Equivalence of definition of absolute convergence

Question

Definition

$$\sum_{n \in \mathbb{N}}a_n =s, s \in \mathbb{R} \Leftrightarrow \forall \varepsilon \gt 0 \space \exists \left( F\subseteq \mathbb{N}\wedge \left|F\right| \in\mathbb{N} \right) \left[ \forall G \subseteq \mathbb{N} \wedge F\subseteq G \wedge \left| \sum_{n \in G}a_n -s\right| \lt \varepsilon \right]$$ In words, Definition 1 claims that a real value $s$ is the sum of the series $a_n$ iff we can make a subsum over finite $F$ so that everything left is sufficiently small for any small $\varepsilon$.

Absolute convergence

$$\sum_{n=1}^{\infty }\left| a_n\right|=s,s \in \mathbb{R}$$

Question

Do these state the same for every series?

Observations

It seemed profitable to think about reordering, but was especially messy when $G$ becomes infinite. Direct proof through definitions seemed too inconsistent to finish. Basic convergence test fails horribly.

Idea

I do have a feeling that this equivalence somehow connects to the axiom of choice for infinite $G$.

Answer 1

On notation: if $G \subseteq \Bbb N$ is infinite, then $\sum_{n \in G} a_n$ represents the limit when the partial sums are added in increasing order of the indices.

I claim that the following statements about the series $\sum a_n$ are equivalent:

1. There exists $s \in \Bbb R$ such that for every $\epsilon > 0$, there is a finite set $F \subset \Bbb N$ such that for every $G$ with $F \subseteq G \subseteq \Bbb N$, $$\left|\sum_{n \in G} a_n - s\right| < \epsilon$$
2. Every subseries of $\sum a_n$ converges.
3. $\sum a_n$ is absolutely convergent.

Proof: 1) $\implies$ 2): Let $H \subseteq \Bbb N$ and $\epsilon > 0$. Then there exists a finite $F$ such that if $F \subseteq G$, then $\left|\sum_G a_n - s\right| < \frac \epsilon 2$. If $M \ge N > \max(F)$ and we set $G_1 =\{n \in \Bbb N\mid n < N\}$ and $G_2 = G_1 \cup \{n \in \Bbb N\mid n \le M \text{ and }n \in H\}$, then

$$\left|\sum_{n\in G_1} a_n - s\right| < \frac \epsilon 2\\\left|\sum_{n \in G_2} a_n - s\right| < \frac \epsilon 2$$ So $$\left|\sum_{\underset{n \in H}{n=N}}^M a_n\right| = \left|\left(\sum_{n\in G_2} a_n - s\right) - \left(\sum_{n\in G_1} a_n - s\right)\right| \le \left|\sum_{n\in G_2} a_n - s\right| + \left|\sum_{n\in G_1} a_n - s\right| < \epsilon$$ Which is the Cauchy condition on the subseries $\sum_{n\in H} a_n$. Therefore it must converge.

2) $\implies$ 3): Let $P = \{n \in \Bbb N\mid a_n \ge 0\}$ and $N = \{n \in \Bbb N\mid a_n < 0\}$. Then both $\sum_{n\in P} a_n$ and $\sum_{n\in N} a_n$ converge. But $$\sum |a_n| = \sum_{n\in P} a_n - \sum_{n \in N} a_n$$ so it also converges (to fill in the details I am skipping, restrict all three series to $n \le K$ and let $K \to \infty$).

3) $\implies$ 1): Let $\epsilon > 0$ and let $s = \sum a_n$. Then there is an $M$ such that $$\left|\sum_{n\le M} a_n - s\right| < \epsilon/2$$ Since $\sum |a_n|$ converges, there is an $N$ such that $\sum_{n > N} |a_n| < \epsilon/2$. Let $F = \{n \in \Bbb N\mid n \le \max\{M,N\}\}$. If $G \supseteq F$ and $H = G \setminus F$, then $$\left|\sum_{n\in H}a_n\right| \le \sum_{n\in H} |a_n| \le \sum_{n > N} |a_n| < \epsilon/2$$ and $$\left|\sum_{n\in G}a_n - s\right| = \left|\sum_{n\in F}a_n - s + \sum_{n\in H}a_n\right| \le \left|\sum_{n\in F}a_n - s\right| + \left|\sum_{n\in H}a_n\right| < \epsilon$$ Q.E.D.