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.
Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Best Answer
First of all, let me comment on the "correct" notion of Cauchy completeness for an ordered field $F$. Given an ordinal $\xi$, say a $\xi$-indexed sequence $(x_\alpha)_{\alpha<\xi}$ is Cauchy if for each $\epsilon>0$ in $F$ there exists $\alpha<\xi$ such that $|x_\beta-x_\gamma|<\epsilon$ for all $\beta,\gamma\geq \alpha$. Similarly, say $(x_\alpha)$ converges to $x\in F$ if for each $\epsilon>0$ there exists $\alpha<\xi$ such that $|x_\beta-x|<\epsilon$ for all $\beta\geq \alpha$. Then $F$ can be called Cauchy-complete if all Cauchy ordinal-indexed sequences in $F$ converge to some element of $F$. Any ordered field has a Cauchy completion obtained by formally adjoining limits to all Cauchy ordinal-indexed sequences (which can in fact be taken to all be indexed by the cofinality of $F$).
(There are many equivalent formulations of this notion, such as using nets or filters instead of ordinal-indexed sequences. Another equivalent formulation is that a Dedekind cut $(L,R)$ in $F$ is always filled by an element of $F$ as long as either for all $\epsilon>0$ there exists $x\in L$ such that $y<x+\epsilon$ for all $y\in L$ or for all $\epsilon>0$ there exists $x\in R$ such that $y>x-\epsilon$ for all $y\in R$.)
In particular, for instance, consider the field $\mathbb{Q}(x)$ ordered such that $x$ is infinitely large. This field is not Cauchy-complete, but let $F$ be its Cauchy completion. (Since $\mathbb{Q}(x)$ is countable, you can construct $F$ by just taking ordinary $\mathbb{N}$-indexed Cauchy sequences.) Note that every element of $\mathbb{Q}(x)$ is either infinitely large or is infinitesimally close to some element of $\mathbb{Q}$: a rational function $f(x)/g(x)$ is infinitely large if $\deg f>\deg g$, infinitesimal if $\deg f<\deg g$, and if $\deg f=\deg g$ then it is infinitesimally close to $a/b$ where $a$ and $b$ are the leading coefficients of $f$ and $g$. Any element in the completion $F$ can be approximated arbitrarily closely by elements of $\mathbb{Q}(x)$, which means any element of $F$ is infinitesimally close to an element of $\mathbb{Q}(x)$ since $\mathbb{Q}(x)$ has infinitesimals. Thus every element of $F$ is either infinitely large or infinitesimally close to an element of $\mathbb{Q}$. In particular, for instance, $F$ cannot contain a square root of $2$, so $\mathbb{R}$ does not embed in $F$.
(If you insist on talking about completeness in terms of a metric, this example can also be given a metric: say $d(f,g)=2^n$ where $n\in\mathbb{Z}$ is the difference between the degrees of the numerator and denominator of $f-g$. This defines a metric on $\mathbb{Q}(x)$ compatible with its order topology and the metric completion with respect to this metric can naturally be identified with the Cauchy completion.)