$\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$ The average power of an integer $m$ is given by
$$
\ap(m):=\log_{\rad(m)}(m)=\frac{\log(m)}{\log(\rad(m))},
$$ where $\rad(m)=\prod_{p|m}p$.

$\ap(m) \ge 2$ if and only if $\rad(m)^2 \le m$ , which is obviously satisfied by the set of powerful integers $P_2$ defined by the relation "$p|m$ implies $p^2|m$". Clearly $$ \sum_{P_2} \frac{1}{m}=\prod_p(1+\frac{1}{p^2}+\frac{1}{p^3}+ \ldots ) =\prod_p\left(1+\frac{1}{p(p-1)}\right)=1.9436\ldots $$ converges. From numerical computation, it seems that we may have $$ \sum_{ap(m) \ge 2} \frac{1}{m} =2.11\ldots\;. $$ Does it actually converge? Is there a similar expression as for $P_2$? Note $\ap(48) \ge 2$ but $48=2^4.3 \notin P_2$