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I need to prove that for $n \ge 3$

$\frac{n}{\phi(n)}=O(\mathrm{loglog}n)$

But I don't know how to start, I'll be grateful if you could give me some hints. Thanks.

Mugenen
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    $\prod_{p | n} (1+p^{-1}) = O(\log \log n)$ it follows from $\psi(2x)-\psi(x) = O(\log {2x \choose x}) = O(x)$ so that $\psi(x) = O(x)$ and (by partial summation) $\sum_{p \le x} p^{-1} =O( \sum_{n \le x} (\psi(n)-\psi(n-1)) \frac{1}{n \log^2 n}) O(\sum_{n \le x} \psi(n) (\frac1{n\log^2 n}-\frac1{(n+1)\log^2 (n+1)})= O(\log \log x)$ – reuns Jun 25 '19 at 09:44
  • Do you know what phi(n)/n looks like? Can you see the connection with Riemann Zeta function? – Astaulphe Jun 25 '19 at 09:45

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If $n=p^a$ then you can compute easily $\varphi(n)$ and $n/\varphi(n)$, and $a$ is a kind of log. If $n$ is arbitrary, use the fact that $\varphi$ is multiplicative. This is not the final answer, but it is a start..

Marco Farinati
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