Timeline for Is there a known condition for partial sums of a decreasing positive sequence to take all values up to the total sum?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 19, 2021 at 22:03 | vote | accept | John Bentin | ||
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
| May 26, 2020 at 19:14 | comment | added | Martin Sleziak | I will add a link to my on answer on Mathematics which contains some references: Surjective Function from a Cantor Set. | |
| May 25, 2020 at 20:25 | history | edited | Will Brian | CC BY-SA 4.0 |
deleted 14 characters in body
|
| May 25, 2020 at 20:14 | comment | added | John Bentin | (+1) Thank you for a beautiful answer! | |
| May 25, 2020 at 20:07 | history | edited | Will Brian | CC BY-SA 4.0 |
added 20 characters in body
|
| May 25, 2020 at 20:05 | comment | added | Will Brian | @JohnBentin: Yes -- this is the "necessity" argument for that half of my condition. But it should still be included in the answer because, as Iosif Pinelis points out in his comment, the other half of my condition is not enough by itself. It's the conjunction of the two statements that is both necessary and sufficient. | |
| May 25, 2020 at 19:45 | comment | added | John Bentin | $\lim_{n\to\infty}a_n=0$ is entailed by the original condition, because $a$ can be arbitrarily close to zero. | |
| May 25, 2020 at 19:42 | history | edited | Will Brian | CC BY-SA 4.0 |
added 233 characters in body
|
| May 25, 2020 at 19:40 | comment | added | Will Brian | @IosifPinelis: I was assuming that $\lim_{n \rightarrow \infty} a_n = 0$. I just failed to notice that this wasn't part of the question. I'll edit to clarity. As for the rest of it, it looks like maybe we had pretty much the same idea at the same time! | |
| May 25, 2020 at 19:37 | comment | added | Iosif Pinelis | Your condition holds when e.g. $a_n=1+1/(n+1)$, whereas then $a=1$ is not a partial sum. | |
| May 25, 2020 at 19:16 | history | answered | Will Brian | CC BY-SA 4.0 |