absolutely convergent series is convergent using completeness axiom for $\mathbb{R}$

Question

Given the completeness axiom for $\mathbb{R}$, stating that each nonempty bounded subset of $\mathbb{R}$ must have a least upper bound and a greatest lower bound.

How to deduce directly from this axiom that every absolutely convergent series of real number is convergent? Is there a direct way how to do it?

I know how to do it using Cauchy completeness (see here https://proofwiki.org/wiki/Absolutely_Convergent_Series_is_Convergent ), but in lecture we haven't discussed that, for $\mathbb{R}$, Cauchy completeness is equivalent to the completeness axiom.

Answer 1

First note that (and define) $$-S_N := -\sum_i^N |a_i| \le T^-_N := \sum_{i; a_i <0}^N a_i \le A_N := \sum_i^N a_i \le \sum_{i; a_i >0}^N a_i =: T^+_N \le \sum_i^N |a_i| = S_N$$ $T^+_N $ and $T_N^- $ are obviously monotonic sequences, with $T_N^+ $ decreasing and $T_N^+$ increasing. So $$T^- := \inf\{T_N^- : N\in \mathbb{N}\} \text{ and } T^+ := \sup\{T_N^+ : N\in \mathbb{N}\} $$ both exist (by your version of the completeness axiom and the assumption that the series is absolutely convergent, which just means that $S_N$ is bounded) and it is not difficult to see (using monotonicity -- this I leave to you) that then $$T^- =\lim_N T^-_N \text{ and } T^+ =\lim_N T^+_N $$

both exist. Assuming you know that the sum of two convergent series is convergent with the limit being equal to the sum of the limits, you get the convergence of $A_N$, because for each $N\in \mathbb{N}$

$$A_N = T_N^+ + T_N^-$$

This is just the claim.

Answer 2

Based on the least upper bound property it is proved in this question that every bounded sequence in $\mathbb R$ has a convergent subsequence.

It is the Bolzano-Weierstrass theorem.

If a Cauchy sequence has a convergent subsequence then it is convergent.

Let $a_n\in\mathbb R$ with $B:=\sum_{n=1}^{\infty}|a_n|<\infty$ and let $s_n:=\sum_{i=1}^na_i$.

Then $|s_n|\leq\sum_{i=1}^n|a_i|\leq B\in[0,\infty)$ so that $\{s_n\mid n\in\mathbb N\}\subseteq[-B,B]\subseteq\mathbb R$.

So we are dealing with a bounded sequence and conclude that $(s_n)$ is convergent.