Euler's Phi Function Worst Case

For each positive integer $k$, let $P_k$ denote the product of the first k primes. Show that $\varphi(P_k) = \theta(P_k / \log \log P_k)$ is the worst case, in the sense that $\varphi(n) = \Omega(n / \log \log n)$.

Any help would be appreciated. Thanks

Solution 1:

The overall result is due to Landau, 1903, Theorem 328 in Hardy and Wright, that $$ \liminf \; \frac{e^\gamma \phi(n) \log \log n}{n} = 1, $$ proof in section 22.9.

Oh, well. Note that the statement you think you are trying to prove is incorrect. Nicolas showed that if the Riemann Hypothesis is true, then whenever $n = 2 \cdot 3 \cdot 5 \cdots p_k$ is a primorial, then $\frac{e^\gamma \phi(n) \log \log n}{n} < 1.$ If the Riemann Hypothesis is false, then $\frac{e^\gamma \phi(n) \log \log n}{n} < 1$ for infinitely many primorials $n$ and $\frac{e^\gamma \phi(n) \log \log n}{n} > 1$ for infinitely many primorials $n.$ As the Rosser and Schoenfeld lower bound from my earlier post and Wikipedia holds for all numbers greater than 2, this says that the ratios depicted that are below 1 are awfully close to it.

For more precise information, see both my answers at Is the Euler phi function bounded below?

sources: NICOLAS and a comment at PLANAT