# Estimating the sum of reciprocals of products of two primes

It's rather well-known that $$\sum_{p \leq X} \frac{1}{p} \sim \log \log X,$$ where this is a sum over the positive integer primes. Can we efficiently estimate the sum $$\sum_{p,q \leq X} \frac{1}{pq}$$ where both $p,q$ range over positive primes up to $X$?

• "Efficiently" is so undefined. Of course we can estimate it... but more important that asymptotic equivalence is the evaluation of the error. – Masacroso Jul 16 '15 at 5:45
• I guess we could do $\displaystyle \left(\sum\limits_{p \le \sqrt{x}} \frac{1}{p}\right)^2 \le \sum\limits_{p,q \leq x} \frac{1}{pq} \le \left(\sum\limits_{p \le x} \frac{1}{p}\right)^2$ and say $\displaystyle \sum\limits_{p,q \leq x} \frac{1}{pq} = (\log\log x)^2 + \mathcal{O}(\log \log x)$ – r9m Jul 16 '15 at 5:51
• Indeed, the interesting part is the error estimation – davidlowryduda Jul 16 '15 at 13:44

$$\sum_{p,q\leq X}\frac{1}{pq}=\sum_{q\leq X}\frac{1}{q}\sum_{p\leq X}\frac{1}{p}=\left(\log\left(\log\left(X\right)\right)\right)^{2}+O\left(\log\left(\log\left(X\right)\right)\right)$$ from $$\sum_{p\leq X}\frac{1}{p}=\log\left(\log\left(X\right)\right)+O\left(1\right).$$ You can get a better approximation using $$\sum_{p\leq X}\frac{1}{p}=\log\left(\log\left(X\right)\right)+M+O\left(\log^{-1}\left(X\right)\right)$$ where $M$ is the Meissel–Mertens constant.