Prove that $\lim_{N \to +\infty}\sum^N_{n=1}{|\sin n^\alpha| \over n^\beta} = +\infty$

Question

I want to prove that $$\lim_{N \to +\infty}\sum^N_{n=1}{|\sin n^\alpha| \over n^\beta} = +\infty \quad \forall\alpha \gt 0, \; \forall \beta \le 1, \; \alpha, \beta \in \Bbb R.$$ In other words, I want to show that, under the specified restrictions on the parameters, the series diverges to $+\infty$ and is not oscillating or irregular, for instance. I have verified that the series indeed does this by looking at its graph for various parameter choices.

I observed that, $\forall \alpha \gt 0$:

• $0 \lt \beta \le 1 \implies \lim_{n \to +\infty}a_n=0$, so the necessary condition for convergence is met. To prove positive divergence, I tried studying the associated improper integral: $$\int^{+\infty}_1{|\sin x^\alpha| \over x^\beta}\ dx = \int^{\sqrt[\alpha]{\pi}}_1{|\sin x^\alpha| \over x^\beta}\ dx + \sum^{+\infty}_{k=1} \int^{\sqrt[\alpha]{(k+1)\pi}}_{\sqrt[\alpha]{k\pi}}{|\sin x^\alpha| \over x^\beta}\ dx.$$ The integrand is $\ge 0$ within the intervals considered, so the integral on the left diverges to $+\infty$. Since the integral test only works if the sequence is monotone decreasing, I'm not sure I can apply this result to show that also $\sum^{+\infty}_{n=1}a_n = {+\infty}$;
• $\beta \le 0 \implies \nexists \lim_{n \to +\infty}a_n$, so the series is not convergent. Still, I want to show that the stronger statement above is true.

An idea that occurred to me, but that I wasn't able to formalize, was that $n \in \Bbb N \implies n^\alpha \neq \pi \; \forall \alpha \gt 0$, since $\pi$ is a transcendental number and as such cannot be construed out of any combination of real numbers, which in turn implies that $|\sin{n^\alpha}| \gt 0 \; \forall \alpha \gt 0.$ (However, I do not have the knowledge necessary to back up this claim.) Wouldn't this imply that $a_n$ is bounded below by some constant $k \gt 0$, meaning that the series would diverge to $+\infty$?

If anybody could help shed some light, I'd much appreciate it.

Answer 1

This is a a partial solution for the case $\alpha=1$.

Here are some special cases in which the answer is no. The general case where $\alpha\in[0,\infty)\setminus\mathbb{N}$ seems to be more involved.

• ($\alpha=1$). Notice $|\sin n|>\sin^2(n)$ and $$\sin^2(n)=\frac{1-\cos 2n}{2}$$ Then $$ \frac{|\sin n|}{n^\beta}\geq \frac12\frac{1}{n^\beta}-\frac12\frac{\cos 2n}{n^\beta} $$

The series $\sum_n\frac{\cos 2n^\alpha}{n^\alpha}$ converges by Dirichlet-Abel's test; and $\sum_n\frac{1}{n^\beta}$ diverges for $0<\beta\leq 1$.

• More general, when $\alpha\in\mathbb{N}$, the sequence $\{e^{ in^\alpha}:n\in\mathbb{N}\}$ is equidistributed in $\mathbb{S}^1$ and so, $$\frac{1}{n}\sum^n_{k=1}|\sin k^\alpha|\xrightarrow{n\rightarrow\infty}c_\alpha>0$$ which means that for $0<\beta<1$ $$\frac{1}{n^\beta}\sum^n_{k=1}|\sin k^m|=n^{1-\beta}\frac{1}{n}\sum^n_{k=1}|\sin k^\alpha|=\left\{\begin{array}{lcr} \infty & \text{if} & 0<\beta<1\\ c_\alpha &\text{if}& \beta=1 \end{array}\right. $$ This means that $\sum_n\frac{|\sin n^\alpha|}{n^\beta}$ diverges (other wise $\frac{1}{n^\beta}\sum^n_{k=1}|\sin n^\alpha|\xrightarrow{n\rightarrow\infty}0$ by Kronecker's lemma).

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Edit:

It is given as an excercise in Kuipers, L. and Nienderreiter, Uniform distributions of Sequences, Dover Publications, New York, 2006, pp. 31 Ex. 3.9 that the sequence $\{a n^\sigma:n\in\mathbb{N}\}$, $a>0$ and $\sigma\in[0,\infty)\setminus\mathbb{N}$ is equidistributed in $(0,1)$; This implies that if $\alpha\in[0,\infty)\setminus\mathbb{N}$, $\frac1n\sum^n_{k=1}|\sin k^\alpha|=c_\alpha>0$ in which case, we also have that $\sum_n\frac{|\sin n^\alpha|}{n^\beta}=\infty$, $0<\beta\leq 1$.