It is not necessarily true that $\sum_{i=1}^\infty a_i = \infty$, even if $a_i>0 \forall i$. One simple example would be the geometric series.
However, this sum is taken for natural value $i$ i.e. it's summing countable infinite many values. Therefore, I wonder if this is true if $\sum_{r \in R} a_r$ can converge, where $R$ is the real number set?
I don't even know how to write the sum down, so is this "sum" even "well-defined"?
Let $(a_i)_{i\in I}$ such that $a_i>0$ for all $i\in I$ and $S:=\sum_{i\in I}a_i<+\infty$, we will show that $I$ is countable. For $n\in\mathbb{N}^*$, let $I_n=\{i\in I\ |\ a_i\geqslant\frac{1}{n}\}$, since $a_i>0$ for all $i\in I$, we have $$ I=\bigcup_{n\in\mathbb{N}^*}I_n $$ If $I$ is not countable, then at least one $I_n$ is uncountable, let $p\in\mathbb{N}^*$ such that $I_p$ is not countable. There exists $J\subset I_p$ a finite set such that $\text{card}(J)\geqslant p(S+1)$ and since $a_i>0$ for all $i\in I$ we have : $$ S\geqslant\sum_{i\in J}a_i\geqslant\sum_{i\in J}\frac{1}{p}=\frac{\text{card}(J)}{p}\geqslant S+1>S $$ which is not.