# 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

• I think you mean $\phi(n) \gg n/\log\log n$, not $\phi(n) = \Omega(n/\log\log n)$. The first statement says that $\phi(n)$ is always at least that large; the second statement says only that $\phi(n)$ is infinitely often that large. – Greg Martin Mar 7 '13 at 1:55
• What does the phrase "is the worst case, in the sense" mean here? The symbols $\theta$ and $\Omega$ are well-defined; neither does "being the worst case" make any sense applied to $\theta$, nor is it clear (to me) how it could be imbued with any sense by the $\Omega$ statement. Please explain. – joriki Mar 7 '13 at 1:56
• It's a little hard to decide what your question is. Are you saying that you already know that $\phi(P_k) = O(P_k/\log\log P_k)$, and you're looking for a proof that $\phi(n) \gg n/\log\log n$ for any $n$? – Greg Martin Mar 7 '13 at 1:56
• What you stated is correct Greg – Matt Mar 7 '13 at 2:03
• Any ideas? I am still struggling on this one. Thanks – Matt Mar 7 '13 at 2:50

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.