Prove that $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$

suppose that $\phi(n)$ is Euler function. prove that, $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$

(if $A_n=\{1 \leq m \leq n \mid m \in \Bbb N ; \gcd(n,m)=1\}$ then $\phi(n)=|A_n|$)

I think : If $p$ is prime then $\phi(p)=p-1$. for any $ \varepsilon > 0 $ there is a prime number $p$ such that $1-\varepsilon \leq \frac{p-1}{p} \leq 1$. for other elements i don't know what can i do.

Solution 1:

This hinges on the fact that $$\sum_{p\text{ prime}}\frac1p=+\infty$$ which in turn implies that $$\prod_{p\text{ prime}}\frac{p-1}p=0\,.$$ Once you know this fact, the proof is easy. Let $p_n$ be the $n$-th prime. Let $\epsilon>0$. There is some prime number $q=p_{N}$ such that $\frac{q-1}{q}\geq 1-\epsilon$. Then consider the sequence of numbers $$k_n=\prod_{i=0}^n\frac{p_{N+i}-1}{p_{N+i}}=\frac{\phi\left(\prod_{i=0}^np_{N+i}\right)}{\prod_{i=0}^np_{N+i}}$$ Then $k_n$ tends to zero, and the difference between consecutive terms is $\leq \epsilon$. Hence any real number $\in[0,1]$ can be approximated by some $\frac{\phi(n)}n$ up to arbitrary $\epsilon>0$.