most recent 30 from http://mathoverflow.net 2013-05-24T14:39:13Z http://mathoverflow.net/feeds/question/38756 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38756/convergence-of-the-harmonic-series-in-larger-fields Asaf Karagila 2010-09-15T01:33:22Z 2010-09-29T04:22:17Z

<p>The Harmonic series is well known and its divergence was proven back in the middle ages.</p> <p>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.</p> <p>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.</p> <p>My question if we take some $\mathcal{F}$ to be a model of RCF in which $\mathbb{R}$ is embedded and that the type <code>$p(x) = \{ x &gt; n | n\in\mathbb{N}\}$</code> 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 $\mathcal{F}$.</p> <p>This is the part where I get confused. What is $x$ and what is the conditions required for it to exist in the model?</p>

http://mathoverflow.net/questions/38756/convergence-of-the-harmonic-series-in-larger-fields/38770#38770 Richard Borcherds 2010-09-15T03:37:42Z 2010-09-15T03:37:42Z

<p>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. </p> <p>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. </p>