Conjecture: $\lim_{x\to\infty} \frac{256}{163x}\sum_{n=1}^{\lfloor x\rceil} |\sin n|=1$

Question

I conjecture: $$\lim_{x\to\infty} \frac{256}{163x}\sum_{n=1}^{\lfloor x\rceil} |\sin n|=1$$

Is this provable? If this is false, then can I have a function $f$ such that:

$$\lim_{x\to\infty} f(x)\sum_{n=1}^{\lfloor x\rceil} |\sin n|=1$$

Answer 1

We will show that $$\lim_{N \to +\infty} \frac{\pi}{2N} \sum_{n = 1}^N | \sin n| = 1.$$ Then your statement (with $256/163$ replaced by $\pi/2$) will follow since $\lfloor x \rceil /x \to 1$ as $x \to + \infty$.

Let $f(x)$ be any left-continuous step function on $[0,\pi]$ satisfying $f(x) \leq \sin x$. Say $0 = x_0 < x_1 < \dots < x_r = \pi$ and $f(x) = a_k$ on $[x_{k-1},x_k)$. Then $$\sum_{n = 1}^N | \sin n| \geq \sum_{n=1}^N f(n \operatorname{mod} \pi) = \sum_{k=1}^r a_k S_{k,N}$$ where $S_{k,N}$ is the number of elements in the set $\{n \operatorname{mod} \pi\ \mid n = 1, \dots N\} \cap [x_{k-1},x_k).$ Since the sequence $n \operatorname{mod} \pi$ is equidistributed in $[0,\pi]$, we have $S_{k,N}/N \to (x_k - x_{k-1})/\pi$ as $N \to +\infty$. Thus $$\frac{\pi}{2N} \sum_{n = 1}^N | \sin n| \geq \frac{\pi}{2}\sum_{k=1}^r a_k \frac{S_{k,N}}{N} \xrightarrow[N \to +\infty]{} \frac{1}{2} \sum_{k=1}^r a_k (x_k - x_{k-1}) = \frac{1}{2} \int_0^{\pi} f(x) \, dx.$$ Therefore $$\liminf_{N \to +\infty} \frac{\pi}{2N} \sum_{n = 1}^N | \sin n| \geq \frac{1}{2} \int_0^{\pi} f(x) \, dx.$$ The supremum of the right-hand side as $f$ is allowed to vary among left-continuous step functions $\leq \sin x$ is $\frac{1}{2} \int_{0}^{\pi} \sin x \, dx = 1$, hence
$$\liminf_{N \to +\infty} \frac{\pi}{2N} \sum_{n = 1}^N | \sin n| \geq 1.$$

Arguing similarly with step functions $g(x) \geq \sin x$, we find that $$\limsup_{N \to +\infty} \frac{\pi}{2N} \sum_{n = 1}^N | \sin n| \leq 1.$$

Combining the two inequalities, we have proved that $$\lim_{N \to +\infty} \frac{\pi}{2N} \sum_{n = 1}^N | \sin n| = 1.$$