A well-known result on Egyptian fractions states that any positive rational number can be written as a sum of finitely many distinct unit fractions.
For each prime $p$, let $p'$ be the first prime after $p$. Then $$\{p+p':\ p\ \text{is prime}\}=\{p_k+p_{k+1}:\ k=1,2,3,\ldots\},$$ where $p_n$ denotes the $n$th prime. By the Bertrand Postulate proved by Chebyshev, for any positive integer $k$, we have $p_{k+1}<2p_k$ and hence $2p_k<p_k+p_{k+1}<3p_k$. Thus the series $$\sum_p\frac1{p+p'}=\sum_{k=1}^\infty\frac1{p_k+p_{k+1}}$$ diverges.
Motivated by Question 316474, I make the following conjecture.
Conjecture. For each positive rational number $r$, there is a finite set $S$ of primes such that $$r=\sum_{p\in S}\frac1{p+p'}.$$
Let $r$ be any positive rational number. As the series $\sum_p\frac1{p+p'}$ (with $p$ prime) diverges, there is a unique prime $q$ such that $$\sum_{p<q}\frac{1}{p+p'}\le r<\sum_{p\le q}\frac1{p+p'}.$$ Thus $$0\le r_0:=r-\sum_{p<q}\frac1{p+p'}<\frac1{q+q'}\le\frac1{2+3}=\frac15.$$ If $r_0=\sum_{p\in S}\frac1{p+p'}$ for some finite set $S$ of primes, then all those $p\in S$ are greater than $q$, and $$r=\sum_{p<q}\frac1{p+p'}+\sum_{p\in S}\frac1{p+p'}.$$ Therefore it suffices to consider the conjecture only for $r<1/5$.
Note that \begin{gather*}\frac16=\frac1{3+5}+\frac1{11+13},\ \ \frac18=\frac1{3+5},\\\frac19=\frac1{5+7}+\frac1{17+19},\ \ \frac1{10}=\frac1{5+7}+\frac1{29+31}. \end{gather*} Also, $$\frac14=\frac1{3+5}+\frac1{5+7}+\frac1{11+13}$$ and $$\frac13=\frac1{3+5}+\frac1{5+7}+\frac1{7+11}+\frac1{11+13}+\frac1{17+19}.$$
QUESTION. Any ideas towards solving the conjecture? Can one find an explicit desired representation for $r\in\{1,1/2,1/7,1/11\}$?
Your comments are welcome!
