@article {MATHEDUC.06428009,
author = {Clark, Pete L. and Diepeveen, Niels J.},
title = {Absolute convergence in ordered fields.},
year = {2014},
journal = {American Mathematical Monthly},
volume = {121},
number = {10},
issn = {0002-9890},
pages = {909-916},
publisher = {Mathematical Association of America (MAA), Washington, DC},
doi = {10.4169/amer.math.monthly.121.10.909},
abstract = {Let $\mathbb{F}$ be an ordered field. The authors show that the relation between convergence and absolute convergence of an infinite series $\sum_{n=1}^\infty a_n$ in $\mathbb{F}$ is closely connected to the sequential (i.e., Cauchy) completeness of the field $\mathbb{F}$. { indent=8mm \item{(i)} If $\mathbb{F}$ is sequentially complete and Archimedean, then $\mathbb{F}$ is isomorphic to $\mathbb{R}$; then, in this case, it is well known (from calculus) that every absolutely convergent series in $\mathbb{F}$ is convergent in $\mathbb{F}$, but the converse is not true, that is, $\mathbb{F}$ has a convergent series that is not absolutely convergent (e.g., let $a_n=(-1)^n/n$). \item{(ii)} If $\mathbb{F}$ is sequentially complete and non-Archimedean, then $\sum_{n=1}^\infty a_n$ converges in $\mathbb{F}$ if and only if it converges absolutely in $\mathbb{F}$. \item{(iii)} If $\mathbb{F}$ is not sequentially complete ($\mathbb{F}$ may be Archimedean, that is, isomorphic to a proper subfield of $\mathbb{R}$, or non-Archimedean), then $\mathbb{F}$ has an absolutely convergent series that is not convergent and $\mathbb{F}$ has a convergent series that is not absolutely convergent. }},
reviewer = {Khodr Shamseddine (Winnipeg)},
msc2010 = {I35xx},
identifier = {2016a.00806},
}