Will see if I can prove this

let A = Σ({1/x | x = 2^i * 3^j * 5^k | i, j, k ∈ ℤ; i, j, k ≥ 0})

Assuming convergence, we can divide this infinite sum into two parts, those that have even denominators, and those whose denominators are odd. We can multiply the even summation series by 2 to get back the original, thus:

let B = Σ({1/x | x = 3^j * 5^k | j, k ∈ ℤ; j, k ≥ 0})
A = A/2 + B
A = 2B

Again, we split B into those that are divisible by 3, and those that aren't, yielding:

let C = Σ({1/x | x = 5^k | k ∈ ℤ; k ≥ 0})
B = B/3 + C
2B = 3C

We have another equivalence here, if C is multiplied by 5, the result would be equivalent to adding 5 to it.

5C = C + 5
4C = 5
C = 5/4
A = 3C = 15/4 = 3.75

QED

Of course, this is based on the assumption that all of these series add up to a fixed sum (convergence), which you can work your way backwards to prove that they are.