Equivalence of Supremum and the Limit My professor gave the following example in class: 
Let $ a_1, a_2, $... be a sequence in nonnegative extended real numbers. 
Then $ \sum\limits_{i=1}^{n}a_i := \lim\limits_{n\to\infty}\sum\limits_{i=1}^{n} a_i:= \sup \limits_{F\subseteq N \\for~\ all~\ finite} \sum\limits_{i \in F} a_i $ 
I am trying to prove why these two definitions are equivalent to each other. In other words I want to prove that 
$$ A=\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}a_i \iff A = \sup \limits_{F\subseteq N \\for~\ all~\ finite} \sum\limits_{i \in F} a_i $$
Possible solution: I am trying to claim that $ \sum\limits_{i=1}^{n}a_i $ is a a nondeacresing function $ f: N \rightarrow [0,\infty] $ but I have difficulty to justify that for any $ n \in N $, $ f(n) $ is well defined. i.e. f assigns only one nonnegative number to this sum no matter how we sum up terms. Which axiom on real numbers make sure this is  the case. 
Any hints or even the full solution would be highly appreciated. 
 A: Denote by $\mathbb{N}$ the set of positive integers, and denote by $\mathcal{F}$ the collection of all finite subsets of $\mathbb{N}$. Define
$$
\begin{align}
\alpha &:= \lim_{n\rightarrow\infty} \sum_{i = 1}^n a_i, \\
\beta &:= \sup_{F \in \mathcal{F}} \sum_{i \in F} a_i.
\end{align}
$$
Since, for every $i \in \mathbb{N}_1$, $a_i \geq 0$, $\alpha$ and $\beta$ are well-defined, and $\alpha, \beta \in [0,\infty]$ (in particular, it may be that $\alpha = \infty$ or that $\beta = \infty$). Another consequence of the non-negativity of the $a_i$'s is the the partial sums in the definition of $\alpha$ form a non-decreasing sequence.
We wish to show that $\alpha = \beta$.
We firstly show that $\alpha \leq \beta$. It suffices to show that, for every $n \in \mathbb{N}$,
$$
\sum_{i = 1}^n a_i \leq \beta.
$$
Let then $n \in \mathbb{N}$. Define $F := \{1, 2, \dots, n\}$. Then $F \in \mathcal{F}$, and
$$
\sum_{i = 1}^n a_i = \sum_{i \in F} a_i \leq \beta.
$$
Next, we show that $\alpha \geq \beta$. It suffices to show that, for every $F \in \mathcal{F}$,
$$
\sum_{i \in F} a_i \leq \alpha.
$$
Let then $F \in \mathcal{F}$. If $F = \emptyset$, then, by convention
$$
\sum_{i \in F} a_i = 0,
$$
and we're done. Otherwise, define $n^* := \max F$ (since $F$ is finite, the maximum is well-defined). Then
$$
\sum_{i \in F} a_i \leq \sum_{i = 1}^{n^*} a_i \leq \alpha.
$$
Q.E.D.
