Upper bounds for Euler's totient function on a set of numbers with unbounded number of prime divisors

Question

If we take an infinite set $S$ of positive numbers with the property that the number of prime divisors of the elements is unbounded above, then can we make $\phi(n)/n$ arbitrarily small for infinitely many $n \in S$? Can we do this by simply taking the number of prime divisors sufficiently large? I ask this in light of the following little calculation: First it is equivalent to show we can make $n/\phi(n)$ arbitrarily large for infinitely many $n \in S$. Now $$ \dfrac{n}{\phi(n)} = \prod_{p \mid n} \left(\dfrac{1}{1 - \frac{1}{p}} \right) = \prod_{p \mid n}\left( 1 + p^{-1} + p^{-2} + \cdots \right) = \sum_{\substack{d = 1 \\ \operatorname{rad}(d) \mid n}}^{\infty} \dfrac{1}{d}, $$ where $\operatorname{rad}(d)$ denotes the product of all the distinct prime divisors of $d$. So as the number of prime divisors of $n$ goes to infinity we get the divergent harmonic series. What are your thoughts on this? Thanks.

Answer 1

By making the primes that occur in the factorizations of elements of $S$ grow sufficiently fast, we can force $\frac{\varphi(n)}{n}$ to be bounded away from $0$ on $S$.

For example, let $q_1,q_2,\dots$ be an infinite increasing sequence of primes such that $q_{k+1}\gt 2q_k$, and let $S$ be the set of all products $q_1q_2\cdots q_m$.