(Absolute) convergence of series vs. complete monoid structure on $[0\,..\infty]$ This question is phrased specifically for real numbers. Although I'm more interested in a generalized version of this question I doubt I would get an answer for that anytime soon.
Consider the following "infinite sum" operation on the set of nonnegative extended reals $[0\,..\infty]$:
$$\sum_{i\in I} x_i := \sup_{\tilde{I} \in K_I} \sum_{i\in \tilde{I}} x_i$$
where $K_I = \{\tilde{I}\subseteq I : \tilde{I}\text{ is finite} \}$.

Given a sequence $(x_i)_{i\in \mathbb{N}}$ of non-negative reals is it true that:

$$\sum_{i\in \mathbb N} x_i = \sum_{i=1}^\infty x_i, \text{ if } \sum_{i\geq 1} x_i \text{ converges }$$
 $$\sum_{i\in \mathbb N} x_i = \infty, \text{ if } \sum_{i\geq 1} x_i \text{ diverges }$$
 A: As Jeff pointed out in the comments this was actually not difficult to answer:

Suppose $\sum_{i\geq 1} x_i$ converges. Given $\tilde{I}\in K_{\mathbb{N}}$ let $\sigma$ be a permutation of $\mathbb{N}$ such that $\tilde I = \{\sigma(1),\dots, \sigma(n)\}$.
Then $\sum_{i\in \tilde I} x_i \leq \sum_{i=1}^n x_{\sigma(i)} + \sum_{i=n+1}^m x_{\sigma(i)}$ for all $m \in \mathbb{N}$ (since $x_i \geq 0$ for all $i$), hence: $$\sum_{i\in \tilde I} x_i \leq \sum_{i=1}^\infty x_{\sigma(i)} = \sum_{i=1}^\infty x_{i} $$
Now suppose we have an $s$ with $\sum_{i\in \tilde I} x_i \leq s$ for all $\tilde I \in K_{\mathbb{N}}$. Then $\sum_{i=1}^n x_i \leq s$ for all $n$, hence $\sum_{i=1}^\infty x_i \leq s$.
Therefore (by definition of what a supremum is): 
$$\sum_{i=1}^\infty x_{i} = \sum_{i\in \mathbb N} x_i$$
Suppose $\sum_{i\geq 1} x_i$ diverges. Then the set $\{ \sum_{i\in \tilde I} x_i : \tilde I \in K_{\mathbb N}\}$ is unbounded (otherwise $\sum_{i\geq 1} x_i$ would converge), hence:
$$\sum_{i \in \mathbb{N}} x_i = \infty$$
