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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"?

Gareth Ma
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1 Answers1

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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.

Tuvasbien
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  • I am sorry but can you explain the line "There exists $J \in I_p$ a finite set such that card(J) $\geq$ p(S+1)"? Why is that? Oh wait I think i get it - because $p(S+1)$ is a finite number ($S$ is finite by assumption), and since $I_p$ is not countable, $J$ can be arbitrarily big? – Gareth Ma Apr 12 '20 at 18:18
  • Also I'll wait for @Reveillark's answer and compare, which is apparently with measure theory. – Gareth Ma Apr 12 '20 at 18:18
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    $I_p$ is not countable thus it contains an infinite number of elements. – Tuvasbien Apr 12 '20 at 18:20
  • Wait I think I am being stupid here, but can't you do the same argument for a countable $I_p$? There still exists a finite set with the property, correct? And the lines following are still true. (So where is the "uncountable" property actually used) – Gareth Ma Apr 12 '20 at 18:28
  • If $I$ is infinite but countable, it is possible that all the $I_n$ are finite, take $a_i=\frac{1}{i}$ for $i\in\mathbb{N}^*$ for instance. But indeed if at least one $I_n$ is countable, it is the same. – Tuvasbien Apr 12 '20 at 18:32
  • I see, so for a countable set you can always ensure that? That for some definition of $I_n$, perhaps not necessarily the one given, that all $I_n$ is finite? Actually no wait I am being stupid again $I_n={n}}$ works? – Gareth Ma Apr 12 '20 at 18:33
  • Yes $I_n={n}$ works. – Tuvasbien Apr 12 '20 at 18:35
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    Thank you for your time. – Gareth Ma Apr 12 '20 at 18:36