Asymptotic Growth of Euler Phi Function

Theorem
Let $\phi$ be the Euler's function

For any $\epsilon > 0$ and sufficiently large $n$ we have:


 * $n^{1-\epsilon} < \phi(n) < n$

Proof
It is clear that $\phi(n) < n$ for all $n$, so it is sufficient to prove that:


 * $\displaystyle \lim_{n \to \infty} \frac{n^{1-\epsilon}}{\phi(n)} = 0$

By Multiplicative Function that Converges to Zero on Prime Powers it is sufficient to prove that


 * $\displaystyle \lim_{p^k \to \infty} \frac{p^{k(1-\epsilon)}}{\phi(p^k)} = 0$

as $p^k$ ranges through all prime powers.

By Euler Phi Function of a Prime we have $\phi(p^k) = p^k - p^{k-1}$ for a prime power $p^k$.

Therefore

Therefore


 * $\displaystyle \lim_{p^k \to \infty} \frac{p^{k(1-\epsilon)}}{\phi(p^k)} \leq \lim_{p^k \to \infty}\frac{2}{p^{k\epsilon} } = 0$