Denoting by $p\_i$ the $i$-th prime, is it known that $\displaystyle \sum\_{i=1}^\infty \frac{1}{p\_{i+1}^2-p\_i^2}$ converges?


Can one compute a few digits based on euristic considerations or plausible conjectures about distributions of primes and prime gaps? I think it may be a bit less that 0.63, but I'm not at all confident.



What can one say about $\sum\limits_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$?
In the paper [On the sum of the reciprocals of the differences between consecutive primes, Ramanujan J., 47,427–433(2018)] by me, I proved under the Hardy–Littlewood prime-pair conjecture that
$$\sum\_{n\le X}\frac{1}{p\_{n+1}-p\_n}\sim \frac{X\log\log X}{\log X},$$
and without the Hardy–Littlewood prime-pair conjecture, one has
$$
\sum\_{n\le X}\frac{1}{p\_{n+1}-p\_n}\ll \frac{X\log\log X}{\log X}.
$$
Therefore, by using Abel’s summation formula, one can prove that the conjecture is true unconditionally.


In fact, this problem has been investigated by Erdős and Nathanson [On the sum of the reciprocals of the differences between consecutive primes. In: Chudnovsky, D.V., Chudnovsky, G.V., Nathanson, M.B. (eds.) Number theory: New York Seminar 1991–1995, pp. 97–101. Springer, New York (1996)]. They proved
$$
\sum\_{n\ge 2}\frac{1}{(p\_{n+1}-p\_n)n(\log\log n)^c}<+\infty,
$$
for all $c>2$. Then by noting that $p\_n\sim n\log n$, one can give an alternative proof.


