Timeline for Sum with the fractional part function
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 21 at 13:16 | comment | added | GH from MO | Yes, $(\lambda^n+1)$ is also very similar to Fibonacci. I guess any linear binary recurrence sequence leads to a result similar to the one at mathoverflow.net/q/467934. | |
| Apr 21 at 8:00 | comment | added | Babar | @GH from MO Thanks! I was applying your arguments on Fibonacci but I had not seen the idea of intervals that Henri Cohen pointed out. Unless I am mistaken, I also manage to show that: $\sum_{k=1}^{n}\left\{ \frac{\lambda^{n}+1}{\lambda^{k}+1}\right\} =\log(2)n+O \left(n^{1/2}\right)$ where $\lambda\geq2$ is an integer. | |
| Apr 20 at 23:22 | comment | added | GH from MO | Henri Cohen's argument yields an error term of $O(n^{1/2})$. Namely, by summing up the contributions of the intervals $(n/(m+1),n/m]$ for $m\in\{1,\dotsc,M\}$, you can see that the sum is $n\log 2+O(M+N/M)$. Now choose $M=\lfloor\sqrt{n}\rfloor$ to conclude that the sum is $n\log 2+O(\sqrt{n})$. This is much the same as (in fact simpler than) my response for mathoverflow.net/q/467934, which is not surprising as the sequence $(2^n+1)$ is very similar to the Fibonacci sequence | |
| Apr 20 at 22:28 | comment | added | Babar | Thanks this confirms the assertion for the main term. But I think we need to be finer to have the error term in $O(n^{1/2})$. My real guess is that it is in $O(n^{1/4+\epsilon})$ as for the Dirichlet divisor problem. | |
| Apr 20 at 20:17 | comment | added | Henri Cohen | For $m$ even the sum for $n/(m+1)<k\le n/m$ is a $O(1)$, and for $m$ odd it is $n(1/m-1/(m+1))+O(1)$, so indeed the sum is $n\log(2)+O(1)$ (at least if the $O(1)$ form a convergent series). | |
| Apr 20 at 19:30 | history | asked | Babar | CC BY-SA 4.0 |