Show $\forall\varepsilon>0\,\lim_{n\to\infty}\frac{\#\{\text{positive divisors of n}\}}{n^\varepsilon}=0$ Show that $\forall\varepsilon>0,$
$$\lim_{n\to\infty}\frac{\#\{\text{positive divisors of  n}\}}{n^\varepsilon}=0$$
I'm trying to solve this problem for a long time, but I'm really stuck I have totally no idea where to start.
I tried replacing $\varepsilon$ by $\frac{1}{k}$ where $k$ is a natural number, and show the statement is true for all $k$ by induction, but I haven't succeeded and it doesn't seem promising.
If you give me any advice or comment, I would greatly appreciate.
Thank you.
 A: I will give you a plot of demonstration I have come up with, although I am not sure about its formality.
The limit we aim to prove is equivalent to prove that, if we are given primes $p_i ;  i\in [N]$, then $$\lim_{\alpha_i \to \infty \text{ for some i}} \frac{\prod_{i \in [N]}(\alpha_i + 1)}{\prod_{i \in [N]}p_i^{\alpha_i e}}= 0$$, where we have used that the number of divisors of the number $n = \prod_{i \in [N]}{p_i}^{\alpha_i}$ equals $\prod_{i \in [N]}(\alpha_i + 1)$.
If you are able to prove the last statement formally, I guess you will not have problems to finish this demonstration.
A: The solution for this problem falls out from the following fact.
\begin{equation}
e^{O(\frac{log(n)}{loglog(n)})}=d(n)
\end{equation}
So,
\begin{equation}
\lim_{n->\infty}\frac{d(n)}{n^\epsilon}=\frac{e^{\frac{Clog(n)}{loglog(n)}}}{n^\epsilon}
\end{equation}
For some constant C.
Therefore,
\begin{equation}
\lim_{n->\infty}\frac{e^{\frac{Clog(n)}{loglog(n)}}}{n^\epsilon}=\lim_{n->\infty}\frac{n^\frac{C}{loglog(n)}}{n^\epsilon}=\lim_{n->\infty}n^{\frac{C}{loglog(n)}-\epsilon}
\end{equation}
You can add a nice proof regarding epsilon and deltas to show from here that
\begin{equation}
\lim_{n->\infty}\frac{C}{loglog(n)}-\epsilon=-\epsilon\\and\\
\lim_{n->\infty}n^{-\epsilon}=0\\
implies\\
\lim_{n->\infty}n^{\frac{C}{loglog(n)}-\epsilon}=0
\end{equation}
This completes our proof.  Many websites have the first point written and I believe you may be able to find out more about it from other stack exchange posts.
