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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

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  • $\begingroup$ 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. $\endgroup$ – Greg Martin Dec 30 '14 at 6:06
  • $\begingroup$ Yes I want to know the method and to know the convergence value when p=$\frac{1+\sqrt{5}}{2} $\endgroup$ – jjoyk Dec 30 '14 at 6:57
  • $\begingroup$ 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. $\endgroup$ – Eric Stucky Dec 30 '14 at 7:33
  • $\begingroup$ 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 $\endgroup$ – jjoyk Dec 30 '14 at 8:41
  • $\begingroup$ 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). $\endgroup$ – Glen O Jan 8 '15 at 10:42
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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.

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  • $\begingroup$ How to find $V_n$ for $p=\frac{1+\sqrt{5}}{2}$ $\endgroup$ – jjoyk Dec 30 '14 at 7:02
  • $\begingroup$ Don't know. Maybe look at the continued fraction. Probably something in the Fibonacci Quarterly. $\endgroup$ – marty cohen Dec 31 '14 at 3:20

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