Reciprocal-sum convergence from a counting bound ($A(x)\ll x/(\log x{)}^{1+\eta }⟹{\sum }_{n\in A}1/n<\infty$)

If $A(x)$ counts a set of positive integers and $\eta >0$, Abel summation gives ${\sum }_{n\in A,n\le X}1/n=A(X)/X+{\int }_1^{X}A(t)t^{-2}dt$. A bound $A(t)\ll t/(\log t{)}^{1+\eta }$ for $t\ge 3$ makes the integral convergent, since the substitution $u=\log t$ gives ${\int }_{\log 3}^{\infty }{u}^{-1-\eta }du<\infty$.