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Clearly if $p$ is prime, the sequence $\frac{\phi(p)}{p} \rightarrow 1$. In general, however, if $s_n \in S \subseteq \mathbb{N}$, we are not even guaranteed of the existence of: $\displaystyle \lim_{n \to \infty} \frac{\phi(s_n)}{s_n}$.

My question is this: Does there exist an infinite sequence $S \subseteq \mathbb{N}$ such that $\displaystyle \lim_{n \to \infty} \frac{\phi(s_n)}{s_n}=1$ and at most finitely many $s_i$ are prime?

My intuition tells me no, but I'm not sure. Having the above limit exist for some sequence alone is quite a strong statement, so having it equal one and contain finitely many primes is pretty restrictive. Admittedly, that isn't an argument and I've been having trouble finding one. Any thoughts would be appreciated.

EDIT: It just occurred to me that if $s_i=p_i^2$ for prime $p_i$, then the above holds. Reformulating, is there such an $S$ with finitely many prime powers?

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Certainly. Let $\{p_i\}_{i\in\mathbb{N}}$ enumerate the prime numbers, and take $s_n=p_np_{n+1}$. We have $$\lim_{n\rightarrow\infty}\frac{\phi(p_np_{n+1})}{p_np_{n+1}}=\lim_{n\rightarrow\infty}\frac{p_np_{n+1}-p_n-p_{n+1}+1}{p_np_{n+1}}=1.$$

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  • $\begingroup$ Awesome. Thanks. $\endgroup$
    – M10687
    Apr 14, 2016 at 6:18
  • $\begingroup$ And informally, if $n$ has "only large factors" then $\phi(n)$ is "approximately $n$". A product of consecutive primes is a good way to ensure a limited number of large factors while avoiding prime powers. $\endgroup$
    – Steve Jessop
    Apr 14, 2016 at 9:47
  • $\begingroup$ I am reminded of the Q of showing that $\lim \sup_{n\to \infty}\phi (n+1)/\phi(n)=1$ and $\lim \inf_{n\to \infty}\phi (n+1)/\phi (n)=0.$.... Let $Q_n$ be the product of the first $n$ primes. Using $n/p_n\to 0,$ where $p_n$ is the $n$th prime, and using $\prod_{j=1}^n(1-1/p_j)\to 0,$ we can show that $\phi (Q_n)/Q_n \to 0$ and that $\phi ( Q_n\pm 1)/( Q_n\pm 1)\to 1.$ $\endgroup$
    – DanielWainfleet
    Apr 14, 2016 at 13:04

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