The sum of the reciprocals of Chen primes converges

Question

It is well know that the The sum of the reciprocals of all primes diverges. How to prove that The sum of the reciprocals of Chen primes converges.

Answer 1

I used the similar method in this answer. I found recently that this method was already also outlined in a comment by Denis in the MO posting I commented about:

The appearances of $C>0$ are absolute constants that are not necessarily the same.

For the twin primes, it is well known that for some absolute constant $C>0$, $$ |\{p: x/2<p\leq x, p+2 \textrm{ is prime } \}|\leq C \frac x{\log^2 x}. $$ The following is Halberstam & Richart "Sieve Methods", Corollary 5.8.4 in page 179.

Let $q$ and $h$ be integers, and let $y$ and $x$ be real numbers satisfying $$ h\neq 0, \ \ (q,h)=1, \ \ 2|qh, \ \ 1\leq q<y\leq x. $$ Then $$ |\{p: x-y<p\leq x, (p-h)/q = p' \}| $$ $$ \leq 16 \prod_{p>2}\left(1-\frac1{(p-1)^2}\right)\prod_{2<p|qh}\frac{p-1}{p-2}\frac{y/q}{\log^2(y/q)}\left(1+O\left(\frac{\log\log 3|h|y}{\log(y/q)}\right)\right). $$

In here, $p$ and $p'$ are primes, and the implied constants in $O$-term is absolute.

Take $h=-2$, $q\leq \sqrt x$, and $y=x/2$. Then we have a simpler expression $$ |\{p: x/2<p\leq x, (p+2)/q=p'\}|\leq C \frac{x/q}{\log^2 x}. $$

Then we sum up for all primes $q\leq \sqrt x$. By $\sum_{q\leq \sqrt x} \frac1q \ll \log\log x$, we obtain $$ |\{p: x/2<p\leq x, p \textrm{ is Chen's prime} \}|\leq C \frac{x\log\log x}{\log^2 x}. $$ Diadically adding this together, we have $$ |\{ p\leq x: p \textrm{ is Chen's prime} \}|\leq C \frac{x\log\log x}{\log^2 x}. $$ The convergence of the reciprocal can be proved by the similar method as in Brun's constant. Let $\pi_C(t)$ be the counting function for Chen's primes up to $t$. Then $$\begin{align} \sum_{p \textrm{ is Chen's prime}, p\leq x} \frac1p&=\int_{2-}^x \frac1t d\pi_C(t)\\ &=\frac{\pi_C(t)}t \Bigg\vert_{2-}^x + \int_{2-}^x \frac{\pi_C(t)}{t^2}dt\\ &=\int_{2-}^{\infty}\frac{\pi_C(t)}{t^2} dt +o(1). \end{align}$$ The integral $\int_{2-}^{\infty}\frac{\pi_C(t)}{t^2} dt$ converges due to $$ \pi_C(t)=O\left(\frac{t\log\log t}{\log^2 t}\right),$$ and $$ \int_2^{\infty} \frac{\log\log t}{t\log^2 t} dt $$ converges.