Prove $X$ is a complete vector space $\iff$ $\Big[\quad\sum_{n=1}^\infty \| x_n \| \implies \sum_{n=1}^\infty x_n$ converges$\quad\Big]$.
I have seen a proof of this already in a lecture but I was wondering if there is a published proof of this theorem in any book on real analysis? This was proved in the context of $L^p$ spaces as a preliminary to show $L^1$ is a complete space.
On
Hint: By the Cauchy Criterion $\sum a_n$ is convergent if and only if, for every $\epsilon > 0$ there exists $n_0 \in \mathbb N$ such that $$\bigg\|\sum_{k=n+1}^{n+p} a _k\bigg\| < \epsilon$$
whenever $n > n_0$ and $p \in \mathbb N$.
Using the above and
$$\|s_{m+p} - s_m\| = \bigg\|\sum_{k=n+1}^{n+p} a _k\bigg\| \leq \sum_{k=n+1}^{n+p}\|a_k\| < \epsilon$$
this may be of some help to show $ (\implies)$.
I am not sure whether you can find a proof of this result in Real Analysis book but you can find it in basic books on Functional Analysis like "Beginning Functional Analysis" by "K.Saxe" and in most of the Functional Analysis books it is an exercise. Nevertheless here is a proof that you might have already seen.
Let $\sum_{k = 1}^{\infty}x_k$ be a series in $X$. First let us assume that $X$ is a Banach space and $\sum_{k = 1}^{\infty}\|x_k\|$ converges. We need to show that $\sum_{k = 1}^{\infty}x_k$ is convergent. Let $(s_n)$ be the sequence of partial sums. Then $(s_n)$ is a Cauchy sequence since \begin{equation} \|s_{n+m} - s_n\| = \biggl{\|}\sum_{k = n+1}^{n+m}x_k\biggr{\|} \leq \sum_{k = n+1}^{n+m}\|x_k\|. \end{equation} Thus, $\sum_{k = 1}^{\infty}x_k$ is convergent as $X$ is a Banach space.
Conversely, let us assume that absolute convergence implies convergence. Let $(z_n)$ be a Cauchy sequence in $X$. Then there exists a subsequence $(z_{n_k})$ of $(z_n)$ such that \begin{align*} \|z_{n_{k+1}} - z_{n_k}\| < \dfrac{1}{2^k} & & (k \geq 1). \end{align*} Let $x_k = z_{n_{k+1}} - z_{n_k}, k \geq 1$. Then $\sum_{k = 1}^{\infty}\|x_k\|$ is convergent. Hence by hypothesis it follows that $\sum_{k = 1}^{\infty}x_k$ is convergent, i.e., $(z_{n_{k}})$ is convergent. Since, $(z_n)$ is a Cauchy sequence which has a convergent subsequence, it follows that $(z_n)$ is convergent.