Timeline for 1+2+3+4+… and −⅛
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2023 at 3:43 | vote | accept | James Propp | ||
| Feb 25, 2023 at 20:36 | comment | added | Caleb Briggs | You can probably find a sensible way to represent what is going on with this approach: terrytao.wordpress.com/2010/04/10/… The idea is that your changes to the sum are invisible in the formal sum expression, but are captured by the smoothing term. | |
| Feb 25, 2023 at 20:04 | comment | added | LSpice | @GeraldEdgar, re, surely the same objection can be made to any method of summing divergent series? It is precisely because the familiar mathematics of absolutely convergent series don't guarantee a priori that all these rearrangements will produce the same answer that one is interested in why, despite not ‘having’ to be the same, they nonetheless are the same. | |
| S Feb 25, 2023 at 13:51 | history | suggested | user3840170 | CC BY-SA 4.0 |
typography in the title, \frac in the body
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| Feb 25, 2023 at 9:58 | review | Suggested edits | |||
| S Feb 25, 2023 at 13:51 | |||||
| Feb 25, 2023 at 8:14 | history | became hot network question | |||
| Feb 25, 2023 at 3:04 | answer | added | KConrad | timeline score: 16 | |
| Feb 25, 2023 at 3:01 | answer | added | Oscar Lanzi | timeline score: 29 | |
| Feb 25, 2023 at 1:31 | review | Close votes | |||
| Feb 26, 2023 at 15:39 | |||||
| Feb 25, 2023 at 1:20 | comment | added | Gerald Edgar | I think the "larger story" is: the series diverges. So all of these grouping steps fail. Note: $x-1=9x$ has solution $x=+\infty$, which is more nearly correct than $x=-1/8$. | |
| Feb 25, 2023 at 0:46 | comment | added | Gro-Tsen | @SamHopkins The alternating version is Abel-summable, so pretty much any summation technique should give the same result. And it doesn't seem to be part of a general phenomenon, because $t + 2t^2 + 3t^3 + \cdots$ doesn't appear to be summable using the technique in the question ($t=1$) and in your comment ($t=-1$). | |
| Feb 25, 2023 at 0:30 | comment | added | Sam Hopkins | Silly comment. Let $z = 1 - 2 + 3 - 4 + \cdots$. Your method gives $z-1 = (-2+3-4) + (5-6+7) + (-8+9-10) + \cdots = -3 + 6 - 9 + \cdots = -3z$ so that $4z=1$, i.e., $z=\frac{1}{4}$. This is the "correct" value (en.wikipedia.org/wiki/…) | |
| Feb 25, 2023 at 0:22 | history | edited | LSpice | CC BY-SA 4.0 |
Uniform dots
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| Feb 25, 2023 at 0:13 | history | asked | James Propp | CC BY-SA 4.0 |