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Does the sum of the reciprocals of the differences of the squares of consecutive prime numbers converge?

$$\sum_{n=1}^{\infty}\frac{1}{p_{n+1}^{2}-p_{n}^{2}}$$


Also here on mathoverflow


My first thought was that it should diverge, simply because $\sum\frac{1}{n}$, $\sum\frac{1}{p_{n}}$ and $\sum\frac{1}{\left(n+1\right)^{2}-n^{2}}$ all diverge to $\infty$ , but then I calculated it and it looks nothing like the $3$ sums above...


As noted in the comments, these two questions are roughly the same, but they contradict each other, and both have accepted answers:

This one says it converges

This one says it diverges


n Partial sum
$1$ $0.200000000000000$
$10$ $0.378601953601953$
$100$ $0.451166931565077$
$1000$ $0.487337524558420$
$10000$ $0.508934950162835$
$100000$ $0.523674386928273$
$1000000$ $0.534504303888848$
$10000000$ $0.542844304872399$
$100000000$ $0.549517981000105$
$1000000000$ $0.555004286269487$
$4000000000$ $0.557867480026235$
François Huppé
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    Idk but note that the sum of the reciprocals of the twin primes does (to Brun's constant). – suckling pig May 23 '25 at 11:57
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    Since $$p_{n+1}^2 - p_n^2 = (p_{n+1}-p_n)(p_{n+1}+p_n) = g_n(p_{n+1}+p_n) > g_np_n,,$$ the average order of the prime gaps $g_n$ is $\log p_n \sim \log n$, and $\sum \frac{1}{p\log p}$ converges, I expect that your series converges. But I'm not sure I can prove it. Will think about that. – Dermot Craddock May 23 '25 at 12:48
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    In 2019, by Rosser's theorem, I proved https://oeis.org/A306658. Anyway, let $p_n$ denote the $n$-th prime, we can also use more powerful tools/improved bounds, such as Dusart's inequality (stating that $p_n > n \cdot (\log(n) + \log(\log(n))-1)$), but Rosser's Th. would be enough to confirm the convergence of the given series. – Marco Ripà May 23 '25 at 14:05
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    According to this answer the series diverges. However, I doubt whether that answer is correct. – Dermot Craddock May 24 '25 at 15:04
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    I found an answer that it converges. https://mathoverflow.net/a/230201/520102 There must be mistakes somewhere in my previous comment :( – Tong Lingling May 24 '25 at 20:13
  • I have some naive reasoning, but it is not a proof: The mean of the prime gaps is $$\frac{p_n-p_1}{n-1}>\frac{n\log(n)-2}{n-1} \to \log(n)$$ So we can write this as something like $$\sum_{n=1}^{\infty}\frac{1}{p_{n+1}^2-p_n^2} < \frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{2p_n\log(n)+\log(n)^2}<\frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{2n\log(n)^2+\log(n)^2}$$ and the series on the far right converges. Should I post this as an answer? – Debalanced May 25 '25 at 02:14

1 Answers1

2

(Credit to Stanley Yao Xiao on mathoverflow)

The series converges.

Let $0 < \delta < 1$, and let $S_1(\delta)$ be the subset of natural numbers satisfying $p_{n+1} - p_n > (\log n)^\delta$ and let $S_2(\delta)$ be the complement of $S_1(\delta)$ in the natural numbers.

We have

\begin{align*} \sum_{n \in S_1(\delta)} \frac{1}{(p_{n+1} - p_n)(p_{n+1} + p_n)} & \ll \sum_{n \in S_1(\delta)} \frac{1}{(\log n)^\delta p_n}\\ & \ll \sum_{n \in S_1(\delta)} \frac{1}{n (\log n)^{1 + \delta}}, \end{align*} which converges.

On the other hand, using an upper bound sieve, we find that

$$\displaystyle \sum_{\substack{n \in S_2(\delta)}} \frac{1}{p_n} < \infty. $$

This implies

\begin{align*} \sum_{n \in S_2(\delta)} \frac{1}{(p_{n+1} - p_n)(p_{n+1} + p_n)} & \ll \sum_{n \in S_2(\delta)} \frac{1}{p_n} < \infty, \end{align*} as required.

François Huppé
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