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a(n+m) <= a(n)+a(m)+f(n+m) with f(x)=x/log(x)

Source: Miklós Schweitzer 2016, Problem 4

November 2, 2016
algebraSequencesMiklos Schweitzerfunction

Problem Statement

Prove that there exists a sequence a(1),a(2),…,a(n),…a(1),a(2),\dots,a(n),\dots of real numbers such that a(n+m)≤a(n)+a(m)+n+mlog⁡(n+m) a(n+m)\le a(n)+a(m)+\frac{n+m}{\log (n+m)} for all integers m,n≥1m,n\ge 1, and such that the set {a(n)/n:n≥1}\{a(n)/n:n\ge 1\} is everywhere dense on the real line.
Remark. A theorem of de Bruijn and Erdős states that if the inequality above holds with f(n+m)f(n + m) in place of the last term on the right-hand side, where f(n)≥0f(n)\ge 0 is nondecreasing and ∑n=2∞f(n)/n2<∞\sum_{n=2}^\infty f(n)/n^2<\infty, then a(n)/na(n)/n converges or tends to (−∞)(-\infty).
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