For arbitrarily small number $\epsilon$ > 0, is there a natural number $n$ such that $\frac{\phi(n)}{n}< \epsilon$? For arbitrarily small number $\epsilon$ > 0, is there a natural number n such that $\frac{\phi(n)}{n}< \epsilon$ ?  $\phi(x)$ is the Euler's totient function, representing number of positive integers up to a given integer $x$ that are relatively prime to $x$. I do think the answer would be yes but I'm stuck with the proof. Any insights?
 A: Let $p_n$ represent the $n$th prime. Let $q_n=\prod\limits_{i=1}^np_i$. I claim that $\lim\limits_{n\to\infty}\frac{\phi(q_n)}{q_n}=0$.
Note that $\frac{\phi(q_n)}{q_n}=\prod\limits_{i=1}^n1-\frac1{p_i}$. So, we want to prove that $$\prod\limits_{n=1}^\infty1-\frac1{p_n}=0$$ Can you finish from here?
Hint: $$\prod\limits_{n=1}^\infty\left(\frac1{1-\frac1{p_n}}\right)=\sum\limits_{n=1}^\infty\frac1n=\infty$$
A: Remember that
$$\frac{\varphi(n)}n=\prod_{p|n}\left(1-\frac1p\right).$$
So, the minimum values of $\varphi(n)/n$ will occur when lots of different primes divide $n$. Because of this, it makes sense to set
$$n=p_1p_2\cdots p_k,$$
where $p_i$ is the $i$th prime. So, it basically suffices to show that the product
$$\prod_{i=1}^{\infty}\left(1-\frac1{p_i}\right)$$
is $0$ (tends to $0$ as the upper limit tends to $\infty$). 
One thing that can be helpful to prove this is reciprocating it: by this, you want to show that
$$\prod_{i=1}^\infty \left(\frac1{1-\frac1{p_i}}\right)$$
diverges. The identity
$$\frac{1}{1-r}=\sum_{k=0}^\infty r^k$$
might be helpful.
