Asymptotic Growth of Euler Phi Function

Theorem
Let $\phi$ be the Euler $\phi$ function.

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


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

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


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

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 Prime Power we have:
 * $\phi \left({p^k}\right) = p^k - p^{k-1}$

for a prime power $p^k$.

Therefore:

Therefore:


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