On the convergence of $\sum_{n = 1}^\infty\frac{\sin\left(n^a\right)}{n^b}$

Question

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.

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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).

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

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

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

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

• $a = 2 \land b = \frac{3}{4}$:

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

Answer 1

If $a > 0$ then the series converges if $b > \max(a,1-a)$.

The general principle is that the series $\sum_{n=1}^\infty f(n)$ and the integral $\int_1^\infty f(x) \, dx$ converge and diverge together if $\int_1^\infty |f'(x)| \, dx < \infty$. This is proved here.

In this case, $f(x) = \sin x^a /x^b$ and

$$\int_1^\infty |f'(x)| \, dx= \int_1^\infty \left| \frac{-b\sin x^a}{x^{b+1}} + \frac{a\cos x^a}{x^{b-a+1}} \right| \, dx \\ \leqslant \int_1^\infty \frac{b}{x^{b+1}} \, dx + \int_1^\infty \frac{a}{x^{b-a+1}} \, dx, $$

so the integrals on the RHS converge and theorem is applicable if $b > 0$ and $b > a$.

We have

$$\int_1^\infty \frac{\sin x^a}{x^b}\, dx = \frac{1}{a}\int_1^\infty \frac{\sin u}{u^{(b+a-1)/a}}\, du, $$

which converges by the Dirichlet test when $a > 0$ and $b+a-1 > 0 \implies b > 1 - a.$

If $b \leqslant 1 - a$ then the integral diverges. In this case we have an integral of the form $\int_1^\infty u^\alpha \sin u \, du$ where $\alpha \geqslant 0$. Divergence is obvious if $\alpha = 0$ since $\int_1^c \sin u \, du = \cos 1 - \cos c$.

For $\alpha > 0$ we have for any positive integer $k$,

$$\left|\int_{2k\pi}^{2(k+1)\pi} u^\alpha \sin u \, du \right| = \int_{2k\pi}^{2(k+1)\pi} u^\alpha \sin u \, du \geqslant (2 k \pi)^\alpha\int_{2k\pi}^{2(k+1)\pi} \sin u \, du = 2(2k\pi)^\alpha.$$

Since the RHS tends to $\infty$ as $k \to \infty$, the Cauchy criterion is violated and the improper integral must diverge.

Hence, the series diverges when $0 < a < b \leqslant 1 - a$.