Let $p_{n,r}$ be $r$-th smallest $n$-power free number. For example $p_{2,7}$ is the $7$-th square free number and $p_{5,7}$ is the $7$-th smallest $5$-th power free number.
Let $q_{n,r}$ be the $r$-th number which contains (or is divisible by) a $n$-th power. For example $q_{2,7}$ is the $7$-th smallest number which contains a square and $q_{5,7}$ is the $7$-th smallest number which contains a fifth power.
We define $\alpha_n$ for $n \ge 2$ as the ratio of the sum of the $n$-th power free number to the sum of the $n$-th power containing numbers i.e.
$$ \alpha_n = \lim_{r \to \infty}\frac{p_{n,1}+p_{n,2}+\ldots + p_{n,r}}{q_{n,1}+q_{n,2}+\ldots + q_{n,r}}. $$
Question: Is it true that
$$ \sum_{n = 2}^{\infty} \frac{\alpha_n}{n} = 1 - \gamma $$
where $\gamma$ is the Euler-Mascheroni constant?
Motivation: I ran a program and the sum seem to converge to 0.422785 which is close to $1-\gamma$.
Update. The equality is true. Since $$Q_{n}\left(x\right)=\frac{x}{\zeta\left(n\right)}+O_{n}\left(x^{1/n}\right) $$ we have, using Abel's summation, that $$ \sum_{p_{n}\leq N}p_{n}=Q_{n}\left(N\right)N-\int_{1}^{N}Q_{n}\left(t\right)dt $$ $$=\frac{N^{2}}{2\zeta\left(n\right)}+O_{n}\left(N^{1+1/n}\right). $$ In the same spirit we get $$ \sum_{q_{n}\leq N}q_{n}=\left(1-Q_{n}\left(N\right)\right)N-\int_{1}^{N}\left(1-Q_{n}\left(t\right)\right)dt $$ $$=\frac{N^{2}}{2}\left(1-\frac{1}{\zeta\left(n\right)}\right)+O_{n}\left(N^{1+1/n}\right) $$ so we have $$\alpha_{n}=\frac{p_{1,n}+p_{2,n}+\dots}{q_{1,n}+q_{2,n}+\dots}=\lim_{N\rightarrow\infty}\frac{\sum_{p_{n}\leq N}p_{n}}{\sum_{q_{n}\leq N}q_{n}}=\zeta\left(n\right)-1 $$ hence $$\sum_{n\geq2}\frac{\alpha_{n}}{n}=\sum_{n\geq2}\frac{\zeta\left(n\right)-1}{n}=\color{red}{1-\gamma}$$ as wanted (for a proof of the last idenity see here).