Is it true that \[\sum_{n\in A}\frac{1}{n}<\infty?\]

OPEN

Let $A$ be an infinite set such that there are no distinct $a,b,c\in A$ such that $a\mid (b+c)$ and $b,c>a$. Is there such an $A$ with
\[\liminf \frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N^{1/2}}>0?\]
Does there exist some absolute constant $c>0$ such that there are always infinitely many $N$ with
\[\lvert A\cap\{1,\ldots,N\}\rvert<N^{1-c}?\]

Is it true that \[\sum_{n\in A}\frac{1}{n}<\infty?\]

Asked by Erdős and Sárközy [ErSa70], who proved that $A$ must have density $0$. They also prove that this is essentially best possible, in that given any function $f(x)\to \infty$ as $x\to \infty$ there exists a set $A$ with this property and infinitely many $N$ such that
\[\lvert A\cap\{1,\ldots,N\}\rvert>\frac{N}{f(N)}.\]
(Their example is given by all integers in $(y_i,\frac{3}{2}y_i)$ congruent to $1$ modulo $(2y_{i-1})!$, where $y_i$ is some sufficiently quickly growing sequence.)

An example of an $A$ with this property where \[\liminf \frac{\lvert A\cap\{1,\ldots,N\}\rvert}{N^{1/2}}\log N>0\] is given by the set of $p^2$, where $p\equiv 3\pmod{4}$ is prime.

For the finite version see [13].