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?



Convergence of the harmonic series in larger fields
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. 


