On the sets $\{\dfrac {\phi(n+1)}{\phi(n)} : n\in \mathbb Z^+\}$ and $\{\dfrac {\phi(n)}{n} : n\in \mathbb Z^+\}$

Question

How to show that the set $\{\dfrac {\phi(n+1)}{\phi(n)} : n\in \mathbb Z^+\}$ is dense in $\mathbb R^+$ ? Also that the set $\{\dfrac {\phi(n)}{n} : n\in \mathbb Z^+\}$ is dense in $(0,1)$ ?

Answer 1

Let $B=\{\frac{\phi(n)}{n}:n\in\mathbb Z^+\}$. Let $m\#$ denote the product of all primes $\le m$ (primorial) and let $p_m$ denote the $m$th prime. For $1\le k\le m$ let $$f(m,k)=\frac{\phi(\frac{p_m\#}{p_k\#})\cdot p_k\#}{p_m\#}=\prod_{p_k<p<\le p_m}\left(1-\frac1p\right).$$ Note that $$ \ln f(m,k)=\sum_{p_k<p<\le p_m}\ln\left(1-\frac1p\right)<-\sum_{p_k<p<\le p_m}\frac 1p. $$ From the divergence $\sum \frac 1p$, we conclude that for any $k$, $\lim_{m\to\infty}f(m,k)=0$. Pick $k_0$ such that $1-\frac1p>\frac ab$ for all $p>p_{k_0}$. The pick $m$ big enough such that $f(m,k_0)<b$. If $f(m,k_0)>a$, we are done. Otherwise consider $f(m,k_0)<f(m,k_0+1)<\ldots <f(m,m)=1$: At each step, the numbers differ by a factor $<\frac ba$, hence some $f(m,k)$ must fall between $a$ and $b$. This shows that $B\cap (a,b)$ is nonempty for any open subinterval of $[0,1]$, i.e., $B$ is dense in $[0,1]$.