# Convergence of the harmonic series in larger fields

The Harmonic series is well known and its divergence was proven back in the middle ages.

I've taken an introductory course in model theory so I know a bit about RCF and some properties of it. We did not explore it thoroughly though and haven't seen many interesting examples.

However, I do know that we can take some real closed field which is large enough (i.e. has cofinality $>\aleph_0$) and then the harmonic series will possibly converge.

My question if we take some $\mathcal{F}$ to be a model of RCF in which $\mathbb{R}$ is embedded and that the type $p(x) = \{ x > n | n\in\mathbb{N}\}$ is realized, $$x = \sum_{n \in \mathbb{N}^+} \frac{1}{n}$$ then $\forall y\in\mathbb{R}(x>y)$ then obviously $x$ is an upper-bound for the real numbers in the field we've chosen. However since $x$ is a non-Archimedean number, it is also clear that $x-1$ is an upper bound of the real numbers in $\mathcal{F}$.

This is the part where I get confused. What is $x$ and what is the conditions required for it to exist in the model?

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What is {x > n|n in N} supposed to be? Is it {x : x > n and n in N}? Is it {n : x > n and n in N}? –  Ricky Demer Sep 15 '10 at 2:47
Ricky: A type is a set of (usually first-order) formulas. The type $p(x)$ has 1 free variable $x$. The formulas that form this type are all the statements $x>1$, $x>2$, $x>3$, etc. –  Andres Caicedo Sep 15 '10 at 3:41
@Ricky: It is a collection of formulae with a free variable $x$, saying that $p(x)$ is realized meaning that there is some element in $\mathcal{F}$ for which all the formulae in $p(x)$ are true. –  Asaf Karagila Sep 15 '10 at 3:44

## 1 Answer

Real closed fields are not complete (unless they are isomorphic to the reals), so the fact that some increasing sequence is bounded does not imply that it has a supremum.

If x is the sum of the harmonic series, then we seem to get x=1+ 1/3 + ...+ 1/2+1/4+...>1/2+1/4...+1/2+1/4+..=x/2+x/2 = x, suggesting that x does not exist in any real closed field.

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When you say complete, do you mean metrically? Because that'd be obvious since we define metrics using the reals. Or do you mean in the sense that it is a complete order (i.e. all the Dedekind cuts are realized)? –  Asaf Karagila Sep 15 '10 at 3:47
@AK: I think he means that the order is "bounded complete": every set which is bounded above has a least upper bound. –  Pete L. Clark Sep 15 '10 at 3:57
@Pete: So there are no real closed fields which are Dedekind-closed except the real numbers? –  Asaf Karagila Sep 15 '10 at 4:07
@AK: Yes, I believe so. (This is not really my area of core expertise.) See for instance emis.de/proceedings/TopoSym2001/20.pdf –  Pete L. Clark Sep 15 '10 at 5:37