# Maximum of $\frac{\phi(i)}i$

For a given $N$, is there an approach to find the maximum $\frac{\phi(i)}i$ ($2\le i\le N$)?

Like for $n=2$, $\frac{\phi(2)}2=\frac12$ is maximum

for $n=3$, $\frac{\phi(3)}3=\frac23$ is maximum

for $n=4$, $\frac{\phi(3)}3=\frac23$ is maximum

From wikipedia's definition : $\frac{\phi(n)}n = \prod_{p|n}\left (1 - \frac1p\right)$.

So, I guess $p$ should be very big for this approach. Still, I can't find a generalised approach. Please help.

• What do you mean "maximum": $\,\frac{\phi(3)}{3}=\frac23\,$ , just like that. It's not a maximum but just a value. A maximum can be searched from a set with at least two values, otherwise it is trivial. – DonAntonio May 4 '13 at 10:31
• There is a set.The set consists of all phi(i)/i for 2≤i≤n – mohan May 4 '13 at 10:32
• Oh, I see...now – DonAntonio May 4 '13 at 10:33
• Your guess that $\max\{\frac{\phi(i)}i\mid 2\le i\le n\}$ equals $\frac24=\frac12$ when $n=4$, is wrong. Obviously, $\frac23$ is bigger. – Hagen von Eitzen May 4 '13 at 10:37
• Oh ya,edited !! – mohan May 4 '13 at 10:39

Let $n\ge 2$ be given and $p$ the biggest prime $\le p$. Then $\frac{\phi(p)}p=1-\frac1p$. If $2\le i\le n$, then $i$ has $m\ge1$ prime factors $q_k\le p$ ($1\le k\le m$), hence $$\frac{\phi(i)}i=\prod_{k=1}^m\left(1-\frac1{q_k}\right)\le \left(1-\frac1p\right)^m\le 1-\frac1p.$$ Thus $1-\frac1p$ is indeed the maximum value: If $n\ge2$, then $$\max\left\{\frac{\phi(i)}i\biggm| 2\le i\le n\right\} =1-\frac1{\max\{p\mid p\le n, p\text{ prime}\}}$$