what value series $\sum_{r=1}^\infty\frac{\lfloor rp \rfloor}{2^r}$ converges?

I have been wondering about series of $$S=\sum_{r=1}^\infty\frac{\lfloor rp \rfloor}{2^r}$$ where p is a constant positive real number and $\lfloor\cdot\rfloor$ is floor function.

I know it converges because $\frac{\lfloor rp \rfloor}{2^r}\leq \frac{ rp }{2^r}$ where $r$ is positive integer and $p$ is constant positive real number; since series $\sum_{r=1}^\infty\frac{rp}{2^r}$ is a A.G.P series with common ratio less than $1$, the given series converges.

But to what value our series $S$ converges? How to find it? What is the method?

Actually I have to find convergence value for $p=\frac{1+\sqrt{5}}{2}$.

Also what would happen if it was ceiling function and nearest integer function??

I couldn't find any method to find value of convergence,any hint would be appreciable,thanks

• Are you interested in the value of the infinite series $\sum_{r=1}^\infty \lfloor rp\rfloor/2^r$? This seems likely, since you're talking about convergence. Dec 30 '14 at 6:06
• Yes I want to know the method and to know the convergence value when p=$\frac{1+\sqrt{5}}{2} Dec 30 '14 at 6:57 • Do you have any reason to suspect there is a simple answer to this question? In general, sums containing floor-brackets tend to be very difficult. Dec 30 '14 at 7:33 • Actually I was doing a question with nearest integer function and golden ratio,so I was wondering what would happen if it was floor function.besides I just want to know what is the general method for this type of question Dec 30 '14 at 8:41 • It might help to use$\lfloor rp\rfloor = rp - \frac12 + \frac1\pi \sum_{k=1}^\infty \frac{\sin 2\pi krp}k$(for irrational$p$... for rational$p$, other techniques must be used). Jan 8 '15 at 10:42 1 Answer As a start, based on similar sums I vaguely recall, write, where$\{z\}$is the fractional part of$z$,$\begin{array}\\ S_n &=\sum_{r=1}^{n}\frac{\lfloor rp \rfloor}{2^r}\\ &=\sum_{r=1}^{n}\frac{rp-\{rp\}}{2^r}\\ &=\sum_{r=1}^{n}\frac{rp}{2^r}-\sum_{r=1}^{n}\frac{\{rp\}}{2^r}\\ &=p\sum_{r=1}^{n}\frac{r}{2^r}-\sum_{r=1}^{n}\frac{\{rp\}}{2^r}\\ &= pU_n - V_n \end{array} U_n$is a standard sum. For$V_n$, if you look at the binary representation of$p$, you might be able to see when that fractional part changes. That's all I can think of for now. • How to find$V_n$for$p=\frac{1+\sqrt{5}}{2}\$ Dec 30 '14 at 7:02
• Don't know. Maybe look at the continued fraction. Probably something in the Fibonacci Quarterly. Dec 31 '14 at 3:20