Timeline for A conjecture relating an integral and a sum, the floor function and squares
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2022 at 19:06 | comment | added | user475930 | That series1/(n-1)^n+1 is in the estimation in the sum as well, but there are an infinite amount of correction terms that bring it down below one, mainly because Stieltjes(1) is negative. | |
| Feb 7, 2022 at 15:53 | comment | added | user475930 | The computed limit for the sum is x + 0.9885435... and is terribly hard to compute due to needing the Stieltje constants. Going to make a new question on here with my 'proof' of the sum. | |
| Feb 3, 2022 at 3:00 | comment | added | user475930 | Thank you for not spoiling the fun :D That was supremely fun. | |
| Jan 26, 2022 at 22:45 | vote | accept | CommunityBot | ||
| Jan 26, 2022 at 22:26 | comment | added | LSpice | @TheHoyt, whether or not begging is appropriate, you made 5 requests in about 20 minutes. @Random may or may not help, but at least it is appropriate to give them the courtesy of a bit of time to consider and respond, if they choose to do so. | |
| Jan 26, 2022 at 21:41 | comment | added | Random | By the way, Wolfram Alpha says that already for $t = 1.01$ we have $\sum_{n = 2}^{\infty} \frac{1}{(n t - 1)^{n + 1}} > 1$, that is already for $x = 10^4$ the floor of the integral is equal to $x + 1$. | |
| Jan 26, 2022 at 21:31 | history | answered | Random | CC BY-SA 4.0 |