I'm trying to calculate this sum: $F(N,p) = \sum_{k=1}^\infty \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor $ with $ N \in \mathbb N^*$, $ p \in \mathbb P$ and $\beta >2$
Clearly, $ F(N,p)$ is finite.
We know that $ \forall x \in \mathbb R , \lfloor x \rfloor = x - \{x\} $ and that for all $ p $ and $ N $, $\left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor \notin \mathbb Z $ (thanks to $\beta>2$)
We can use the fourrier series of $\{\}$: $ \forall x \notin \mathbb Z ,\{x\} = \frac12-\frac{1}{\pi}\sum_{l=1}^\infty \frac{sin\left(2\pi lx\right)}{l}$
Then $$F(N,p) = \left(\frac{2N+1}{2}+\frac1\beta\right)\frac{1}{p-1} + G_\infty(N,p)$$ with $$G_{a}(N,p) = \frac1\pi\sum_{k=1}^a\sum_{l=1}^\infty\frac{sin\left(2\pi l\left(\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}\right)\right)}{l}$$
And now, I don't know how to proceed rigorously. I tried somethings below but I don't know if it's correct
I try to use the fact that for every closed interval set without integer, the fourier series of fractional part converge uniformly to swap sums. In fact, it's easy to see that there can only be a finite number of terms in the sum of fractional part that are not equals to their value (values greater than 1...) and that $\lim_{k\to\infty}\left\{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}\right\}=\frac12$ It's what make me think that swapping was possible. Then I used the series expansion of sin function to get something like this:
$$G_\infty(N,p)=\frac1\pi\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{k=1}^\infty\sum_{m=0}^\infty\frac{(-1)^m\left(2\pi p^{-k}\left(\frac{2N+1}{2}+\frac1\beta\right)\right)^{2m+1}}{(2m+1)!}$$
And then
$$G_\infty(N,p)=\frac1\pi\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{m=0}^\infty\frac{(-1)^m\left(2\pi l \left(\frac{2N+1}{2}+\frac1\beta\right)\right)^{2m+1}}{(2m+1)!(p^{2m+1}-1)}$$
What do you think?
EDIT 1 :
My assumptions were wrongs because there is no absolute convergence. However, using this, and not summing over infinity, I can swap sums symbols.
First :
$$ \forall k> \left\lfloor\frac{ln(2N+1)}{ln(p)}\right\rfloor=k_0,\ \ \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor = 0 $$
So
$$ F(N,p) = \sum_{k=1}^{k_0} \left\lfloor{\frac{2N+1}{2p^k}+\frac12+\frac{1}{\beta p^k}}\right\rfloor = \left(\frac{2N+1}{2}+\frac1\beta\right)\frac{1-p^{-k0}}{p-1} + G_{k_0}(N,p) $$
Using the proof here, and the fact that the Fourier series of fractional part is uniformly convergent for every close real interval without integer:
For clarity I pose $g$ as
$$ g_k(x)=\sum_{m=0}^\infty\frac{(-1)^m\left(2\pi x \left(\frac{2N+1}{2}+\frac1\beta\right)\right)^{2m+1}\left(1-p^{-k(2m+1)}\right)}{(2m+1)!(p^{2m+1}-1)} $$
Then:
$$ G_{k}(N,p)=\frac1\pi\sum_{l=1}^{\infty} \frac{(-1)^lg_k(l)}{l} $$
Then I applied Ramanujan's master theorem :
$$ f_k(x)=\sum_{i=0}^\infty \frac{\psi_k(i)(-x)^i}{i!} $$$$ \int_0^\infty x^{s-1}f_k(x)dx = \Gamma(s)\psi_k(-s) $$
With some variable change, and posing : $$ \psi_k(x)=\frac{\Gamma(x+1)\left(1-p^{-k(2x+1)}\right)}{\Gamma(2x+2)\left(p^{2x+1}-1\right)} $$ $$ \alpha = 2\pi \left(\frac{2N+1}{2}+\frac1\beta\right) $$
We get :
$$ \int_0^\infty x^{2(s-1)}g_k(x)dx=\frac{1}{2\alpha^{2s-1}}\Gamma(s)\psi_k(-s) $$
And finally $$ \lim_{s\to\frac12} \frac{1}{2\alpha^{2s-1}}\Gamma(s)\psi_k(-s) = \int_0^\infty \frac{g_k(x)}{x}dx = \frac{\pi k}{2} $$
I suppose it's quite a good result but do you have any idea of how to evaluate $G_k(N,p)$ using this ?
