calculus - Are the partial sums for $sum_{n=1}^{infty}sin(n^a)$ bounded for $ageq1$ and unbounded for $0

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.

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 $a>1$.