Cfrgtkky

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

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.

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

If $a>1$ is not an integer, the same argument by Terry Tao still works, since we have equidistribution of $n^a$ mod $2pi$, and any sum or difference of $(n+i)^a$ with $1leq i leq h$ modulo $2pi$. 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 $hrightarrow infty$ by assumption, but $mathrm{Var}(X_1+cdots+X_h)sim h/2$ as $hrightarrowinfty$.