Trying to show that $\phi(n) > c_1 \frac{n}{\text{log log }n}$ for some constant $c_1 > 0$ where $\phi(n)$ is the euler phi function.

I was wondering if I could use something like

\begin{align} \phi(n)\geq \frac{n}{e^{\gamma}\log \log n}+O\left(\frac{n}{(\log \log n)^2}\right) \end{align}

to show this, but I'm not sure. Could anyone maybe give me some advice or tips on how to do this? I have quite a few Analytical Number Theory texts and notes with me, so even suggestions to some theorems would be great. Thanks.

  • $\begingroup$ Is the formula you give one that you already know or one that you are conjecturing? $\endgroup$ – Laertes Oct 18 '14 at 17:44
  • $\begingroup$ one that I already know. $\endgroup$ – Pablo Oct 18 '14 at 17:47
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    $\begingroup$ Then your result follows, since the first term dominates all other terms as $n \to \infty$, so that there always exists some $\epsilon$ such that $\phi(n)\geq \frac{n}{e^{\gamma}\log \log n}(1+\epsilon)$ where $\lim_{n \to \infty} \epsilon=0$ and for all $n>N$ for some $N$, $\epsilon$ is monotonic decreasing, from which the result follows. $\endgroup$ – Laertes Oct 18 '14 at 17:53
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    $\begingroup$ See the answer here and here. $\endgroup$ – Dietrich Burde Oct 18 '14 at 18:47
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    $\begingroup$ $c_1={1+\epsilon \over e^\gamma}$ for the highest value of $\epsilon$ for any $n$. One value that appears to work for $c_1$ is $1\over 9$, although I haven't rigorously tested that. $\endgroup$ – Laertes Oct 19 '14 at 17:02

The best theorem to look up is, I think, the following effective lower bound for $\phi(n)$

Theorem: For all $n>2$ we have $$ \frac{\phi(n)}{n}>\frac{1}{e^{\gamma}\log \log n+\frac{3}{\log \log n}}. $$

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  • $\begingroup$ If I prove this result, can I then say that $\frac{\phi(n)}{n} > \frac{1}{\text{loglog }n}\left(\frac{1}{e^\gamma+3}\right)$? $\endgroup$ – Pablo Oct 19 '14 at 10:55
  • $\begingroup$ Also do you maybe have a link to a proof of this result? $\endgroup$ – Pablo Oct 19 '14 at 11:28
  • $\begingroup$ Yes, it is the same link as above: see the answer with the link "LOWER" here. $\endgroup$ – Dietrich Burde Oct 19 '14 at 15:49

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