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

  • $\begingroup$ 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. $\endgroup$ 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} $\endgroup$ 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. $\endgroup$ 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 $\endgroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.