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

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).

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

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

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

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

• $$a = 2 \land b = \frac{3}{4}$$:

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

• Actually, consider $a+1<b\le0$. That is, the series will converge for $a<b-1$ for any $a,b\in\mathbb R$. – Simply Beautiful Art Jul 28 '17 at 11:00
• For $a<0$, you can use the limit comparison with the sequence $\frac{1}{n^{b-a}}$ to conclude your sequence is convergent for $(b-a)>1$ and divergent for $(b-a)\leq 1$. – Peter Košinár Jul 28 '17 at 22:22
• For $b = 1$ and all $a > 0$ see here. That was not easy and the region where $0 < b < 1$ and $a > 1$ probably no easier to analyze. – RRL Jul 30 '17 at 0:33
• @TeM One thing to remark is that the solution by David Speyer is for all "integers" $a>0$ and $b=1$. In such case we indeed have convergence for $b>1-\epsilon(a)$ for some function $\epsilon(a)$ satisfying $\epsilon(a)\rightarrow 0$ as $a\rightarrow\infty$. – Sungjin Kim Jul 7 '19 at 13:51
• @TeM although David Speyer’s solution is for integer $a>0$, $b=1$, the series is convergent for indeed ”ALL” $a>0$, $b=1$. The method of mine uses Van Der Corput’s method. The application or Van Der Corput for integer $a$ is Weyl’s inequality in David Speyer’s answer. The method can be applied to real $a>0$ too. More about the method are outlined in chapter 4 and 5 of Titchmarsh’s book. – Sungjin Kim Jul 8 '19 at 16:25

The series diverges if $$0, $$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 $$0 and $$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

Let $$0. We apply the above with $$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

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 $$k\in\mathbb{Z}\cap [2,\infty)$$, $$k-1, and $$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

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 $$N\geq 1$$, $$\sum_{N Here, the implied constant depends only on $$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}$$.

• Fantastic, now also what is suggested here (at.yorku.ca/cgi-bin/…) is clear; unfortunately, not having a purely mathematical preparation, some concepts escape me. Thank you! – TeM Jul 4 '19 at 16:32
• Thank you very much for the demonstrations provided! – TeM Jul 9 '19 at 6:48

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 $0 < a < b \leqslant 1 - a$.

• It's a pretty good thing, I have to sleep on and I'll tell you! For now thank you! – TeM Jul 28 '17 at 23:25
• You're welcome. This should fill some of the remaining area. – RRL Jul 28 '17 at 23:38
• Also when $0 < a < b$ and the integral test applies, the integral $\int_0^\infty \frac{\sin x^a}{x^b} \, dx$ diverges when $b \leqslant 1-a$ and, hence, the series diverges as well. – RRL Jul 29 '17 at 16:21
• Here we have an integral of the form $\int_1^\infty u^\alpha \sin u \, du$ with $\alpha \geqslant 0$. This obviously diverges, but I added a proof to the answer. – RRL Jul 29 '17 at 20:00
• Now I understand, thank you for your patience! I've updated the image above. – TeM Jul 29 '17 at 21:35