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Feb 19, 2021 at 22:03 vote accept John Bentin
May 26, 2020 at 7:02 comment added John Bentin @MiloBrandt : Your answer showed the sufficiency of the bound $a_n \leq \sum_{m > n}a_m$, but not the necessity (which wasn't needed to answer the question).
May 26, 2020 at 1:56 comment added Milo Brandt I wrote an answer a while ago on Math Stack Exchange that answered exactly this: here. I think it's entirely in line with Will Brian's answer.
May 26, 2020 at 1:29 history became hot network question
May 25, 2020 at 19:16 answer added Will Brian timeline score: 17
May 25, 2020 at 19:15 answer added Iosif Pinelis timeline score: 6
May 25, 2020 at 18:54 comment added Will Brian OK, thanks -- I'll see if my idea works, and write an answer if it does :)
May 25, 2020 at 18:54 comment added John Bentin @WillBrian : Yes, your sum condition doesn't involve indices comprising arbitrary subsets of $\Bbb N$, only terminal segments of $\Bbb N$. So it's not just a rewrite of the originally stated property.
May 25, 2020 at 18:39 comment added Will Brian I think it's necessary and sufficient to have $a_n \leq \sum_{m > n}a_m$ for all $n$. Is this the kind of condition you have in mind? I'm not sure what you mean by "not involving arbitrary partial sums".
May 25, 2020 at 18:32 comment added Will Brian Even deleting a single term (other than the first term) from the sequence $a_n = 1/2^n$ makes it no longer have your property.
May 25, 2020 at 17:25 history asked John Bentin CC BY-SA 4.0