In the process of touching up some notes on infinite series, I came across the following "result":
Theorem: For an ordered field $(F,<)$, the following are equivalent:
(i) Every Cauchy sequence in $F$ is convergent.
(ii) Absolutely convergent series converge: $\sum_n |a_n|$ converges in $F$ $\implies$ $\sum_n a_n$ converges in $F$.
But at present only the proof of (i) $\implies$ (ii) is included, and unfortunately I can no longer remember what I had in mind for the converse direction. After thinking it over for a bit, I wonder if I was confusing it with this result:
Proposition: In a normed abelian group $(A,+,|\cdot|)$, the following are equivalent:
(i) Every Cauchy sequence is convergent.
(ii) Absolutely convergent series converge: $\sum_n |a_n|$ converges in $\mathbb{R}$ $\implies$ $\sum_n a_n$ converges in $A$.
For instance, one can use a telescoping sum argument, as is done in the case of normed linear spaces over $\mathbb{R}$ in (VIII) of this note.
But the desired result is not a special case of this, because by definition the norm on a normed abelian group takes values in $\mathbb{R}^{\geq 0}$, whereas the absolute value on an ordered field $F$ takes values in $F^{\geq 0}$.
I can show (ii) $\implies$ (i) of the Theorem for ordered subfields of $\mathbb{R}$. Namely, every real number $\alpha$ admits a signed binary expansion $\alpha = \sum_{n = N_0}^{\infty} \frac{\epsilon_n}{2^n}$, with $N_0 \in \mathbb{Z}$ and $\epsilon_n \in \{ \pm 1\}$, and the associated "absolute series" is $\sum_{n=N_0}^{\infty} \frac{1}{2^n} = 2^{1-N_0}$.
Because an ordered field is isomorphic to an ordered subfield of $\mathbb{R}$ iff it is Archimedean, this actually proves (ii) $\implies$ (i) for Archimedean ordered fields. But on the one hand I would prefer a proof of this that does not use the (nontrivial) result of the previous sentence, and on the other hand…what about non-Archimedean ordered fields?
Added: The article based on this question and answer has at last appeared:
Clark, Pete L.; Diepeveen, Niels J.;
Absolute Convergence in Ordered Fields.
Amer. Math. Monthly 121 (2014), no. 10, 909–916.
If you are a member of the MAA, you will be frustrated if you try to access it directly: the issue is currently advertised on their website but the articles are not actually available to members. The article is available on JSTOR and through MathSciNet. Anyway, here is an isomorphic copy. Thanks again to Niels Diepeveen!
Best Answer
Proving this for arbitrary ordered fields is a little trickier than for Archimedean fields, partly because there are no concrete sequences -- other than eventually constant ones -- that are guaranteed to converge or even to be Cauchy sequences and partly because we lack the embedding in a completely ordered field.
These problems can be overcome by constructing all the necessary sequences and series from the one Cauchy sequence we assume to exist. To begin with, note two important facts. First, for a Cauchy sequence to converge, it is sufficient that some subsequence converges. Second, any sequence has a strictly increasing subsequence, a strictly decreasing subsequence or a constant subsequence. For this problem, the latter case is trivial and the first two can be reduced to each other by negation, so we need to prove only one of them.
Let $K$ be an ordered field in which every absolutely convergent series is convergent. If $\{a_n\}$ is a strictly increasing Cauchy sequence in $K$, then $\{a_n\}$ converges.
Proof:
Let $b_n = a_{n+1} - a_n$. Then ${b_n}$ is positive and converges to $0$, so it has a strictly decreasing subsequence $\{b_{n_k}\}$. Let $c_k = b_{n_k} - b_{n_{k+1}}$. We now have a convergent series with positive terms $\sum_{k=1}^\infty c_k = b_{n_1}$. As $\{a_n\}$ is a Cauchy sequence, it has a subsequence $\{a_{m_k}\}$ such that $a_{m_{k+1}} - a_{m_k} < c_k$ for all $k$.
Now consider the series $\sum_{i=1}^\infty d_i$ where $d_{2k-1} = a_{m_{k+1}} - a_{m_k}$ and $d_{2k} = a_{m_{k+1}} - a_{m_k} - c_k$. Note that $-c_k < d_{2k} < 0 < d_{2k-1} < c_k$, so we can pair off terms to get $$ \sum_{i=1}^\infty |d_i| = \sum_{k=1}^\infty (d_{2k-1} - d_{2k}) = \sum_{k=1}^\infty c_k = b_{n_1} $$ By the hypothesis on $K$ we may conclude that $\sum_{i=1}^\infty d_i$ converges and $$ \sum_{i=1}^\infty d_i + \sum_{k=1}^\infty c_k = \sum_{k=1}^\infty (d_{2k-1} + d_{2k} + c_k) = 2 \sum_{k=1}^\infty (a_{m_{k+1}} - a_{m_k}). $$ Because a Cauchy sequence with a convergent subsequence converges, we have $$ \lim_{n \to \infty} a_n = a_{m_1} + \sum_{k=1}^\infty (a_{m_{k+1}} - a_{m_k}) = a_{m_1} + \frac{1}{2}\left(b_{n_1} + \sum_{i=1}^\infty d_i \right) $$
To the question "how did I come up with this?": there are not many things that could possibly work. The problem is set in an environment where none of the power tools of analysis work. Basic arithmetic works, inequalities work, some elementary properties of sequences and series work, but if you want to take a limit of something it'd better be convergent by hypothesis or by construction.
One more or less obvious attack is by contraposition: assume that there is a divergent Cauchy sequence and try to construct a divergent, absolutely convergent series. Such a series must be decomposable into a positive part $a$ and a negative part $b$, where $a+b$ diverges and $a-b$ converges. This is possible in several ways by taking $a$ and $b$ to be linear combinations of known convergent and divergent series.
A complication is that the terms of the convergent series must dominate those of the divergent series, as they must control the signs. I wasted a lot of time trying to get the convergent series to do this, which is very hard, perhaps impossible. Then I turned to the proof for vector spaces for inspiration, and saw that it was in fact very easy to adjust the divergent series instead, as the partial sums are a Cauchy sequence. I also adopted the overall structure of that proof, which is why the final version is not by contraposition.