TITLE: About the convergence of the series $\sum_{k\ge1}\frac1{m_kp_k+1}$
QUESTION [5 upvotes]: As usual, let it be $p_k$ the $k$-th prime.
For each $p_k$ one can define
$$m_k=\min\,\{m\in N:\,mp_k+1\;\;is\;prime\}$$
and consider the following series
$$\sum_{k\ge1}\frac1{m_kp_k+1}\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
The series (1) seems to converge but I am not sure this is true. Up to the $10^7$-th prime, the summation gives the value $1.0278802480162903...$
Could it be found a formal way to decide the convergence of such series?
Many thanks.

REPLY [2 votes]: The random model for the primes predicts that the $m_k$ are approximatively independent random variables with $$Pr[m_k <\log^2\log p_k]\approx \sum_{d=0}^{\log^2 \log p_k}(1- \frac1{\log p_k})^d \frac1{\log p_k}$$ $$\approx 1-(1- \frac1{\log p_k})^{\log^2 \log p_k}\approx \frac{\log^2 \log p_k}{\log p_k}$$
so that $$\Bbb{E}[\sum_k \frac1{m_kp_k+1}]\le \sum_k (\frac{C\log^2 \log p_k}{\log p_k}\frac1{p_k+1}+\frac1{p_k \log^2 \log p_k+1})<\infty$$ which means that $$Pr[\sum_k \frac1{m_kp_k+1}=\infty]=0$$
It is probably out of reach to prove it does really converge. If someone can prove it diverges  they will earn the Fields medal.