Does a positive constant $\nu$ exist so that $\varphi(n)>\nu\cdot n$ for all $n$?

Question

Does a positive constant $\nu$ exist so that $\varphi(n)>\nu\cdot n$ for all $n$? Clearly this problem is exactly the same as asking if $\prod\limits_{i=1}^\infty \frac{p_i-1}{p_i}=0$. This is because $\varphi(n)=n\prod\limits_{p|n}\frac{p-1}{p}$. And if $n$ has $k$ prime divisors this is clearly greater than $n\prod_{i=1}^k\frac{p_i-1}{p_i}$.

So the problem breaks down to figuring out to what the sequence $a_n=\prod\limits_{i=1}^n\frac{p_i-1}{p_i}$ converges. (It clearly converges since it is decreasing and bounded). In fact we only need to figure out if it converges to zero or something else.

Answer 1

If $\prod_{k=1}^n\left(1-\frac{1}{p_i}\right)$ converges to a non-zero value, then:

$$\prod_{k=1}^n \frac{1}{1-\frac{1}{p_i}}$$ converges. Then write $$\frac{1}{1-\frac{1}{p_i}}=\sum_{j=0}^\infty \frac{1}{p_i^j}$$

Then $$\prod_{k=1}^n\frac{1}{1-\frac{1}{p_i}} = \sum_{m} \frac{1}{m}$$

where the sum is over the natural numbers $m$ that have no prime factors other than $p_1,\dots,p_n$.

In particular, then, $$\prod_{k=1}^n\frac{1}{1-\frac{1}{p_i}} > \sum_{m=1}^{n} \frac{1}{m}$$

And $\lim_{n\to\infty} \sum_{m=1}^{n}\frac{1}{m}=+\infty$.

Answer 2

Since $\displaystyle \sum_i \dfrac{1}{p_i} = \infty$, $\displaystyle \prod_i \left(1 - \dfrac{1}{p_i}\right) = 0$.

Answer 3

Rosser and Schoenfeld (1962) prove that, for $n \geq 3,$ $$ \frac{n}{\phi(n)} \leq e^\gamma \log \log n + \frac{2.50637}{\log \log n}, $$ where the constant $2.50637$ is the value that gives equality at the primorial $$ 223092870 $$ and nowhere else.

Meanwhile, Nicolas (1983) that the condition that $$ \frac{N}{\phi(N)} > e^\gamma \log \log N $$ for all primorials $N$ is equivalent to the Riemann Hypothesis.

Here we go, Hardy and Wright, theorem 328, page 267 in my edition, result of Landau that $$ \liminf \frac{\phi(n) \log \log n}{n} = e^{-\gamma} $$ That is, there is a sequence of numbers with $$ \phi(n) \approx n \left( \frac{e^{-\gamma} }{ \log \log n} \right) $$ and no such constant $\nu$ exists.