Showing $\tau(n)/\phi(n)\to 0$ as $n\to \infty$ I was wondering how to show that $\tau(n)/\phi(n)\to 0$, as $n\to \infty$. Here $\tau(n)$ denotes the number of positive divisors of n, and $\phi(n)$ is Euler's phi function.

Here's a hint: let $Q(n)$ denote the largest prime power that divides $n$. Then prove:


*

*$\displaystyle \frac{\tau(n)}{\phi(n)} \le 2 \frac{\tau(Q(n))}{\phi(Q(n))} \le \frac4{\log2} \frac{\log Q(n)}{Q(n)}$;

*$Q(n) \to \infty$ as $n\to \infty$.
For #1, you'll want to use the fact that $\tau(n)/\phi(n)$ is multiplicative, as well as the explicit evaluations of them on prime powers.

