Is there a strictly increasing sequence $\{ a_k\}$ of positive integers such that $\sum 1/{a_k}$ is finite and minimal with respect to this property, in a sense that if $\{ b_k \}$ is another sequence of such, then $a_k \leq b_k$ for all but finitely many values of $k$??
Motivation came from comparing the series $\sum 1/n$ and $\sum 1/{n^2}$, the first of which is divergent but the second of which is not; the sequence for the first is "smaller" than that of the second.
If this question need be motified to make my question clearer and/or make it mathematically answerable, please comment/suggest adit as well.
No, there is not. To have $\sum \frac 1{a_k}$ converge we must have $\frac 1{a_k}\to 0$. Now define $$b_k=\begin {cases}a_k&a_k \lt 1000\\a_k-1 & a_k \ge 1000 \end {cases}$$ Only finitely many of the $a_k$ are less than $1000$, so only finitely many $a_k\le b_k$ but $$\sum \frac 1{b_k}\lt 2\sum \frac 1{a_k}$$ is convergent.
Even more so, we know that $\sum \frac 1n$ diverges, as does $\sum \frac 1{n \log n},$ and $\sum \frac 1{n \log n \log \log n}$ and so on, while $\sum \frac 1{n \log^2 n}, \sum \frac 1{n\log n (\log \log n)^2}$ and so on converge.