TITLE: On the convergence of $\sum_{n = 1}^\infty\frac{\sin\left(n^a\right)}{n^b}$
QUESTION [31 upvotes]: Given the infinite series
$$\begin{aligned}\sum_{n = 1}^{\infty}\end{aligned} \frac{\sin\left(n^a\right)}{n^b}$$
with $a,\,b \in \mathbb{R}$, study when it converges and when it diverges.


Easy cases

$\forall\,a \in \mathbb{R}$ we have $\left|\frac{\sin\left(n^a\right)}{n^b}\right| \le \frac{1}{n^b}$ so the series $\color{green}{\text{converges}}$ for $b > 1$.

If $a \le 0$ we have $\frac{\sin\left(n^a\right)}{n^b} \le \frac{1}{n^{b-a}}$ so the series $\color{blue}{\text{diverges}}$ for $b \le a + 1$ and $\color{green}{\text{converges}}$ for $b > a + 1$.

If $a > 0 \, \land \, b \le 0$ we have $\not\exists \begin{aligned}\lim_{n \to \infty} \frac{\sin\left(n^a\right)}{n^b} \end{aligned}$ so the series  $\color{blue}{\text{diverges}}$.

If $a = 1\, \land \, b > 0$ the series $\color{green}{\text{converges}}$ by Abel-Dirichlet's test.


Hard cases

If $0 < a < 1\, \land \, 0 < b \le 1-a$ the series $\color{blue}{\text{diverges}}$ by proof of i707107.

If $0 < a < \frac{1}{2}\, \land \, a < b \le 1-a$ the series $\color{blue}{\text{diverges}}$ by proof of RRL.


If $0 < a < 1\, \land \, b > 1-a$ the series $\color{green}{\text{converges}}$ by proof of i707107.

If $a > 0\, \land \, b > \max(a,\,1-a)$ the series $\color{green}{\text{converges}}$ by proof of RRL.


If $k \in \mathbb{Z}_{\ge 2}, $ $k-1 < a < k\, \land \, b > 1 - \frac{k-a}{2^k-2}$ the series $\color{green}{\text{converges}}$ by proof of i707107.

If $a > 0 \, \land \, b = 1$ the series $\color{green}{\text{converges}}$ by proof David Speyer (+ i707107 in the comments).

If $a = 2 \, \land \, 0 < b \le \frac{1}{2}$ the series $\color{blue}{\text{diverges}}$ (Theorem 2.30 by Hardy&Littlewood).




$\color{red}{\textbf{Open cases}}$

$a = \frac{3}{2} \land b = \frac{1}{4}$:


$a = \frac{3}{2} \land b = \frac{1}{2}$:


$a = \frac{3}{2} \land b = \frac{3}{4}$:


$a = 2 \land b = \frac{3}{4}$:


$a = \frac{5}{2} \land b = \frac{1}{2}$:

REPLY [10 votes]: The series diverges if $0<a<1$, $b>0$ and $a+b\leq 1$
This idea is already discussed in the link provided at $a=b=1/2$ case. 
Let $k$ be a positive integer. If 
$$
2\pi k + \pi/6\leq n^a\leq 2\pi k + 5\pi/6,
$$
then $\sin (n^a)\geq 1/2$. 
The number of $n$'s in this interval is $\asymp k^{(1/a) -1}$. In the interval, we have $n^b\asymp k^{b/a}$. Thus, the contribution of $n$'s to the sum in this interval is $\asymp k^{(1/a)-(b/a)-1}$. 
If $(1/a)-(b/a)-1 \geq 0$, then we have divergence. Note that the partial sum of the series fails to be Cauchy sequence. 
The series converges if $0<a<1$ and $a+b>1$
The case particularly requires a good estimation of the sum by an integral. A crucial lemma for this case is Lemma 4.8 of The Theory of the Riemann Zeta-function written by Titchmarsh. 

Let $f(x)$ be a real differentiable function in the interval $[a,b]$, let $f'(x)$ be monotonic, and let $|f'(x)|\leq \theta <1$. Then
  $$
\sum_{a<n\leq b}e^{2\pi i f(n)} =  \int_a^b e^{2\pi i f(x)} \ dx + O(1).
$$

Let $0<a<1$. We apply the above with $f(x)=\frac{x^a}{2\pi}$. The assumptions of the lemma are satisfied for $x\geq X_0>0$. Then we obtain
$$
\sum_{X_0<n\leq N} e^{in^a}=\int_{X_0}^N e^{ix^a} \ dx + O(1).
$$
Apply change of variable $x^a=t$, then 
$$
\int_{X_0}^N e^{ix^a} \ dx = \int_{X_0^a}^{N^a} e^{it} \frac1a t^{\frac1a -1} \ dt.
$$
Now, integration by parts shows that the above integral is $O(N^{1-a})$. Thus, we have for $t\geq 1$, 
$$
\sum_{n\leq t} \sin(n^a) = O(t^{1-a}).
$$
Then the partial summation yields
$$
\sum_{n=1}^N \frac{\sin(n^a)}{n^b} = O(N^{1-a-b})+b\int_{1-}^N \frac{\sum_{n\leq t} \sin(n^a)}{t^{b+1}} \ dt.
$$
Hence, we have the convergence when $b>1-a$. 
The series converges if $k\in\mathbb{Z}\cap [2,\infty)$, $k-1<a<k$, and $b>1-\frac k{2^k-2}+\frac a{2^k-2}$
We apply Theorems 5.11-5.13 of The Theory of the Riemann Zeta-function written by Titchmarsh. These theorems are proved by Van Der Corput's method. If the  method is applied to integer $a$, we obtain Weyl's inequality, which was used in David Speyer's answer. The method is outlined in Chapter 5 of Titchmarsh's book.

Let $f(x)$ be real and have continuous derivatives up to $k$-th order, where $k\geq 2$. Let $\lambda_k\leq f^{(k)}(x)\leq h\lambda_k$ (or the same for $-f^{(k)}(x)$). Let $b-a\geq 1$, $K=2^{k-1}$. Then 
  $$
\sum_{a<n\leq b} e^{2\pi i f(n)}=O(h^{\frac2K}(b-a)\lambda_k^{\frac1{2K-2}})+O((b-a)^{1-\frac2K}\lambda_k^{-\frac1{2K-2}}).$$

For the series in question, use $f(n)=\frac{n^a}{2\pi}$ over diadic intervals $(N,2N]$, and $\lambda_k\asymp N^{a-k}$. Then for $N\geq 1$, 
$$
\sum_{N<n\leq 2N} \sin(n^a)=O(N^{1-\frac{k-a}{2^k-2}}).
$$
Here, the implied constant depends only on $a$. Then adding up the results for diadic intervals, we obtain for $t\geq 1$, 
$$
\sum_{n\leq t} \sin(n^a)=O(t^{1-\frac{k-a}{2^k-2}}).
$$
Applying partial summation, we obtain 
$$
\sum_{n=1}^N \frac{\sin(n^a)}{n^b} = O(N^{1-\frac{k-a}{2^k-2}-b})+b\int_{1-}^N \frac{\sum_{n\le t} \sin(n^a)}{t^{b+1}}\ dt.
$$
Then the convergence of the series follows if $b>1-\frac k{2^k-2}+\frac a{2^k-2}$.