# Yet another question on sums of the reciprocals of the primes

I recall reading once that the sum $$\sum_{p \,\, \small{\mbox{is a known prime}}} \frac{1}{p}$$ is less than $$4$$.

Does anybody here know what the ultimate source of this claim is?

• How do you define a "known" prime? – Wojowu Mar 9 at 17:04
• I think that by "known prime" it is meant a number whose primality has already been established (maybe with the aid of the computer). For example, $2^{82 \, 589 \, 933} -1$ is a "known prime"; further, even though Bertrand's postulate guarantees the existence of many prime numbers in the interval $(2^{43 \, 112 \, 609}-1,2^{57 \, 885 \, 161}-1)$, as far as I know there are no "known primes" in that range. – José Hdz. Stgo. Mar 9 at 17:32
• Taking a (very generous) estimate that testing primality of a number with $>20$ digits takes one microsecond, in a thousand years we could have only checked around $10^{16}$ such numbers. Blindly estimating using the second Mertens theorem will give the result. – Wojowu Mar 9 at 17:35
• My previous comment was wrong, though I am going to leave it. Firstly, I have mixed up the calculation using Mertens' theorem, which gave me that we would have to sum primes up to $10^30$. Second, I have severely underestimated the power of sieving primes. – Wojowu Mar 9 at 19:03
• I don't know the ultimate source, but I know where I first heard it: a talk given by Matiyasevich, maybe 30 years ago. Although as I recall, his claim was that the sum of the reciprocals of the known primes was less than $5$ – and would always remain so. – Gerry Myerson Mar 9 at 21:58

It's well known that the sum of the reciprocals of the primes below $$n$$ tends to $$\log \log n + M$$, where $$M$$ is a small constant (the Meissel-Mertens constant). That is to say:

$$\sum\limits_{\small{\mbox{prime}} \, p \, < \, n} \frac{1}{p} = \log \log n + M + o(1)$$

This allows us to determine an approximate lower bound on the number of primes we would need to include in the series in order to surpass $$4$$. Specifically, the number of primes is minimised if we take an initial segment, and we would need to go up to:

$$e^{e^{4 - M}} \approxeq 1.80 \times 10^{18}$$

assuming the $$o(1)$$ term can be neglected.

Sieving up to this point with a Segmented Sieve of Eratosthenes (which parallelises quite easily) would not take particularly long at all, especially if you optimise by only checking numbers that are $$\pm 1 \mod 6$$. Sebah and Gourdon 2002 were able to compute the sum of reciprocals of twin primes up to $$10^{16}$$, and computing power has advanced considerably since then.

To give a comparison, the first SHA1 collision involved $$9.2 \times 10^{18}$$ hash computations, which is orders of magnitude more work than would be required to sieve the primes up to $$1.80 \times 10^{18}$$.

• Huh, in my rough calculation I have mixed up $-M$ and $+M$ in the formula, thus getting a bound a dozen orders of magnitude larger, which lead to my comment above. However, your answer clearly indicates that it is entirely possible to find enough primes to get the reciprocal above $4$. – Wojowu Mar 9 at 18:49
• Nicely et al. have analyzed all the primes up to $2^{64}$ in their efforts to study prime gaps, you can read up on it here. They have looked at all the primes in this interval, so by your calculation, which has pushed the sum of reciprocals of known primes to around $4.05$, thus disproving the claim OP has asked about. – Wojowu Mar 9 at 19:02
• See also ams.org/journals/mcom/2014-83-288/S0025-5718-2013-02787-1 (Oliveira e Silva, Herzog and Pardi). They had to compute primes up to $4\cdot 10^{18}$. – H A Helfgott Mar 9 at 19:25
• "and computing power has advanced considerably since then": [citation needed]. 2002 is close to the bend to near constant computing power versus time (Figures A.1 and A.2 here ). It is an endless source of frustration in my research that computing power has barely advanced in the last 15 years. Parallelizing independent (to avoid memory contention) calculations has advanced, but individual calculations ... not so much. – Eric Towers Mar 9 at 23:41