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? 
Please, let me thank you in advance for your insightful replies...
 A: 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 $1.80 \times 10^{18}$.
