For an infinite sequence $x_1, x_2, \ldots \in (0, 1)$, define $$A_k = \limsup_{n \to \infty} \left\lvert \sum_{j \le n} e(k x_j) \right\rvert,$$ where $e(x) = e^{2\pi i x}$.

Question 1:

Is it true that $\limsup_{k \to \infty} A_k = \infty$?

Erdős [Er64b] remarks it is "easy to see" that $\limsup_k \sup_n |\sum_{j \le n} e(k x_j)| = \infty$. Erdős [Er65b] later found a "very easy" proof that $A_k \gg \log k$ for infinitely many $k$. Clunie [Cl67] proved that $A_k \gg k^{1/2}$ for infinitely many $k$, which implies the answer is yes (Tao independently found a proof). This is Problem 7.21 in [Ha74].

Erdős [Er64b] remarks it is "easy to see" that for every infinite sequence $x_1, x_2, \ldots \in (0, 1)$, $$\limsup_{k \to \infty} \sup_n \left\lvert \sum_{j \le n} e(k x_j) \right\rvert = \infty.$$

Erdős [Er65b] proved that, for every infinite sequence $x_1, x_2, \ldots \in (0, 1)$, $A_k \gg \log k$ for infinitely many $k$.

Clunie [Cl67] proved that, for every infinite sequence $x_1, x_2, \ldots \in (0, 1)$, $A_k \gg k^{1/2}$ for infinitely many $k$. (Tao independently found a proof.)

Linear upper bound (weakened). A first weakened version of Clunie's A_k ≤ k: there exists a sequence $x \in (0,1)$ with $A_k \le 2k$ for all $k \ge 1$. The witness is the (shifted) van der Corput sequence. For the tighter $A_k \le k + 1$ bound see linear_upper_bound_clunie.

Linear upper bound (tight via Clunie phase tracking). Tighter than linear_upper_bound_weak: there exists a sequence $x \in (0,1)$ with $A_k \le k + 1$ for all $k \ge 1$, via the (shifted) van der Corput sequence. Whether the $+1$ can be eliminated to recover Clunie's exact $A_k \le k$ under the strict $\mathrm{Ioo},0,1$ hypothesis is open in this formalization (see linear_upper_bound).

Clunie [Cl67] proved that there exists an infinite sequence ${z_\nu}$ on the unit circle with $A_\nu \le \nu$ for all $\nu \ge 1$. Translating $z_\nu = e(x_\nu)$, the natural domain of $x_\nu$ is the half-open unit interval $\mathrm{Ico},0,1$, matching the original [Er64b]/[Cl67] statement (any unit complex number is allowed, including $z = 1$, i.e. $x = 0$).

Note: erdosproblems.com/987 phrases the problem with the open interval $x_\nu \in (0, 1)$, which excludes $z = 1$ and is strictly stronger than what [Er64b]/[Cl67] state; we align with the original papers here. The shifted-vdc $\le k + 1$ variant under the open interval is preserved as linear_upper_bound_clunie.

An internal OpenAI model (see [APSSV26b, §3]) proved that there exists an infinite sequence $x_1, x_2, \ldots \in (0, 1)$ such that $\sup_n \left\lvert \sum_{j \le n} e(k x_j) \right\rvert \ll (k \log k)^{1/2}$ for all $k \ge 1$ (in particular $A_k \ll (k \log k)^{1/2}$).

Note: the bound is restricted to $k \ge 2$ since $\log 1 = 0$ would make the RHS vanish at $k = 1$, while the LHS $|\sum_{j < n} e(x_j)|$ can equal $1$ (e.g. for $n = 1$).

Question 2 (parts.ii): Is it possible for $A_k = o(k)$? Yes — there exists a sequence $(x_n) \in (0, 1)$ and a bound $b(k) = o(k)$ with $A x k \le b k$ eventually. A corollary of sqrt_log_upper_bound (which gives a $\sqrt{k \log k}$ bound) plus the asymptotic $\sqrt{k \log k} = o(k)$.

Liu [Li69] showed that, for any $\epsilon > 0$, $A_k \gg k^{1 - \epsilon}$ infinitely often under the additional assumption that there are only a finite number of distinct points. Clunie observed in the Mathscinet review of [Li69] that under this assumption in fact $A_k = \infty$ infinitely often (the version stated here).