Sum of the reciprocals of the products of primes Q. Find an upper bound for the following sum of an infinite series:
$$\sum(\frac{1}{2\times3}+\frac{1}{5\times7}+\frac{1}{11\times13}+...)$$
The denominator is a series of prime numbers.
I am aware that the sum of an infinite series of reciprocals of primes diverges, but I am told that the summation above converges.
 A: A similar sum, which also includes the other half of the products like 1/(3*5), 1/(7*11) and so on, is discussed in https://oeis.org/A210473 and I have posted estimates of that sort in a PDF file in a URL in https://oeis.org/A209329 .
A: Here is another hint.  Your sum of "adjacent" prime reciprocals is obviously less than:$$\frac1 {2^2}+\frac 1 {4^2}+\frac 1 {6^2}+...=\frac 1 4.\frac {\pi^2} 6=\frac {\pi^2} {24}=0.411 (approx.)$$  With only a bit more cunning I am sure that this upper bound could be improved significantly.
Here is another much better method:  Your sum is less than the following sum:$$\frac 1 {2.3}+\frac1 {5.7}+\frac1 {11.13}+\frac1 {17.19}+\frac1 {23.25}+...+\frac1 {(6n-1)(6n+1)}+...$$  This is easily shown to be (by treating the first term alone and then the rest as a simple infinite series using the cotangent expansion): $$\frac1 6+\frac1 {12}(6-\sqrt3\pi)=0.213216826...$$  This gives a much tighter upper bound for the prime series of fractions (the first four terms are the same!).  By actual numerical calculation, the original infinite sum involving prime numbers is just a little over 0.21, slightly less than this upper bound just calculated.
