# How to estimate $\sum_{p\leqslant x}\sum_{q\leqslant x}\frac{1}{p+q}$?

How to estimate $$\sum_{p\leqslant x}\sum_{q\leqslant x}\frac{1}{p+q}, \qquad\qquad(1)$$ where $$p$$, $$q$$ are prime numbers.

We have the Mertens' formula $$\sum_{p\leqslant x} \frac{1}{p} = \log\log x+ B + O\left( \frac{1}{\log x} \right),$$ where $$p$$ is prime number, and $$B=\gamma - \sum_{p} \left( \log \left( \frac{1}{1-1/p} \right) - \frac{1}{p} \right)$$ is the Mertens constant, $$\gamma$$ is Euler constant.

I guess the main term of (1) is $$\dfrac{x\log\log x}{\log x}$$, but I don't prove it, Can you help me?

In fact, your sum (let's denote it $$S$$) satisfies $$\frac x{2\log^2 x}\, (1+o(1)) \le S\le \frac{2x}{\log^2 x}\, (1+o(1)), \tag{\ast}$$ so that $$x\log\log x/\log x$$ cannot be the main term.
The upper bound is easy to prove observing that $$\frac1{p+q} \le \frac1{2\sqrt{pq}}$$ by the AM-GM inequality. It follows that $$S \le \frac12\,\sum_{p,q\le x} \frac1{\sqrt{pq}} = \frac12\left(\sum_{p\le x} \frac1{\sqrt p}\right)^2.$$ The sum in the RHS is known to satisfy $$\sum_{p\le x} \frac1{\sqrt p} = \frac{2\sqrt x}{\log x}\,(1+o(1)).$$ Combining the last two estimates, we get the upper bound in ($$\ast$$).
For the lower bound, just notice that $$S \ge \sum_{p,q\le x}\frac1{2x} = \frac{(\pi(x))^2}{2x} = \frac x{2\log^2 x}\, (1+o(1))$$ by the prime number theorem.
• @reuns: What do you mean "need"? I probably can, using partial summation; and I also can avoid using the PNT, but then the lower bound will be relaxed to $\Omega(x/\log^2(x))$. But the OP was asking for an estimate, and have not indicated that he does not want to use the PNT, correct? – W-t-P Apr 20 at 8:00
• I find $\sum_{p\leqslant x}\sum_{q\leqslant x}\frac{1}{p+q}\sim\sum_{2\le n \le x}\sum_{2\le m \le x} \frac{1}{n+m} \frac{1}{\ln n\ \ln m}\sim \sum_{ n \le x}\sum_{ m \le x} \frac{1}{n+m} \frac{1}{\ln^2 x}$ $\sim \frac{1}{\ln^2 x}\sum_{l \le 2x} \frac{\min(l,x)}{l} \sim \frac{x (1+\ln 2)}{\ln^2 x}$ – reuns Apr 20 at 8:08