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.
Best 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 $h\rightarrow\infty$.
$\bullet$ $0<a<1$
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.
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 $a>1$.