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

You write "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". Since any natural number can be written as a product of primes (unique up to reordering the factors) your set A will in this case be all natural numbers. So in this case your product can (in a formal sense) be rewritten as $$ \prod_{primes}\frac{p}{p-1} = \prod_{primes}\frac{1}{1-1/p} = \prod_{primes}\sum_{i=0}^{\infty} \frac{1}{p^i} = \sum_{n=1}^{\infty} \frac{1}{n} $$ To make sense of this formal manipulation, you could look at the function $$ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{primes}\frac{p^s}{p^s-1} \ \ \ \ , \ \ \ \ (s>1)$$ which is known as the Riemann zeta function - and arguably the most famous function in all of mathematics. Your product would then correspond to the value of the Riemann zeta function as 1. As mentioned above the Riemann zeta-function has a pole at one (i.e. it is infinite), but one can could try to look at the asymptotic expansion of the sum $$ \sum_{n=1}^{N} \frac{1}{n}$$ as $N$ goes to infinity. In this case something interesting does happen, namely $$ \lim_{N} (\sum_{n=1}^{N}\frac{1}{n} - \ln(N)) = \gamma$$ where $\gamma$ is the Euler-Mascheroni constant. Now in your case one could argue that one should look not at the asymptotic expansion of $ \sum_{n=1}^{N} \frac{1}{n}$, but rather at the asymptotic expansion of $$ \prod_{p \ prime}^{p < N}\frac{p}{p-1}$$ but I would expect the answer to be the same. In short - even though the product diverges doesn't mean something cool won't happen along the way.

Answer 2

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

Answer 3

An alternative approach: $\frac{p}{p-1}\geq 1+\frac{1}{p}$ gives $$ \prod_{p}\frac{p}{p-1}\geq \prod_{p}\left(1+\frac{1}{p}\right) =\!\!\!\!\!\!\!\! \sum_{\substack{n\geq 1\\n\text{ squarefree}}}\!\!\!\!\!\!\!\frac{1}{n} $$ but since the set of squarefree numbers has a positive density in $\mathbb{N}$ ($\frac{6}{\pi^2}$, which is also the probability that two "random integers" are coprime), by summation by parts it follows that $$\sum_{\substack{1\leq n\leq N\\n\text{ squarefree}}}\!\!\!\!\frac{1}{n}=O(1)+\frac{6}{\pi^2}\sum_{n=1}^{N}\frac{1}{n}=\frac{6}{\pi^2}\log(N)+O(1) $$ so the original product is divergent.