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

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.
