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Jun 18, 2017 at 2:14 comment added Joel David Hamkins For reference, our paper (with the OP as co-author) is available at The rearrangement number (arxiv:1612.07830), and see also my introductory talk. The paper provides various sufficient criteria for a family to have the rearrangement property, such as theorem 11: every co-meager family of permutations successfully tests for absolute convergence. Necessary criteria arise from the lower bounds on the rearrangment number.
Jun 17, 2017 at 22:04 comment added reuns Clearly if $l = \sum_n a_n$ then for any $\sigma \sim Id$ : $l = \sum_n a_{\sigma(n)}$. Intuitively, the converse should hold : that for any $\sigma \not\sim Id$ we can find a series such that $l = \sum_n a_n \ne \sum_n a_{\sigma(n)}$. In that case you really want to find an explicit description of $\text{Sym}(\mathbb{N})/ \sim $ (and you can forget about conditionally convergent series)
Jun 17, 2017 at 21:37 comment added Michael Hardy @reuns : Can you elaborate a bit? Why would that answer this?
Jun 17, 2017 at 21:36 comment added Michael Hardy @reuns : ok, I'm thinking about what your question means. In the meantime, do you see the difference between the following, and the difference betwen their MathJax coding, and why someone might consider the former incorrect?: $$ \begin{align} & \operatorname{Sym}(\mathbb N)/\sim \\ \\ & \operatorname{Sym}(\mathbb N)/{\sim} \end{align} $$
Jun 17, 2017 at 21:32 comment added reuns Isn't the problem to forget about convergent series and really look at $Sym(\mathbb{N})/ \sim \ \ $ where $\sim$ is the equivalence relation defined by Robert Israel ?
Jun 17, 2017 at 19:30 history edited Michael Hardy CC BY-SA 3.0
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Jun 17, 2017 at 19:20 history edited Michael Hardy CC BY-SA 3.0
typo
Jun 17, 2017 at 18:48 history asked Michael Hardy CC BY-SA 3.0