Is there a strictky 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, please comment/suggest adit as well.
No, there is not such a minimizer. Suppose there were a sequence $\{a_k\}$ with the desired property. Clearly there must be a gap at some point, so there is an index $k_0$ such that
$$a_{k_0} > a_{k_0 - 1} + 1.$$
Otherwise, we would have the (divergent) harmonic series.
Now consider the sequence $b_k$ where $b_k = a_k$ for all $k \ne k_0$, and $b_{k_0} = a_{k_0} - 1$.