Timeline for Does the limit of $x_n$, defined by $x_{n+1}=1/(m+1-nx_n)$ exist?
Current License: CC BY-SA 4.0
11 events
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| Sep 27, 2022 at 11:17 | vote | accept | math110 | ||
| Sep 27, 2022 at 11:10 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Sep 27, 2022 at 11:02 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 27, 2022 at 5:50 | comment | added | GH from MO | @PietroMajer I showed below that even $x_n\to 0$ is impossible. So the limit does not exist. | |
| Sep 27, 2022 at 5:49 | comment | added | Pietro Majer | The only possible limit is 0, for if $x_n$ converges to a nonzero limit, $|nx_n| $ goes to infinity and $x_{n+1}$ goes to 0. | |
| Sep 27, 2022 at 5:47 | history | edited | GH from MO |
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| Sep 27, 2022 at 5:47 | answer | added | GH from MO | timeline score: 8 | |
| Sep 27, 2022 at 4:35 | comment | added | Shahrooz | It seems that since we must have $(m+1)^2\geq 4n$, for any fixed $m$, the limit does not exist. | |
| Sep 27, 2022 at 2:13 | comment | added | Sam Hopkins | Is it even clear you don't ever get division by zero? | |
| Sep 27, 2022 at 1:13 | comment | added | math110 | First of all, I think this problem is still somewhat researchable, because there are still a lot of less systematic conclusions about this nonlinear recursive approximation problem, and second, this problem seems easy, but it seems that I have discussed with many people that this problem is difficult and worth studying, and finally my account can not be used in that platform，Thanks | |
| Sep 27, 2022 at 0:47 | history | asked | math110 | CC BY-SA 4.0 |