I want to prove that $\sum_{n=1}^{\infty}a_n=\sum_{n\in\mathbb{N}}a_n$ if $a_n$ is non-negative. Call the first series $A$ and the second series $B$. I'm just assuming that the definition of the latter series is $B:=\sup R$, where $R:=\left\{\sum_{n\in K}a_n\mid K\subset\mathbb{N} \ \text{is finite}\right\}$. I wasn't given one in my textbook, so I'm just assuming that's it.
$(\leq)$: Suppose $\{s_m\}$ is the sequence of partial sums for $\sum_{m=1}^\infty a_m$. Then $s_n=\sum_{j=1}^na_m$. Since this is a finite sum, the order of the terms doesn't matter, and so I can write this as $\sum_{j\in K} a_j$ such that $\operatorname{card}(K)=n$. So each $s_m\in R\implies s_m\leq B \ \forall n\geq 1$. Thus, $A=\lim_{n\to\infty}s_m\leq\lim_{n\to\infty}B=B$
$(\geq)$: Let $x$ be any element of $R$. So for some finite $K\subset\mathbb{N}$, we have $x=\sum_{n\in K}a_n$. Suppose $M$ is the largest element of $K$. Then since the terms $a_n$ are non-negative, we have $x\leq\sum_{j=1}^ma_j$ for all $m\geq M$. Thus, $x\leq\sum_{j=1}^ma_j\leq\lim_{m\to\infty}\sum_{j=1}^ma_j=A$ for every $x\in R$. Thus, $A$ is an upper bound for $R$. Since $B$ is the least upper bound for $R$, we have $A\geq B$.
Does anyone see anything wrong with my argument?