I had the following question related to the sum of the reciprocals of the prime numbers restricted to a certain subset.
Let $T, c > 1$ be real numbers. Let $A=\cup_{k \geq 0} [Tc^{2k},Tc^{2k+1}]$ and $B=\cup_{k \geq 0} [Tc^{2k+1},Tc^{2k+2}]$. It should be true that $\sum_{p \in A} 1/p$ and $\sum_{p \in B} 1/p$ diverge, and it should be a consequence of the prime number theorem, but I was not able to conclude this last fact.
We know by Pigeonhole Principle that at least one of them diverges, but I need to show both of them do.
Write $$\sum _{p\leq X}\frac {1}{p}=\log \log X+B+E(X)$$ and $K=Tc^{2k}$. If $$E(X)=o(1/\log X)\hspace {10mm}(\star )$$ then $$\sum _{K<p\leq Kc}\frac {1}{p}=\log \log (Kc)-\log \log K+\mathcal O\left (E(K)\right )=\log (1+\log c/\log K)+o\left (\frac {1}{\log K}\right )\sim \frac {\log c}{\log K}=\frac {\log c}{\log T+2k\log c}=\frac {1}{2k}$$ so for some $k_0$ $$k\geq k_0\implies \sum _{K<p\leq Kc}\frac {1}{p}\gg \frac {1}{k}$$ so $$\sum _{p\in A}\frac {1}{p}=\sum _{k\geq 0}\sum _{K<p\leq Kc}\frac {1}{p}\geq \sum _{k\geq k_0}\sum _{K<p\leq Kc}\frac {1}{p}\gg \sum _{k\geq k_0}\frac {1}{k}=\infty .$$
I'm not sure myself how to get $(\star )$ but Theorem 6.10 of https://arxiv.org/pdf/1002.0442.pdf gives it.