In which ordered fields does absolute convergence imply convergence?

Solution 1:

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.