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Apr 1 at 21:34 comment added GH from MO I updated my post. The result there is now stronger and more general, while the proof is more detailed and more streamlined.
Apr 1 at 10:01 history edited Babar CC BY-SA 4.0
I added a ps thanks to the anwser.
Mar 31 at 21:25 vote accept Babar
Mar 30 at 20:42 history edited GH from MO
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Mar 30 at 20:02 comment added GH from MO More precisely, I don't understand the "close" votes. See my previous comment.
Mar 30 at 19:31 comment added GH from MO I don't understand the downotes. This is a rather tricky limit, and it has a different value than stated in the original post. See my response below.
Mar 30 at 19:30 answer added GH from MO timeline score: 21
Mar 28 at 23:31 comment added Babar I just asked but as it is a consequence of more general work he does not want to communicate certain results. On the other hand, he finds it normal that people have fun searching. The theme was the Dirichlet divisor problem formulated as: $$\sum_{k=1}^{n}\left\{ \frac{n}{k}\right\} =\left(1-\gamma\right)n+O\left(n^{\theta+\varepsilon}\right) $$ and for which increasing functions $f$ do we also have: $$\sum_{k=1}^{n}\left\{ \frac{f(n)}{f(k)}\right\} =C_{f}n+O\left(n^{\theta+\varepsilon}\right) $$ and $f(n)=F(n)$ doesn't work since there are two limits for $C_f$.
Mar 28 at 21:46 comment added Gerry Myerson Maybe you should ask the lecturer?
Mar 28 at 20:51 comment added Babar I know that we have $\left\{ x\right\} =\frac{1}{2}-\frac{1}{\pi}\sum_{j\geq1}\frac{\sin\left(2\pi jx\right)}{j}$ but I can't see the equidistribution theorem you are mentioning.
Mar 28 at 19:31 comment added Babar I thought about it but I can't get much done with Binet's formula. It seems to me that something more subtle is needed. The speaker had cited this example in relation to the problem of Dirichlet divisors so I thought of mathoverflow rather than MSE.
Mar 28 at 19:03 review Close votes
Apr 3 at 11:57
Mar 28 at 18:46 comment added Steven Stadnicki This might be better suited for math.SE; I haven't gone all the way through it but it seems as though it should be fairly straightforward using the usual Fibonacci identities to simplify the fractional-part term and turn it into a Riemann integral limit.
S Mar 28 at 18:24 review First questions
Mar 29 at 0:23
S Mar 28 at 18:24 history asked Babar CC BY-SA 4.0