What can one say about $\sum\limits_{i=1}^\infty \frac{1}{p_{i+1}^2-p_i^2}$?

Question

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.

Answer 1

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.

Answer 2

Cramer's conjecture gives the probability that $p_{i+1} - p_i$ is $k$ is like $e^{ - k / \log p_i})/ \log p_i$ so using $$\frac{1}{p_{i+1}^2- p_i^2}= \frac{1}{ (p_{i+1} - p_i) (p_{i+1}+p_i)} \approx \frac{1}{ 2 (p_{i+1}-p_i) p_i}$$

the expected contribution from $p_i$ is

$$ \frac{1}{2 p_i} \sum_{k=1}^{\infty} \frac{e^{ - k / \log p_i}}{ k \log p_i} \approx \frac{1}{2 p_i} \frac{ \log \log p_i}{ \log p_i} $$ (here small $k$, say $k< \log p_i^{1-\epsilon}$, dominate, so we can treat $e^{ - k/\log p_i}$ as the constant 1 supported on $k< \log p_i$) and since the probability that $n$ is prime is $\frac{1}{\log n}$, the expected contribution from $n$ is

$$ \frac{\log \log n}{2 n (\log n)^2 } $$

so a first-order heuristic is $$\int_{e^e}^{\infty} \frac{ \log \log x}{ 2 x (\log x)^2} dx =\int_{e}^{\infty} \frac{ \log y}{ 2 y^2} dy = \frac{ - \log y -1 }{ 2y} ]_{0}^{\infty} = \frac{1}{e}$$

and a heuristic incorporating numerical data is

$$ \sum_{i=1}^{n} \frac{1}{ p_{i+1}^2 - p_i^2} + \int_{p_{n+1} }^{\infty} \frac{ \log \log x}{ 2 x (\log x)^2} dx = \sum_{i=1}^{n} \frac{1}{ p_{i+1}^2 - p_i^2} + \frac{1}{2} \left( \frac{ \log \log p_{n+1} + 1}{ \log p_{n+1} } \right) $$

I don't think the corrections for congruences to small moduli, like forcing all primes to be odd, will affect the leading term here, although they should probably introduce lower-order terms.