$\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$ 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 $48=2^4.3 \notin P\_2$



Does the sum of reciprocal of integers with average power at least two converge?
$\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$


I think so. Fix $\rad(m)=p\_1\ldots p\_k=:P$. Denote by $\Omega$ the set of positive integers with all prime divisors in $\{p\_1,\ldots,p\_k\}$. Then $m=PQ$ where $Q\in \Omega$ and $Q\geqslant P$.
For $s\in (0,1)$ we have
$$
\prod\_{p|m}(1-p^{s-1})^{-1}=\sum\_{n\in \Omega} n^{s-1}\geqslant P^s\sum\_{n\in \Omega,n\geqslant P}n^{-1}.
$$
Thus
$$
\sum\_{n\in \Omega,n\geqslant P}n^{-1}\leqslant \prod\_{p|m}p^{-s}(1-p^{s-1})^{-1}=\prod\_{p|m}(p^s-p^{2s-1})^{-1}.
$$
For $s=1/2$ the right hand side reads as $\prod(p^{1/2}-1)^{-1}$. Thus
$$
\sum\_{m:\ap(m)\geqslant 2} m^{-1}\leqslant \sum\_{p\_1,\ldots,p\_k}\prod\_{i=1}^k\frac1{p\_i(\sqrt{p\_i}-1)}=\prod\_p\left(1+\frac1{p(\sqrt{p}-1)}\right)<\infty.
$$


