I know $\sum\_{k=1}^{n} \sin(k)$ is bounded by a constant. How about $\sum\_{k=1}^{n} \sin(k^2)$?



Is $\sum_{k=1}^{n} \sin(k^2)$ bounded by a constant $M$?
No. If one selects a number $k$ at random from $1$ to a large number $n$, then for any fixed $h$, the random variables $\sin((k+1)^2), \dots, \sin((k+h)^2)$ asymptotically have mean zero, variance 1/2, and covariances 0, from standard Weyl sum estimates. Hence the variance of $\sum\_{i=1}^h \sin((k+i)^2)$ is asymptotically $h/2$, which goes to infinity as $h \to \infty$. On the other hand, if the partial sums of $\sin(k^2)$ were bounded, then this variance would have to be bounded also. [Exercise: what part of the above argument breaks down when working with $\sin(k)$ instead of $\sin(k^2)$?]


It may be possible to push this argument to show that the partial sums have to fluctuate by $\gg \sqrt{n}$ infinitely often, but I haven't checked this (certainly a lower bound of $\gg n^\varepsilon$ for some small $\varepsilon>0$ should be possible from the above argument, perhaps contingent on some conjecture about the irrationality measure of $\pi$). Heuristically, the law of the iterated logarithm suggests that the sum can occasionally get as large as $\gg \sqrt{n \log\log n}$, but no larger.


