Positive-Term Generalized Sum Converges iff Supremum

Theorem
Let $(G,\circ,\le)$ be an abelian totally ordered group, considered under the order topology.

Let $\{x_i:i \in I\}$ be an indexed set of positive elements of $G$.

Then the generalized sum $\sum \{ x_i: i \in I \}$ converges to a point $x\in G$ iff $x$ is the supremum of


 * $P := \displaystyle\Bigl\{\sum_{i\in F} x_i: \text{$F\subseteq I$ and $F$ is finite}\Bigr\}$

Forward implication
Suppose that $\sum \{ x_i: i \in I \}$ converges to $x$.

We first show that $x$ is an upper bound of $P$:

Suppose for the sake of contradiction that for some finite subset $F$ of $I$,


 * $\sum_{i\in F} x_i > x$

Then the net of finite sums is eventually less than $\sum_{i\in F} x_i$.

Then since finite sums are monotone, if $F\subseteq F' \subseteq I$ and $F'$ is finite, then $\sum_{i\in F'} x_i \ge \sum_{i\in F} x_i$, a contradiction.

Thus we conclude that $x$ is an upper bound of $P$.

Let $m\in G$ with $mx$, $b$ is greater than any element of $P$, so the net of finite sums is always (hence eventually) less than $b$.

For any $a a$.

Since finite sums are monotone increasing, the net of finite sums is eventually greater than $a$.

Thus $\sum \{ x_i: i\in I \}$ converges to $x$.