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? 


Please, let me thank you in advance for your insightful replies...



Yet another question on sums of the reciprocals of the primes
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](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.464.1118&rep=rep1&type=pdf) 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](http://shattered.io/) 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 $1.80 \times 10^{18}$.


