The set $\{\frac{\varphi(n)}n:n\in \Bbb N\}$

Question

Let $f(n)=\varphi(n)/n$, where $\varphi$ is the totient function. Since $0<\varphi(n)\le n$ and $$\lim_{n\to \infty}f(p_n)=1$$ (where $\{p_n\}$ is the increasing sequence of primes) and $$\lim_{n\to\infty}f(n\#)=0$$ we know that $\limsup f(n)=1$ and $\liminf f(n)=0$. But is the set $\{f(n):n\in\Bbb N\}$ dense in $[0,1]$?

Warning: this is not a problem from a book, so it might be very hard (honestly, I have no idea).

Answer 1

Let $p_n$ denote the $n$-th smallest prime number. For each $\epsilon > 0$, choose $N$ such that $1/p_N < \epsilon$. Now consider $n_k = p_{N+1}\cdots p_{N+k}$ so that

$$ f(n_k) = \left(1 - \frac{1}{p_{N+1}}\right) \cdots \left(1 - \frac{1}{p_{N+k}}\right). $$

This gives

$$ |f(n_k) - f(n_{k-1})| \leq \frac{1}{p_{N+k}} \leq \frac{1}{p_N} < \epsilon. $$

Since $f(n_k) \to 0$ as $k\to\infty$, it follows that for each $x \in [0, 1]$ there is $n_k$ such that $|x - f(n_k)| < \epsilon$. This proves that the set in question is dense in $[0, 1]$.

Answer 2

Yes. As $\sum_p\frac{1}{p}=+\infty$, we get $$\prod_p f(p)=0.$$ Now choose $a\in[0,1]$ and a prime $p_n$ such that $\frac{p_n-1}{p_n}\geq1-a$. Consider the sequence $$u_n=\prod_{k=0}^nf(p_{n+k})=\frac{\varphi\left(\prod_{k=0}^np_{n+k}\right)}{\prod_{k=0}^np_{n+k}}=f\left(\prod_{k=0}^np_{n+k}\right).$$ Then $u_n\to0$ and the difference between consecutive terms is at most $a$, so we're done.