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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. For every sequence z1,z2,…z_1,z_2,\ldots on the unit circle, with s(k,n)=∑j≤nzjks(k,n)=\sum_{j\le n}z_j^k and Ak=lim sup⁡n→∞∣s(k,n)∣A_k=\limsup_{n\to\infty}\lvert s(k,n)\rvert, there are infinitely many kk with Ak>c1log⁡kA_k>c_1\log k, for an absolute constant c1>0c_1>0; in particular lim sup⁡kAk=∞\limsup_kA_k=\infty. This is the Theorem of Section 1 of P. Erdős, Some remarks on number theory, Israel J. Math. 3 (1965), 6-12 ([Er65c] on the problem page), pp. 6-7. With zj=e(xj)z_j=e(x_j) it answers the first question of Problem 987 yes for every sequence in (0,1)(0,1): the paper's AkA_k is the problem's AkA_k, a limit superior over nn, which the paper distinguishes from BkB_k, the supremum over nn of the same sums. Erdős writes that in his 1964 paper ([Er64b] on the problem page) he had observed lim sup⁡kBk=∞\limsup_kB_k=\infty but stated that he could not prove the same for AkA_k, and that he had "overlooked the fact that it is very easy to show" the theorem.

Covers. The first question: lim sup⁡k→∞Ak=∞\limsup_{k\to\infty}A_k=\infty for every sequence, with the rate log⁡k\log k. The rate was raised to k1/2k^{1/2} by Clunie's theorem, which the paper's footnote added in proof already reports; the second question, whether Ak=o(k)A_k=o(k) is possible, is answered by the 2026 construction.

Depends on. Nothing in this wiki; the result rests on the cited paper alone.

Proof (pp. 6-7, read in full). Dirichlet's theorem on simultaneous approximation gives, for any nn complex numbers y1,…,yny_1,\dots,y_n of modulus one, an integer 1≤k≤10n1\le k\le10^n with Re⁡(yik)>1/2\operatorname{Re}(y_i^k)>1/2 for every ii. Applied to the blocks zrn+1,…,z(r+1)nz_{rn+1},\dots,z_{(r+1)n} for r=0,1,2,…r=0,1,2,\dots, this yields one k≤10nk\le10^n that serves infinitely many blocks, each of whose sums then exceeds n/2n/2 in modulus, so Ak≥n/4A_k\ge n/4; since k≤10nk\le10^n, this is Ak>c1log⁡kA_k>c_1\log k, and letting nn grow gives infinitely many such kk. The paper adds that Ak≥ckA_k\ge ck for infinitely many kk may hold, with c≤1c\le1 by a remark of Clunie that a footnote records, asks for the least f(n,c)f(n,c) such that any nn complex numbers with ∣zi∣≥1\lvert z_i\rvert\ge1 have some 1≤k≤f(n,c)1\le k\le f(n,c) with ∣∑i≤nzik∣≥c\lvert\sum_{i\le n}z_i^k\rvert\ge c, notes f(n,1)=nf(n,1)=n from Turán's results, and records in a footnote added in proof that Clunie proved f(n,c)<g(c) nlog⁡nf(n,c)<g(c)\,n\log n and Ak>ck1/2A_k>ck^{1/2}.

Source and dating. The page's basis is the scan in the Rényi Institute's Erdős archive, linked above. It was received on 10 February 1965 and appeared in volume 3, issue 1, which the publisher's record dates March 1965; the day in the page name is a placeholder. The site's commentary credits the proof to Erdős under its key [Er65b], which the site's reference record resolves to his 1965 Wiley lectures, which do not contain the passage; the Israel J. Math. note is the paper that holds it, as the problem page's reference [Er65c] records.

Formalization. The contributor's fork of formal-conjectures linked above states the theorem as erdos_987.variants.log_lower_bound, for sequences in (0,1)(0,1): there is c>0c>0 with clog⁡k≤Akc\log k\le A_k for infinitely many kk. Its proof outline follows Erdős's argument, with simultaneous Dirichlet approximation to denominator 7n7^n, the same block device and a pigeonhole on the blocks; the repository's main branch points the declaration's formal_proof attribute at that file. Not built or audited here, so it adds no evidence kind.

Acceptance. Refereed: Israel J. Math. 3 (1965), 6-12, a journal paper. Reviewed: the site's curator, Thomas Bloom, labels the problem PROVED (LEAN) and credits Erdős in the problem's commentary with the proof that Ak≫log⁡kA_k\gg\log k for infinitely many kk (page last edited 2026-04-09, as of 2026-10-07); Tao's thread comment of 2025-08-31 reports the same solution. The curator is independent of the author.