id: 06428009 dt: j an: 2016a.00806 au: Clark, Pete L.; Diepeveen, Niels J. ti: Absolute convergence in ordered fields. so: Am. Math. Mon. 121, No. 10, 909-916 (2014). py: 2014 pu: Mathematical Association of America (MAA), Washington, DC la: EN cc: I35 ut: convergent series; absolutely convergent series; ordered fields; Archimedean fields; non-Archimedean fields; Cauchy completeness ci: li: doi:10.4169/amer.math.monthly.121.10.909 ab: 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. } rv: Khodr Shamseddine (Winnipeg)