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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.

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  • $\begingroup$ You mean the partial sums of these series are bounded/unbounded. $\endgroup$ – zhw. Nov 10 '18 at 20:12
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    $\begingroup$ It would be good to edit to reflect this. $\endgroup$ – zhw. Nov 10 '18 at 20:26
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    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Nov 10 '18 at 22:11
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    $\begingroup$ 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. $\endgroup$ – Sungjin Kim Nov 11 '18 at 3:51
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    $\begingroup$ @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/… $\endgroup$ – Sungjin Kim Nov 12 '18 at 18:14
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$\bullet$ $a>1$ 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 $h\rightarrow\infty$.

$\bullet$ $0<a<1$

The first part of this answer of mine, shows that $\sum_{\alpha<n\leq \beta} \sin(n^a)$ 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 $$ \sum_{a<n\leq b}e^{2\pi i f(n)} = \int_a^b e^{2\pi i f(x)} \ dx + O(1). $$

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 $0<a<1$, and $a>1$.

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