OPEN
This is open, and cannot be resolved with a finite computation.
Let $k\geq 3$ and $f(k)$ be the supremum of $\sum_{n\in A}\frac{1}{n}$ as $A$ ranges over all sets of positive integers which do not contain a $k$-term arithmetic progression. Estimate $f(k)$.
Is\[\lim_{k\to \infty}\frac{f(k)}{\log W(k)}=\infty\]where $W(k)$ is the van der Waerden number?
Berlekamp
[Be68] proved $f(k) \geq \frac{\log 2}{2}k$. Gerver
[Ge77] proved\[f(k) \geq (1-o(1))k\log k.\]It is trivial that\[\frac{f(k)}{\log W(k)}\geq \frac{1}{2},\]but improving the right-hand side to any constant $>1/2$ is open.
Gerver also proved (see the comments for an alternative argument of Tao) that
[3] is equivalent to $f(k)$ being finite for all $k$.
The current record for $f(3)$ is $f(3)\geq 3.00849$, due to Wróblewski
[Wr84]. Walker
[Wa25] proved $f(4)\geq 4.43975$.
Walker
[Wa25] has shown that it suffices to consider Kempner sets (that is, sets of integers defined as all those whose base $b$ digits are contained in some $S\subset \{0,\ldots,b-1\}$ for fixed $b$ and $S$), in the sense that for any $k\geq 3$ and $\epsilon>0$ there is a Kempner set $A$ lacking $k$-term arithmetic progressions such that\[\sum_{n\in A}\frac{1}{n}\geq f(k)-\epsilon.\]In
[Er80] Erdős asks whether for every $\epsilon>0$ and $k\geq 3$ if $A$ is a set of integers without a $k$-term arithmetic progression such that $\min(A)$ is sufficiently large in terms of $\epsilon$ and $k$ then $\sum_{n\in A}\frac{1}{n}<\epsilon$.