Showing a sequence in $\mathbb{N}$ is Cauchy For a metric $d(x,y) =\left| \frac{x}{1+|x|} - \frac{y}{1+|y|} \right|$
Where $d: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$. 
Prove that $\{ x_n \}$ in $\mathbb{R}$ given by $x_n = n$ is a Cauchy sequence with respect to $d$.
I am not sure of what would be a smart way to prove / disprove this. Here, since $n \in \mathbb{N}$, then we know that as the sequence continues the points $x_n$ and $x_m$ won't get closer than $1$, thus it is not Cauchy?
 A: Consider $d(m,n)$:
$$
d(m,n)=\left|\frac{m}{1+m}-\frac{n}{1+n}\right|=
\left|\frac{m-n}{(1+m)(1+n)}\right|
$$
By the triangle inequality,
$$
d(m,n)\le \frac{m}{1+m}\frac{1}{1+n}+\frac{n}{1+n}\frac{1}{1+m}
\le\frac{1}{1+n}+\frac{1}{1+m}
$$
If $\varepsilon>0$, take an integer $N$ such that $N>2/\varepsilon$; if $m,n>N$ then
$$
d(m,n)\le\frac{1}{1+n}+\frac{1}{1+m}<
\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
Slow motion
\begin{align}
d(m,n)
&=\left|\frac{m}{1+m}-\frac{n}{1+n}\right|
\\[6px]
&=\left|\frac{m-n}{(1+m)(1+n)}\right|
\\[6px]
&=\left|\frac{m}{(1+m)(1+n)}+\frac{-n}{(1+m)(1+n)}\right|
\\[6px]
&\le\left|\frac{m}{(1+m)(1+n)}\right|+\left|\frac{-n}{(1+m)(1+n)}\right|
&&\text{triangle inequality}
\\[6px]
&=\frac{m}{1+m}\frac{1}{1+n}+\frac{n}{1+n}\frac{1}{1+m}
\\[6px]
&\le\frac{1}{1+n}+\frac{1}{1+m}
&&\text{because }\frac{k}{1+k}\le1
\end{align}
A: You have that $\frac{n}{1+n}=1-\frac1{1+n}<1$. Hence, for any $m>n$, we have: $$d(n,m)=|\frac{n}{1+n}-\frac{m}{1+m}|=|\frac1{1+m}-\frac1{1+n}|=\frac1{1+n}-\frac1{1+m}<\frac1{1+n}$$
Hence, for any $\epsilon>0$, choose $N$ such that $\frac1{N+1}<\epsilon$, then you have that $|x_n-x_m|<\epsilon$ for all $m,n>N$. Therefore, the sequence is Cauchy
A: You need to show that, for any small $\epsilon > 0$ given, there is some $N = N(\epsilon)$ such that for any $n, m > N$, you have $d(n,m) < \epsilon $ where $d$ is the distance that you have defined. This is simply the definition of the Cauchy sequence. 
You need to do an explicit calculation with the metric to show that this is indeed the case. 
