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Let $a_i\ge0$ for all $i\in I$ such that $\sum\limits_{i\in I} a_i$ is finite, then $J=\{i\in I:a_i \neq 0\}$ is countable.

Let $A= \sum\limits_{i\in I} a_i$, then $A$ is an unordered sum defined as $$\sum\limits_{i\in I} a_i = \sup \left\{\sum\limits_{i \in I^0}a_i: I^0\text{ is a finite subset of }I\right\}.$$

I think I have worked most of them out. I let $J_n = \{i\in I:a_i\gt A/n \}$, $n\in \Bbb{N}$. So as $n \to \infty$, $J_n \to \{i \in I: a_i \gt 0\}$ and because $i\in I$, $a_i \in [0,\infty)$ then $|J_n|$ is at most $|I|$. And since $|J|\to |J_n|$, that $|J|$ is bounded by $|I|$ thus it is countable? But I just thought it is not that rigorous. I also need to prove $|I|\lt \infty$ right? If so, how?

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    You have no hypothesis that $I$ is countable (if it is there is nothing to prove). Hint: Can you bound $I_n=\{n\mid a_n>1/n\}$? Then deduce that $\bigcup\limits_nI_n$ is at most countable and you are done. – 2017-02-06
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    You have given no def'n of $A_n$ nor $J_n$. – 2017-02-06
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    @Did, so following your method, the bound I can think of for $I_n$ is $|I|$, then it returns to my previous again. – 2017-02-06
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    The bound of $|I_n|$ by $|I|$ is mostly empty since the interesting case is when $I$ is infinite. Rather, try to bound $|I_n|$ using $A$ and $n$, starting from $$\sum_{i\in I_n}a_i\geqslant\frac1n\sum_{i\in I_n}1=\frac1n|I_n|$$ – 2017-02-06
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    @Did Thank you! I got what you mean but just one minor point should $I_n = \{ i| a_i \gt 1/n \}$ ? – 2017-02-07
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    @UserQ Quite so. Well spotted. – 2017-02-07

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I'll present another proof. Find a sequence $F_n$ of finite subsets of $I$ such that $\displaystyle\sum_{i\in F_n} a_i \to A$ as $n\to \infty$. Such a sequence exists (can you prove it ?). Then $\displaystyle\bigcup_{n\in \mathbb{N}} F_n$ is a countable union of finite sets and is thus countable. Assume $J$ isn't countable and let thus $a_k\in J- \bigcup_{n\in \mathbb{N}} F_n$. Thus $a_k>0$. Then take $n$ sufficiently large so that $A- \displaystyle\sum_{i\in F_n} a_i < a_k/2$. Then $F:= F_n\cup\{a_k\}$ is still a finite set and $\displaystyle\sum_{i\in F} a_i > A$, which is a contradiction. So $J$ is countable.

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    No harmonic series in the approach I suggest. – 2017-02-06
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    Oh my bad, it's just that usually the hint that you gave is followed by the harmonic series. I'll edit my answer – 2017-02-06
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    Let me suggest to try to complete the hint I suggest to fully understand it (it seems that at present you do not) and to see that it uses no harmonic series but only the fact that a countable union of finite sets is at most countable. (Still an irrelevant mention of harmonic series in your parenthesis.) – 2017-02-06
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    Yeah I had forgotten that parenthesis. Yes, I see how you do not use the harmonic series, but the last time I saw this exercise (it was in class, a long time ago), I remember someone giving a proof with the harmonic series, proof that started out just like yours. Or maybe I'm misremembering, but I'm pretty sure. – 2017-02-06
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    I doubt that, since the proof I indicate would work equally well using the sets $I'_n=\{i\in I:a_i>1/2^n\}$, but I would be interested to read what precisely you have in mind. – 2017-02-06
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    As you can guess by my answer, the proof with the $I_n$ wasn't the one I had thought of at the time, so I don't quite remember what the person in question had done. If you want, I can ask them, and then tell you (but know that you may be right and I may end up stupidly answering "Oh you were right, no harmonic series") – 2017-02-06