Solution to infinite product $\prod_{p-primes}^{\infty} \frac{p}{p-1}$

Question

I want to find the $\prod_{p-primes}^{\infty} \frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum
$$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots$$
of the reciprocals of the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? Through some working out you can work out that the solution to this question is $\sum_0^\infty \frac{1}{2^n}$$\sum_0^\infty \frac{1}{3^n}$$\sum_0^\infty \frac{1}{5^n}$, simple power series multiplication. But I started to wonder what the sum would be if $A$ was the set of positive integers that have no prime factors other than all primes. I think that the product will converge since $\sum_1^\infty log(a_n)$ seems to converge, but hey I could be wrong.

Answer 1

${p\over p-1}=1+{1\over p-1}>1+\frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.