Is there a proof that no lower bound exists for the totient function?

Question

I read here that there is no lower bound for the totient function. Is there a proof of that?

Answer 1

If you read carefully, the article says not that there is no lower bound, but that there is no linear lower bound — that is, there's no $c$ (and $n_0$) such that for all $n\gt n_0$, $\phi(n) \geq cn$.

There is a relatively straightforward proof of this: consider primorials, that is, numbers of the form $n=2\cdot 3\cdot 5\cdot\ldots\cdot p_k$. Then for these $n$, $\displaystyle\frac{\phi(n)}{n} = \prod_{i=1}^k\left(1-\frac1{p_i}\right)$. This product can be shown to 'diverge' to zero as $i$ goes to $\infty$, essentially because the sum $\sum_i\frac{1}{p_i}$ diverges: $\ln\left(\prod(1-\frac1{p_i})\right) = \sum\ln(1-\frac1{p_i}) = \sum\left(-\frac1{p_i}+O(\frac1{p_i^2})\right)$, and since the latter sum goes to $-\infty$, then the logarithm must go to $-\infty$ likewise, and thus the product goes to $0$.

The product's divergence to zero thus means that for any $\epsilon$ we can find some $n_0$ (a sufficiently large primorial) such that $\frac{\phi(n_0)}{n_0}\lt\epsilon$, or in other words such that $\phi(n_0)\lt\epsilon\cdot n_0$.