Show that the space $(X,d)$ is not complete and prove its completion Let $X = \left\{(x_n):\sum\limits_{n=1}^\infty n|x_n|<\infty\right\}$, $d(x,y) = \sup|x_n - y_n|$. Show that the space $(X,d)$ is not complete. Prove that the space $c_0$ is its completion.
For the solution, I know that a metric space $(X,d)$ is complete if every Cauchy sequence $(x_n)$ in $X$ converges to a point in $X$ so for the first part I need to find a Cauchy sequence that does not converges to a point in $X$.  
For $(x_n) = (1/n^3)_{n=1}^\infty$ and $(y_n) = (1/n^4)_{n=1}^\infty,$  $d(x,y)$ not converges. Thus $X$ is not complete. Is my counter example right?
For the second part, proving $c_o$ is its completion I am having trouble about what kind of approach I should follow.
 A: For the first Q: Note that for  if $x_n=0$ for all but finitely many $n$ then $(x_n)_n\in X.$ 
Let $x[j]=(x_{j,n})_n$ where $x_{j,n}=1/n$ for $n\leq j$ and $x_{j,n}=0$ for $n>j.$
Then $(x[j])_j$ is a Cauchy sequence in $X$. If this sequence converged  to $y=(y_n)_n\in X$, then it would be necessary that $y_n=1/n$ for every $n.$ But $(1/n)_n\not\in X.$   
A: This is an old question, but I noticed that its second part, showing that $c_0$ is the completion, was never addressed.
$c_0$ (which denotes the space of sequences converging to $0$, something I wasn't aware of) is equipped with the $\sup$ norm. We show $X$ is a dense subspace of $c_0$. If $x\in c_0$ and $\epsilon >0$, let $K$ be so large that $\vert x(k)\vert <\epsilon/2$ for $k\geq K$ (the notation $x(k)$ is used in place of $x_k$ for more clarity later, when, for example, $x_m(k)$ will mean the $k$-th element of the $m$-th sequence). The sequence $y$ with term
$$
y(k)=\begin{cases}
x(k), & k<K \newline
0, & k\geq K
\end{cases}
$$
clearly satisfies $d(x,y)\leq\epsilon/2<\epsilon$ and is in $X$.
It is well-known that $c_0$ is complete. How to show that a Cauchy sequence of sequences converges is relevant here; below is my own approach to proving this, which is similar to the one in the linked post, except that the steps are in a different order. (I also availed myself of the notation in that post.)
Take a Cauchy sequence $(x_m)_{m\in\mathbb{N}}$ (each $x_m$ being an element of $c_0$) and define a new sequence $y$ by
$$
y(k) = \lim_{m\rightarrow\infty}x_m(k)
$$
That the above limit actually exists is guaranteed by the fact that for fixed $k$, the number sequence $(x_m(k))_{m\in\mathbb{N}}$ is Cauchy, as $\vert x_m(k)-x_n(k)\vert\leq d(x_m,x_n)$.
Now, we show at once both that $y$ is bounded and that $(x_m)_{m\in\mathbb{N}}$ converges to $y$ in norm in the space of bounded sequences $\mathcal{l}^\infty$, of which $c_0$ is a subspace. Pick $\epsilon >0$ and $N$ so large that $d(x_m,x_n)< \epsilon/2$ for all $m,n\geq N$. Then, for $n\geq N$:
$$
\vert y(k)-x_n(k) \vert = \lim_{m\rightarrow\infty} \vert x_m(k)-x_n(k)\vert \leq \epsilon/2
$$
This is true for all $k$, so $y$ is bounded (remembering that $x_n$ is, of course, itself bounded) and $d(y,x_n)<\epsilon$.
All that's left to prove is $y\in c_0$. Let $\epsilon >0$ and pick $n$ so that $d(y,x_n)<\epsilon/2$. Then, let $K$ be such that if $k\geq K$, then $x_n(k)<\epsilon/2$. For $k\geq K$, calculate:
$$
\vert y(k)\vert\leq \vert y(k)-x_n(k) \vert + \vert x_n(k) \vert\leq d(y,x_n) + \vert x_n(k)\vert<\epsilon
$$
Hence, $\lim_{k\rightarrow\infty}y(k)=0$.
