Infinite sum of non-negative terms is equal to the supremum of the set of all finite sums

Question

Prove that if $a_n \geq 0$, for all $n$, then $$\sum_{n=1}^{\infty} a_n = \sup \bigg\{ \sum_{i \in A} a_i : A \subset \mathbb{N}, A \text{ finite} \bigg\}.$$

My general strategy right now is to prove $\geq$ and $\leq$, both directions.

I was able to prove $\geq$ (because the LHS is an infinite sum, while the RHS is finite, so an infinite sum must be greater or equal to a finite sum), but I can't get the $\leq$ direction.

Any help is appreciated!

Answer 1

Hint: Say $s=\sum_{n=0}^\infty a_n$ and $\displaystyle S=\left\{\sum_{n\in F}a_n:F\in\mathscr P_{finite}(\Bbb N)\right\}$, so you want to show $s=\sup(S)$.

From what you say about the inequality $s\ge\sup(S)$ it seems clear you have the right idea; it could be much better phrased.

To show $s\le \sup(S)$, note first that it's enough to show that $$\alpha<\sup(S)\quad(\forall {\alpha<s}).$$But saying something is less than the least upper bound says it is not an upper bound. So what you need to show is this:

If $\alpha<s$ there exists a finite $F\subset\Bbb N$ with $$\alpha<\sum_{n\in F}a_n.$$