We need to show that any Cauchy sequence in \(\ell^2\) is convergent to another \(\ell^2\) sequence. First, we propose a candidate for the limit of a Cauchy sequence of vectors in \(\ell^2\text{.}\) To do so, we’ll need to take advantage of the fact that vectors in \(\ell^2\) are themselves sequences.
Consider the array we can construct by arranging the vectors \(x^k\) as rows -
\begin{equation*}
\begin{array}{cccccc}
x^1 \amp = \amp (x^1_1, \amp x^1_2, \amp x^1_3, \amp \ldots)\\
x^2 \amp = \amp (x^2_1, \amp x^2_2, \amp x^2_3, \amp \ldots)\\
x^3 \amp = \amp (x^3_1, \amp x^3_2, \amp x^3_3, \amp \ldots)\\
\vdots \amp \vdots \amp \vdots \amp \vdots \amp \vdots \amp \\ \hline
(x^k) \amp \amp (x^k_1) \amp (x^k_2) \amp (x^k_3) \amp
\end{array}
\end{equation*}
We are going to argue that the column sequences in the array above are convergent. Consider the \(j\)-th column sequence \((x^k_j)\text{.}\) Choose \(\eps > 0.\) Because \(x^k\) is a Cauchy sequence of vectors, there exists some \(K\) so that \(k, l > K\) implies that \(\norm{x^k - x^l} \lt \eps\text{.}\) But since
\begin{equation*}
\norm{x^k - x^l} = \left(\sum \abs{x_i^k - x_i^l}^2\right)^{1/2} \lt \eps
\end{equation*}
and each term in the sum is positive, we get \(\abs{x^k_j - x^l_j} \lt \eps\) when \(k, l > K\text{.}\) This shows that \((x^k_j)\) is a Cauchy sequence in \(\C\text{,}\) which is a complete metric space, and so \((x^k_j) \to y_j\) for some limit \(y_j \in \C\) as \(j \to \infty\text{.}\) Let \(y = (y_j)\) be the sequence of column limits. This is our candidate limit in \(\ell^2\) for the sequence of vectors \(x^k\text{.}\)
Let us show that \(y \in \ell^2\text{.}\) To do so, we will show that \(x^k - y\) is in \(\ell^2\) for some \(k\) and use the vector space structure of \(\ell^2\text{.}\) Let \(\eps > 0\) be given. Since \(x^k\) is Cauchy, there exists \(K\) so that \(\norm{x^k - x^l} \lt \eps\) for all \(k, l > K\text{.}\) Noting again that the terms are positive, it is clear that for \(k, l > K\text{,}\)
\begin{equation*}
\sum_{i=1}^N \abs{x_i^k - x_i^l}^2 \leq \sum_{i=1}^{\infty} \abs{x_i^k - x_i^l}^2 \leq \eps^2.
\end{equation*}
We previously showed that the sequences \(x_i^l\) converge to \(y_i\text{,}\) so taking a limit as \(l \to \infty\) on the left-hand side of the inequalty gives
\begin{equation*}
\sum_{i=1}^N \abs{x_i^k - y_i}^2 \leq \eps^2.
\end{equation*}
This statement holds for all \(n \in \mathbb{N}\text{,}\) and so letting \(N\) tend to \(\infty\) gives us
\begin{equation*}
\sum_{i=1}^\infty \abs{x_i^k - y_i}^2 \leq \eps^2,
\end{equation*}
which is to say that \(\norm{x^k - y} \lt \eps\text{.}\) Having shown that \(x^k - y\) is in \(\ell^2\text{,}\) we note that \(y = x^k - (x^k - y) \in \ell^2\text{,}\) which concludes this step of the proof.
The final step is to argue that the sequence \(x^k \to y\text{.}\) In the previous step, we showed that for a given \(\eps\text{,}\) there exists \(K\) so that \(k > K\) implies that \(\norm{x^k - y} \lt \eps\text{.}\) Thus, \(x^k\) converges to the \(\ell^2\) sequence \(y\text{.}\) Since every Cauchy sequence converges to a limit in \(\ell^2\text{,}\) we conclude that \(\ell^2\) is a complete metric space.
