OIES sequence A002473 is the sequence of 7-smooth numbers: positive numbers whose prime divisors are all less than or equal to 7.
Let S be the infinite summation of all the reciprocals of the members of A002473. Does the sum converge, and if so then what is the sum?
In case it is not obvious,
35/8 = (2/1)*(3/2)*(5/4)*(7/6)
For 11-smooth numbers, the answer is (35/8)*(11/10))
For 13-smooth numbers, the answer is (35/8)*(11/10)*(13/12).
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Further, it now occurs to me that if p and q and r are prime, that the set of all numbers whose only prime divisors are 1 and p and q and r is just the piecewise product of all powers of p times all powers of q times all powers of r. Because the reciprocals of each of these converge, the sum of the reciprocals can be obtained by multiplying the sum of reciprocals of powers of p times the sum of the reciprocals of powers of q times the sum of the reciprocals of powers of r. In other words, (p/(p-1))*(q/(q-1))*(r/(r-1)). This is more direct than my previous analytical solution.
Edited on January 23, 2019, 8:38 pm
Beyond 7, and a simpler solution
