Are the partial sums for $\sum_{n=1}^{\infty}\sin(n^a)$ bounded for $a\geq1$ and unbounded for $0<a<1$?

I know that the partial sums of $$\sum_{n=1}^{\infty}\sin(n)$$ are bounded between $$\frac{\cos\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2} \right)}$$ and $$\frac{1+\cos\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)}$$.

On the other hand, the partial sums of $$\sum_{n=1}^{\infty}\sin(\sqrt{n})$$ are unbounded.

I think that the partial sums of $$\sum_{n=1}^{\infty}\sin(n^a)$$ are bounded for $$a \geq 1$$ and unbounded for $$0< a<1$$, but how can I prove this? I think this question involves Euler-Maclaurin sum.

• You mean the partial sums of these series are bounded/unbounded. – zhw. Nov 10 '18 at 20:12
• It would be good to edit to reflect this. – zhw. Nov 10 '18 at 20:26
• The partial sums of $\sum \sin(n^\alpha)$ are not bounded for $\alpha>1$, but $\sum_{n=1}^{N}\sin(n^\alpha)$ can be suitably controlled by a power of $N$ times a power of $\log N$. See Weyl's inequality, giving $$\sum_{n=1}^{N}\sin(n^2) \ll \sqrt{N}\log^2 N,$$ for instance. – Jack D'Aurizio Nov 10 '18 at 22:11
• If you could achieve such bound $\sqrt{N}\log^2 N$, it solves your first question in math.stackexchange.com/questions/215528/…. When I tried it years ago, it did not work out very well, I could obtain $\alpha\leq 7/8$. If you managed to prove such bound, please include your method as an answer. – Sungjin Kim Nov 11 '18 at 3:51
• @JackD'Aurizio You mentioned that the partial sums are unbounded when $a>1$. Would you please include that as an answer? I know Terry Tao solved for $a>1$ integer case in here mathoverflow.net/questions/201250/… – Sungjin Kim Nov 12 '18 at 18:14

1 Answer

$$\bullet$$ $$a>1$$ is an integer

If $$a>1$$ is an integer, then this post in MO by Terry Tao solves that the partial sums are not bounded.

$$\bullet$$ $$a>1$$ is not an integer

If $$a>1$$ is not an integer, the same argument by Terry Tao still works, since we have equidistribution of $$n^a$$ mod $$2\pi$$, and any sum or difference of $$(n+i)^a$$ with $$1\leq i \leq h$$ modulo $$2\pi$$. Let $$X_i$$ be the random variable $$\sin^a (k+i)$$ where $$k=1,\ldots n$$. Assuming the boundedness of partial sums, we end up having a contradiction that the random variables $$X_1, \ldots, X_h$$ such that $$\mathrm{Var}(X_1+\cdots+X_h)$$ is bounded as $$h\rightarrow \infty$$ by assumption, but $$\mathrm{Var}(X_1+\cdots+X_h)\sim h/2$$ as $$h\rightarrow\infty$$.

$$\bullet$$ $$0

The first part of this answer of mine, shows that $$\sum_{\alpha can be arbitrarily large. Thus, unboundedness of the partial sums follows.

For a better estimate, we apply 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

Taking imaginary part from the lemma and $$f(n)=n^a/(2\pi)$$, we have

$$\sum_{n\leq N} \sin(n^a) = \int_{1-}^N \sin(x^a) \ dx + O(1).$$

The change of variable $$x^a=t$$ gives $$\int_{1-}^N \sin(x^a) \ dx=\int_{1-}^{N^a} \frac1a t^{\frac1a-1}\sin t \ dt.$$

Applying the integration by parts to the last integral, we obtain an estimate of $$-\frac1a N^{1-a}\cos(N^a) + O(N^{\max\{0,1-2a\}}).$$ This expression is clearly unbounded. Therefore, the partial sums are unbounded when $$0, and $$a>1$$.