# Does the sum of reciprocal of integers with average power at least two converge?

$$\DeclareMathOperator{\ap}{ap}$$ $$\DeclareMathOperator{\rad}{rad}$$ 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 $$48=2^4.3 \notin P_2$$

• As an explicit formula: $$\sum_{ap(m) \geq 2}=\sum_{P_2}\frac{1}{m}+\sum_{\beta, \gamma \in S^{(2)}_{*} \\(\beta, \gamma)=1} \frac{1}{\beta\gamma^2} \left(\sum_{\alpha > \beta \\ (\alpha, \beta)=1} \frac{1}{\alpha( \text{rad}(\alpha)^2)}\right)$$, here $S^{(2)}_*$ is the set of all squarefree numbers. Commented Mar 22, 2023 at 5:22
• I find it amusing that the right hand side is an explicit formula for the expression on the left hand side (but I have to admit that I don't know the first thing about number theory…)-
– Dirk
Commented Mar 22, 2023 at 6:56

$$\DeclareMathOperator{\ap}{ap}$$ $$\DeclareMathOperator{\rad}{rad}$$
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 $$\sum_{m:\ap(m)\geqslant 2} m^{-1}\leqslant \sum_{p_1,\ldots,p_k}\prod_{i=1}^k\frac1{p_i(\sqrt{p_i}-1)}=\prod_p\left(1+\frac1{p(\sqrt{p}-1)}\right)<\infty.$$