum_{i=N}^\infty \frac{|x_i^k -x_i|}{2^i (1+|x_i^k-x_i|)} \\ 
& \le (N-1)(\frac{\varepsilon}{2^p(1+\varepsilon)})+\varepsilon \\
& \le (N-1)(\frac{\varepsilon}{2^Ñ(1+\varepsilon)})+\varepsilon=:\star
\end{align}$$
So if $\varepsilon\rightarrow 0$ then $\star\rightarrow 0$, and we're done.
$\color{red}{\mathbf {(iv)}}$ Now, I feel this last part has a trick that I can't see, so first they divided the sum in two parts, the way the first part is delimited I don't understand, plus why did we needed the $p>Ñ$ ?
 A: Re: (i). The presence of $2^i$ does not really matter, since it's independent of $k$. We see  that $d(x_i^k,x_i)\to 0$ implies $\dfrac{|x_i^k -x_i|}{(1+|x_i^k-x_i|)} \to 0$.  

(iii) What is he doing here? 

Continuing the proof that $d(x_i^k,x_i)\to 0$.  The idea of the proof is to split the sum in the definition of $d(x_i^k,x_i)$ into "large indices" (tail) and "small indices" (head). The contribution of the tail was handled already, by comparison with geometric series (that was (ii)). Now we deal with the head, which has finitely   many terms. 
I think the later parts of the proof are messed up, including the choice of $\tilde N$. This is what was supposed to happen: 
Choose $N$ so that $$\sum_{i=N}^\infty \frac{1}{2^i}<\frac{\varepsilon}{2} \tag1$$
This takes care of the tail. 
The head of the sum  in $d(x_i^k,x_i)$ contains fewer than $N$ terms. If we can show that each of them is less than $\varepsilon/(2N)$, we are done. To this end, for each $M=1,\dots,N-1$ pick $K_M$ such that $|x_M^k-x_M|<\varepsilon/(2N)$  whenever $k>K_M$. 
Now, if $k>\max(K_1,\dots,K_{N-1})$, then 
$$d(x_i^k,x_i) < \frac{\varepsilon}{2N}+\dots + \frac{\varepsilon}{2N} +\frac{\varepsilon}{2 } <\epsilon$$
which is exactly what the definition of $d(x_i^k,x_i)\to 0$ requires.
