Timeline for Asymptotics of the sum $\sum_{n=1}^\infty \frac{x^n}{n^n}$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2022 at 4:26 | answer | added | Qiaochu Yuan | timeline score: 2 | |
| Sep 14, 2022 at 9:24 | answer | added | Gary | timeline score: 3 | |
| Jan 8, 2019 at 2:11 | answer | added | leonbloy | timeline score: 3 | |
| Dec 26, 2018 at 20:06 | answer | added | Maxim | timeline score: 18 | |
| Dec 26, 2018 at 19:56 | vote | accept | Franklin Pezzuti Dyer | ||
| Dec 26, 2018 at 19:49 | comment | added | Clement C. | @Frpzzd Done -- see my edit. | |
| Dec 26, 2018 at 18:58 | comment | added | Clement C. | @Frpzzd A priori, I can mostly improve the constant $c>1/e$ in the upper bound $\log f(x) \leq cx + o(x)$ (and also improve a bit the $o(x)$ in the lower bound). | |
| Dec 26, 2018 at 16:53 | comment | added | Franklin Pezzuti Dyer | @ClementC. Yes, that would be awesome! It would be nice to know $\log f$ within $O(1/x^2)$ or $O(1/x)$, but that might be a stretch... I am interested to see what magic you can work with it, though. :D | |
| Dec 26, 2018 at 10:51 | comment | added | Clement C. | @Frpzzd How accurate do you want your asymptotics to be? I can improve my upper and lower bounds on the asymptotics of $\log f$ by the same technique as my current answer. Is it worth it? | |
| Dec 26, 2018 at 4:25 | comment | added | Antonio Vargas | Using some heuristic reasoning I guess that $$f(x) \sim e^{x/e} \sqrt{\frac{2\pi x}{e}}.$$ | |
| Dec 25, 2018 at 18:47 | answer | added | Clement C. | timeline score: 13 | |
| Dec 25, 2018 at 18:22 | comment | added | zhw. | why do you think it's dominated by any exponential? | |
| Dec 25, 2018 at 18:16 | history | asked | Franklin Pezzuti Dyer | CC BY-SA 4.0 |