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Nov 29, 2010 at 5:33 vote accept Georgii
Nov 29, 2010 at 5:33
Nov 29, 2010 at 3:15 comment added Pete L. Clark @Andy: no worries.
Nov 29, 2010 at 2:31 comment added Andy Putman Whoops, there were several votes to close already so I didn't read the question as closely as I should have. Sorry!
Nov 29, 2010 at 1:45 answer added Andrés E. Caicedo timeline score: 12
Nov 29, 2010 at 0:46 history reopened José Figueroa-O'Farrill
Joel David Hamkins
Akhil Mathew
Qiaochu Yuan
HJRW
Nov 28, 2010 at 23:28 comment added Pete L. Clark This extension of the Riemann Rearrangement Theorem is due to W. Sierpinski. It was established in a 1911 paper, Sur une propriété des séries qui ne sont pas absolument convergentes. From what I recall the proof is indeed more involved than the usual rearrangement argument. (I am not quite sure why the question was closed.)
Nov 28, 2010 at 17:33 comment added Georgii The statement I wrote about isn't a Riemann's theorem, which is well known. I consider only the permutations of NEGATIVE part of the series. In this case the proof of Riemann's theorem fails.
Nov 28, 2010 at 16:58 history closed Andrés E. Caicedo
Andy Putman
Nate Eldredge
Bill Johnson
Robin Chapman
too localized
Nov 28, 2010 at 15:54 comment added Andy Putman This site is intended for research-level mathematics. The FAQ lists a number of other sites (eg math.stackexchange.com) which are more appropriate for lower-level questions like this. FYI, your question is answered in Rudin's "Principles of mathematical analysis".
Nov 28, 2010 at 15:51 comment added J. M. isn't a mathematician Have you seen en.wikipedia.org/wiki/Riemann_series_theorem already?
Nov 28, 2010 at 15:35 history asked Georgii CC BY-SA 2.5