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)