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A prime $p$ is called a Sophie Germain prime if $2p+1$ is also prime: OEIS A005384. Whether there are an infinite number of such primes is unsolved. My question is:

If there are an infinite number of Germain primes, is the sum of the reciprocals of these primes known to converge, or diverge?

Of course if there are only a finite number of Germain primes, the sum is finite. And a lower bound on any infinite sum can be calculated. But it is conceivable that it is known that the sum either converges or is a finite sum. And maybe even an upperbound is known?

(My connection to this topic is via this question: "Why are this operator's primes the Sophie Germain primes?".)

Joseph O'Rourke
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    I believe one can show that the sum of their reciprocals must converge in the same way that the sum of the reciprocals of twin primes must converge. – Stanley Yao Xiao Apr 29 '22 at 00:16
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    Of course, if the answer is known, then it is known that the sum converges—or else we'd already know that there are infinitely many! – LSpice Apr 29 '22 at 15:29

2 Answers2

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Googling "sum of reciprocals of Sophie Germain primes" brings up the very recent paper:

Wagstaff, Samuel S. jun., Sum of reciprocals of germain primes, J. Integer Seq. 24, No. 9, Article 21.9.5, 10 p. (2021). ZBL1482.11122.

Here's a link to the paper on the journal's website. In particular, Wagstaff proves that the sum of the reciprocals of the Sophie Germain primes is between 1.4898 and 1.8027.

Asaf Karagila
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Joe Silverman
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Here is a general result. For a sequence of nonnegative numbers $\{a_n\}$, let $A(x) = \sum_{n \leq x} a_n$. For example, if $S \subset \mathbf Z^+$ and we set $a_n = 1$ when $n\in S$ and $a_n = 0$ when $n \not\in S$, then $A(x)$ is the number of elements of $S$ that are $\leq x$.

Exercise: If $A(x) = O(x/(\log x)^r)$ for a positive integer $r$ and all $x \geq 2$, then $\sum_{n \leq x} a_n/n$ converges as $x \to \infty$ if $r \geq 2$ and $\sum_{n \leq x} a_n/n = O(\log \log x)$ for $r = 1$.

Example: if $f_1(T), \ldots, f_r(T)$ are polynomials with integer coefficients that fit the hypotheses of the Bateman-Horn conjecture (twin primes are $f_1(T) = T$ and $f_2(T) = T+2$, while Sophie Germain primes are $f_1(T) = T$ and $f_2(T) = 2T+1$), then Bateman and Stemmler showed $60$ years ago that the number of $n \leq x$ such that $f_1(n), \ldots, f_r(n)$ are all prime is $O(x/(\log x)^r)$, where the $O$-constant depends on $f_1, \ldots, f_r$. Therefore if above we take $S$ to be the $n \in \mathbf Z^+$ such that $f_1(n), \ldots, f_r(n)$ are all prime and define $a_n$ to be $1$ or $0$ according to $n \in S$ or $n \not\in S$, then the exercise above says the sum of all $1/n$ for $n \in S$ converges if $r \geq 2$.

So for any sequence of pairs of primes $p$ and $ap+b$ that are expected to occur infinitely often ($p$ and $p+2$, or $p$ and $2p+1$, or $\ldots$), the sum of $1/p$ for such primes converges.

That the sum of the reciprocals of the twin primes converges indicates that this summation is the wrong thing to be looking at. We want a strategy to prove the infinitude of twin primes, and that suggests a better sum. The Bateman-Horn conjecture predicts the number of $n \leq x$ such that $f_1(n), \ldots, f_r(n)$ are all prime is asymptotic to $Cx/(\log x)^r$ where $C$ is a positive constant depending on $f_1, \ldots, f_r$, and if $A(x) \sim cx/(\log x)^r$ as $x \to \infty$ for some $c > 0$ then $\sum_{n \leq x} a_n(\log n)^{r-1}/n \sim c\log\log x$. Therefore we expect (but have never proved) that the sum of $(\log p)/p$ over prime $p \leq x$ such that $p$ and $p+2$ are prime should grow like $c\log\log x$ for some constant $c > 0$, and a similar asymptotic estimate (for a different constant $c$) should hold for the sum of $(\log p)/p$ over all prime $p \leq x$ such that $p$ and $2p+1$ are prime.

KConrad
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