Proof there is no lower bound for the totient function.

Question

I wish to show there is no constant $c$ such that $ c n \leq \varphi (n) $, is this proof correct?.

From Euler's identity for the zeta function we have:

$$ \prod \left( 1 - \frac{1}{p} \right) = \frac{1}{\zeta (1)} $$

where the product is over all primes. We conclude the product approaches $0$. This means, for every constant c greater than $0$ there we can give a certain $k$ such that.

$$ \frac{\varphi(n_k)}{n_k} \leq c $$

where $n_k$ denotes the multiplication of the first k primes. This is as required.

Answer 1

Without using divergent series, your argument can be modified this way:

Define $a_m$ to be the product of the first $m$ prime numbers. Now calculate $\varphi(a_m)/a_m$ and see what happens as $m\to\infty$.