The harmonic series $$\sum_{n\in\mathbb N} \frac{1}{n}$$ is well known to be divergent. Using Cauchy condensation test one immediately sees that even $$\sum_{n\in\mathbb N} \frac{1}{n\log n}$$ is divergent. Let $p_i$ be the $i$th prime. Euler showed in 1737 that $$\sum_{i\in\mathbb N} \frac{1}{p_i}$$ is divergent.
My question: What do we know about the convergence behavior of $$\sum_{i\in\mathbb N} \frac{1}{p_i \log p_i}?$$ More general: Let $(a_n)_{n\in\mathbb N}\subseteq \mathbb N$ be a strictly increasing sequence of natural numbers with $$\sum_{n\in\mathbb N} \frac{1}{a_n}=\infty.$$ What do we know about the convergence behavior of $$\sum_{i\in\mathbb N} \frac{1}{a_i \log a_i}?$$
Convergence depends just on how fast the sequence $\{a_i\}$ increases. For instance, if $a_i=p_i$, by the PNT or just its weaker version, the Chebyshev bound, we have: $$ p_i \gg i \log i,\tag{1}$$ hence: $$ \sum_{i\geq 2}\frac{1}{p_i \log p_i}\leq C\cdot \sum_{i\geq 2}\frac{1}{i \log^2 i}\tag{2}$$ and the RHS is convergent by the Cauchy's condensation test.