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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?

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Hah! I saw that this question has not been answered this morning, and was just about to add a bounty to it myself. (How many experts on non-Archimedean ordered fields do you think there are on Math.SE?) –  Willie Wong Feb 26 '12 at 19:20
    
@Willie: well, as I understand it (i.e., not very well), you are still free to tack on your own bounty. As for your question...well, I am hoping for at least one. –  Pete L. Clark Feb 26 '12 at 22:04
    
I think only one bounty can be active on one question at a time. So if you still don't have an answer after the week, I'll put one on then. –  Willie Wong Feb 26 '12 at 23:03
    
If my answer is correct, you needed only a little more patience. It was nearly done yesterday. –  Niels Diepeveen Feb 27 '12 at 1:40
    
In the most familiar non-Archimedean ordered field, convergence in the sense that is usual in real analysis is rarely if ever thought about at all. The set of finite natural numbers is, after all, an external set. –  Michael Hardy Feb 27 '12 at 1:45
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1 Answer

up vote 21 down vote accepted
+300

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.

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This is fantastic! I just read it over in some detail and it looks completely correct. Because of my unfamiliarity with bounties I will wait a little while before formally accepting it to allow others a chance to vet the answer. But I am very pleased with this. May I include this argument in my notes www.math.uga.edu/~pete/3100supp.pdf? I will attribute it to you. –  Pete L. Clark Feb 27 '12 at 3:16
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By the way, there is some seriously clever stuff going on here: how did you come up with this? –  Pete L. Clark Feb 27 '12 at 3:48
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+1, very nice. The key is the construction of $d$ -- that would seem slightly less magical (to me) if you require $d_{2k}-d_{2k-1}=c_k$ and $d_{2k}+d_{2k-1}=a_{m_{k+1}}-a_{m_k}$, with solution $d_{2k}=\frac12(a_{m_{k+1}}-a_{m_k}+c_k)$ and $d_{2k-1}=\frac12(a_{m_{k+1}}-a_{m_k}-c_k)$; then $\sum_i|d_i|=\sum_kc_k$ and $\sum_i d_i=\sum_k(a_{m_{k+1}}-a_{m_k})$. –  joriki Feb 27 '12 at 8:15
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@Pete L. Clark, Willie Wong: If you have any use for it, I'll be honoured. –  Niels Diepeveen Feb 29 '12 at 8:38
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@Niels: please email me if you get this message. It turns out that a note in the September 2012 issue of the College Math Journal ends by asking precisely the question that I asked and you answered here. So I have prepared a followup note with you as a coauthor. –  Pete L. Clark Aug 29 '12 at 3:37
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