A set $A = \{a_1, a_2 ,..\}$ of positive integers is called large if $\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + ...$ diverges. A small set is any set of the positive integers that is not large
We define $A(n) = \{1,2,..n\} \cap A$ and $a(n) = |A(n)| $ the number of elements in $A$ less or equal to $n$.
We say that $A$ is smaller than $B$ if: $ \lim_{n\to\infty} \frac{a(n)}{b(n)} = 0$
I wondered if there exist a "smallest" large set, or whether we can always find a smaller subset. I suspect the latter, the idea of an absolutely smallest large set feels unsettling. Therefore, I conjecture the following:
For any large set $A$, there exist a subset $A^{\ast} \subset A $ so that $A^{\ast}$ is large and smaller than $A$.
First observe that if $A$ is a finite set of positive integers and $B \subseteq A$ contains every $n$th element of $A$ (starting with the smallest) then $\sum_{b \in B} 1/b \ge (1/n) \sum_{a \in A} 1/a$.
Now let $A$ be a set of positive integers such that $\sum_{a \in A} 1/a = \infty$. We can partition $A$ into segments $A_1, A_2, A_3, \dots$ (with $\max A_n < \min A_{n+1}$ for each $n$) such that $\sum_{a \in A_n} 1/a \ge 1$ for each $n$. We may also arrange that $|A_n|$ is a multiple of $n$. Let $B$ contain every element of $A_1$, every other element of $A_2$, every third element of $A_3$, and so on. Then clearly $B$ has asymptotically zero density in $A$ and $$\sum_{b \in B} 1/b \ge \sum_{n=1}^\infty \frac1n \sum_{a\in A_n} 1/a \ge \sum_{n=1}^\infty \frac1n = \infty.$$