Timeline for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2010 at 7:05 | vote | accept | José Hdz. Stgo. | ||
| Nov 8, 2009 at 18:04 | answer | added | Marko Amnell | timeline score: 1 | |
| Nov 8, 2009 at 16:22 | answer | added | Hugh Thomas | timeline score: 55 | |
| Nov 8, 2009 at 13:54 | answer | added | Gerald Edgar | timeline score: 16 | |
| Nov 8, 2009 at 5:15 | history | edited | José Hdz. Stgo. | CC BY-SA 2.5 |
deleted 288 characters in body
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| Nov 8, 2009 at 5:11 | answer | added | Terry Tao | timeline score: 47 | |
| Nov 8, 2009 at 5:10 | comment | added | Harrison Brown | There are still fairly trivial answers to your question -- for instance, take the set 2, 4, 5, 8, 11, 14, ..., p_n, p_{n+1} + 1, and this satisfies your conditions. I'm not sure what you're trying to ask, here. | |
| Nov 8, 2009 at 5:03 | history | edited | José Hdz. Stgo. | CC BY-SA 2.5 |
added 33 characters in body; deleted 7 characters in body
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| Nov 8, 2009 at 4:56 | comment | added | Rob Harron | btw if you take A to be the set {1} union {p+1 : p odd prime}, then sum of 1/a diverges, A contains no primes, and the kth member is greater than the kth prime for all k>1. | |
| Nov 8, 2009 at 4:48 | history | asked | José Hdz. Stgo. | CC BY-SA 2.5 |