Reciprocal-sum convergence from a counting bound
= Reciprocal-sum convergence from a counting bound
{title2=$A(x)\ll x/(\log x)^{1+\eta}\Longrightarrow\sum_{n\in A}1/n<\infty$}
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 $t\ge3$ makes the integral convergent, since the substitution $u=\log t$ gives $\int_{\log3}^\infty u^{-1-\eta}\,du<\infty$.