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.

Note that for an odd integer $n>1$, we have $\tau(n)/\varphi(n)<1$ (first note it for an odd prime-power, then by multiplicativity to all odd $n$).
For each $\epsilon>0$, consider the set $S_\epsilon$ of those odd $n$'s with $\tau(n)/\varphi(n)>\epsilon$. Write $n=p_1^{\alpha_1}\dots p_k^{\alpha_k}$, where $p_1, \dots, p_k$ are odd. Then
$$\frac{\alpha_1+1}{p_1^{\alpha_1}} \dots \frac{\alpha_1+1}{p_1^{\alpha_1}}>\epsilon$$
and since each term of the product is $<1$, each term must itself be $>\epsilon$. Hence, for each $i$,
$$\frac{1+\alpha_i}{p_i^{\alpha_i}}>\epsilon.$$
Clearly there are finitely combinations of $p$ and $\alpha$ which satisfy this. Hence $S_\epsilon$ is finite. Now for a general integer $n=2^mn'$ where $n'$ is odd and with $\tau(n)/\varphi(n)> \epsilon$, note that $\tau(n)/\varphi(n)<2\tau(n')/\varphi(n')$, so there are finitely many choices for $n'$, and $\tau(n)/\varphi(n)<\frac{m+1}{2^m}$ so there are finitely many choices for $m$.
Hence, for every $\epsilon>0$, there are finitely many integers $n$ with $\tau(n)/\varphi(n)>\epsilon$.
Here is a plot of $\tau(n)/\varphi(n)$ for $n<10^5$:

And also, here is a picture of a baby bear
