A Cauchy sequence in a metric space $(X,d)$ is not necessarly convergent. for example, consider the metric space $(\mathbf{Q},d)$ where d is the usual absolute value. And consider the sequence $x_{n}$, $x_{1}=1$, $x_{n+1}=\frac{1}{2}(x_{n}+\frac{2}{x_{n}})$. One can check that $x_{n}$ is a Cauchy sequence in $\mathbf{Q}$ but has no limit in $\mathbf{Q}$.
There is a notational issue here. You need to show that a Cauchy sequence of elements of $c_0$ converges to an element of $c_0$. It might help to refer to an element of $c_0$ as $x$, and then address a specific element of the sequence as $x(k)$.
So suppose you have $x_n \in c_0$ such that $x_n$ is Cauchy with the distance
given above.
For each $k$, you have $|x_n(k)-x_m(k)| \le d(x_n,x_m)$, so there is some $x(k)$
such that $x_n(k) \to x(k)$. This is the candidate sequence.
Now you must show that $x \in c_0$ and $d(x,x_n) \to 0$.
Let $\epsilon>0$, choose $N$ such that if $m,n \ge N$ then
$d(x_n,x_m) < { 1\over 3} \epsilon$. Now choose $K$ such that
if $k \ge K$, then $|x_N(k) | < { 1\over 3} \epsilon$. Then, for $k \ge K$
\begin{eqnarray}
|x(k)| &\le& |x(k)-x_m(k)| + |x_m(k)-x_N(k)| + |x_N(k)| \\
&<& |x(k)-x_m(k)| + {2 \over 3}\epsilon
\end{eqnarray}
Now choose $m \ge N$ (which will, in general, depend on $k$) such that
$|x(k)-x_m(k)| < {1\over 3} \epsilon$, and we see that
$|x(k)| <\epsilon$. Hence $x (k) \to 0$ and so $x \in c_0$.
Showing that $x_n \to x$ is similar: Let $\epsilon>0$ and choose
$N$ such that if $m,n \ge N$ then
$d(x_n,x_m) < { 1\over 2} \epsilon$. Then, for $m,n \ge N$ we have
\begin{eqnarray}
|x(k)-x_n(k)| &\le& |x(k)-x_m(k)| + |x_m(k)-x_n(k)| \\
&<& |x(k)-x_m(k)| + {1 \over 2} \epsilon
\end{eqnarray}
Now choose $m \ge N$ (which will, in general, depend on $k$) such that
$|x(k)-x_m(k)| < {1 \over 2} \epsilon$, then we see that
$|x(k)-x_n(k)| < \epsilon$ for all $n \ge N$, and so $d(x,x_n) \le \epsilon$.
Best Answer
Not always. A space in which every Cauchy sequence is a convergent sequence is called a complete space, but not every imaginable space is complete.
The set of real numbers is complete, which means that a Cauchy sequence of real numbers will have a real limit. Other sets, like interval $(0,1)$ or the set of rational numbers $\mathbb Q$ are not complete, and the Cauchy sequence of numbers from them do not have to have a limit in these sets.
However, when you have a non-complete set, you can always construct its completion, by adding new elements to this set in such a way that the result will be a complete set. So you can say that every Cauchy sequence of elements of some space has a limit in the completion of this set, but not necessarily in the set itself. For example, the completion of interval $(0,1)$ is interval $[0,1]$, and the completion of $\mathbb Q$ is $\mathbb R$.