In this answer it is explained that a big reason that the nth prime formula discussed with (13) and (14) isn't "useful" is because $\lfloor\frac{x}{b}\rfloor-\lfloor\frac{x-1}{b}\rfloor$ doesn't have "nice analytical properties". This special floor function is $1$ when $x$ divides $b$ and $0$ otherwise. I'm curious, does $$f(x)=\lim\limits_{m\to\infty}\frac{1}{m}\sum_{n=1}^m \cos\left(\frac{2\pi nx}{b}\right)$$ have good analytical properties? In other words, is it "useful"? It essentially performs the same function as the one referenced.
EDIT (Feb 12):
I've found that there is another equivalent expression we can write where, as @user1952009 noted there is for the above expression, there is no inversion of limits in this expression. Observe $$f(x)=\lim\limits_{m\to\infty}\frac{1}{2^m}\sum_{n=1}^{2^m} \cos\left(\frac{2\pi nx}{b}\right)=\lim\limits_{m\to\infty}\cos\left(\frac{\pi x}{b}(1+2^m)\right)\prod_{n=1}^m \cos\left(\frac{\pi x}{b}(2^{n-1})\right)$$ and $m$ is an integer.
Yes.
Weyl's criterion states that a sequence $a_{k}$ is equidistributed modulo 1, meaning its fractional part is uniformly distributed in the region [0,1], if for all non-zero integers $p$ this holds
$$\lim\limits_{m \to \infty} \frac{1}{m} \sum\limits_{k=1}^{m} e^{2\pi i p a_{k}} = 0$$
Your equation is the real part of this expression for the sequence $\frac{x}{b},\frac{2x}{b},\frac{3x}{b},\cdots,\frac{nx}{b},...$ and $p=1$.