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: ∑n|an| converges in F ⟹ ∑nan converges in F.But at present only the proof of (i) ⟹ (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,+,|⋅|), the following are equivalent:

(i) Every Cauchy sequence is convergent.

(ii) Absolutely convergent series converge: ∑n|an| converges in R ⟹ ∑nan converges in A.For instance, one can use a telescoping sum argument, as is done in the case of normed linear spaces over 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 R≥0, whereas the absolute value on an ordered field F takes values in F≥0.

I can show (ii) ⟹ (i) of the Theorem for ordered subfields of R. Namely, every real number α admits a

signed binary expansionα=∑∞n=N0ϵn2n, with N0∈Z and ϵn∈{±1}, and the associated "absolute series" is ∑∞n=N012n=21−N0.Because an ordered field is isomorphic to an ordered subfield of R iff it is Archimedean, this actually proves (ii) ⟹ (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!

**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 {an} is a strictly

increasing Cauchy sequence in K, then {an} converges.

**Proof:**

Let bn=an+1−an. Then bn is positive and

converges to 0, so it has a strictly decreasing

subsequence {bnk}.

Let ck=bnk−bnk+1. We now have a convergent

series with positive terms ∑∞k=1ck=bn1.

As {an} is a Cauchy sequence, it has a subsequence {amk}

such that amk+1−amk<ck for all k.

Now consider the series ∑∞i=1di where

d2k−1=amk+1−amk and

d2k=amk+1−amk−ck.

Note that −ck<d2k<0<d2k−1<ck, so we can pair off

terms to get

∞∑i=1|di|=∞∑k=1(d2k−1−d2k)=∞∑k=1ck=bn1

By the hypothesis on K we may conclude that ∑∞i=1di

converges and

∞∑i=1di+∞∑k=1ck=∞∑k=1(d2k−1+d2k+ck)=2∞∑k=1(amk+1−amk).

Because a Cauchy sequence with a convergent subsequence converges,

we have

lim

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.

**Attribution***Source : Link , Question Author : Pete L. Clark , Answer Author : Niels J. Diepeveen*