Timeline for If $\sum\limits_{n=0}^{\infty} a_n\ $ is a conditionally convergent series, then is $\sum\limits_{n=0}^{\infty} |a_n+a_{n+1}|\ $ convergent?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2021 at 22:44 | comment | added | Adam Rubinson | Yeah sure. I see. | |
| Jan 14, 2021 at 22:42 | comment | added | Clement C. | Oh, that'd just be useful if you require extra conditions (e.g., non-zero terms) -- there'd still be counterexamples even then. | |
| Jan 14, 2021 at 22:41 | comment | added | Adam Rubinson | Yeah but can't the counter-example be generalised by using $0$'s as the interwoven terms? Why do we need the interwoven terms to be terms in a conditionally convergent series? | |
| Jan 14, 2021 at 22:39 | comment | added | Clement C. | @AdamRubinson The counter example can be generalized to 3, or any fixed number of terms, instead of two consecutive; and some other patterns than just consecutive terms as well. | |
| Jan 14, 2021 at 22:37 | comment | added | Adam Rubinson | @Steven I'm not sure I follow. | |
| Jan 14, 2021 at 22:28 | comment | added | Steven Stadnicki | Note that this generalizes pretty broadly; the interwoven terms could e.g. be terms of an absolutely convergent series rather than zero, and they can be laid in whatever pattern is desired, to show that the answer is the same for whatever finite subsequence you might choose to average over. | |
| Jan 14, 2021 at 22:24 | vote | accept | Adam Rubinson | ||
| Jan 14, 2021 at 22:22 | comment | added | Clement C. | There might be more natural/less contrived, but it does the job. :) | |
| Jan 14, 2021 at 22:16 | comment | added | Adam Rubinson | Wow, I didn't think of that at all. | |
| Jan 14, 2021 at 22:14 | history | answered | Clement C. | CC BY-SA 4.0 |