Convergent sequences in metric space $(\mathbb{R}^n,d)$ Given a metric $d : \mathbb{R}^n \times \mathbb{R}^n \mapsto \mathbb{R}$ defined as 
$$ d(x,y) = \begin{cases} 
      \lVert x \rVert + \lVert y \rVert & x\neq y \\
      0 & x = y
   \end{cases}
 $$
How can one describe the convergent sequences in metric space $(\mathbb{R}^n,d)$?
My work:
This metric will return a longer distance between any two points $x, y \in \mathbb{R}^n$ So by the triangle inequality if a sequence converges in $(\mathbb{R}^n,d)$ it must also converge in $\mathbb{R}^n$ with the standard metric $\lVert x - y \rVert$ but the reverse is not necessarily true, the standard metric on $\mathbb{R}^n$ is finer.
What about Cauchy sequences? It's not very clear to me whether $(\mathbb{R}^n,d)$ is a complete metric space.
 A: Note that if for any $k\geq 0$, 
$$
d(x_{n},x_{n+k})\to 0\iff ||x_n||+||x_{n+k}||\to 0\implies ||x_n||\to 0
$$
So, for any Cauchy sequence, we have 
$$
d(x_n,0)=||x_{n}||\to 0
$$
and $x_n\to 0$ in this metric.
Since convergent sequences are always Cauchy, this goes a long way towards describing the convergent sequences, even in terms of the usual Euclidean metric.
A: So assume that $(x_n)$ converges to $x$. It follows that $d(x_n, x)$ is arbitrarly small. Now assume that $(x_n)$ is not eventually constant, i.e. $x_n\neq x$ for infinitely many $n$. Then $d(x_n,x)=\lVert x_n\rVert+\lVert x\rVert\geq \lVert x\rVert$. If $x\neq 0$ then $d(x_n,x)\geq \lVert x\rVert>0$ and thus $x_n$ cannot be convergent to $x$.
It follows that if $x\neq 0$ then $(x_n)\to x$ if and only if $(x_n)$ is eventually constant.
On the other hand following similar reasoning we conclude that $(x_n)\to 0$ in $d$ if and only if $(x_n)\to 0$ in $\lVert\cdot\rVert$.
Similarly if $(x_n)$ is Cauchy then $d(x_n, x_m)=\lVert x_n\rVert +\lVert x_m\rVert$ assuming $(x_n)$ is not eventually constant (which is the trivial case). So $d(x_n,x_m)$ is arbitrarly small if and only if $\lVert x_n\rVert$ is arbitrarly small. Meaning that $(x_n)$ is Cauchy if and only if it converges to $0$ (I omit the case when $(x_n)$ is eventually constant). In particular $(\mathbb{R}^n,d)$ is complete.
