[Math] Best upper bound on the number of divisors of $n$ that are larger than $N$.

Question

I am looking for the best upper bound on $$\sum_{\substack{d | n\\ d \geq N}} 1.$$

I know that
$$
d(n) = \sum_{\substack{d | n}} 1 \leq e^{O(\frac{\log n}{\log \log n})}.
$$

For my application, I need something like
$$\sum_{\substack{d | n\\ d \geq N}} 1 \leq \frac{o(n^{\epsilon})}{\log N} \quad \forall \epsilon > 0.
$$

A reference where the bound can be found or a simple proof would be appreciated.

Thanks.

EDIT: Johan Andersson has pointed out that the third display follows from the second. (Thanks.) I am still interested to learn what the best known bound is.

Answer 1

This is not an answer to what the best known upper bound is, but rather a comment that the (known) average distribution of divisors indicates you might not expect to do any better than the bounds on the divisor function itself. Tennenbaum's "Introduction to Analytic and Probabilistic Number Theory", $\S$ 6.2 p. 207 says

For each integer $n$, let us define a random variable $D_n$ taking the values $\log d/\log n$, as $d$ runs through the the set of the $\tau(n)$ divisors of $n$, with uniform probability $1/\tau(n)$. The distribution function $F_n$ of $D_n$ is then defined by $$ > F_n(u):=\text{Prob}(D_n\le > u)=\frac{1}{\tau(n)}\sum_{d|n,d\le > n^u}1\quad (0\le u\le 1). $$ It is clear that the sequence $\{F_n\}_{n=1}^\infty$ does not converge pointwise on $[0,1]$. However, we shall see the sequence of Cesaro means $$ > G_N(u):=\frac{1}{N}\sum_{n\le N}F_n(u) > $$ is uniformly convergent on $[0,1]$. Remarkably, the limit is the distribution function of a probability law well known to specialists: the arcsine law, with density $1/(\pi\sqrt{u(1-u)})$. Large and small values have high probability: if $D$ is a random variable with this distribution law, we have $$ > \text{Prob}(D<0.01\text{ or > }D>0.99)\approx0.128 $$ This indicates that, on average, an integer has many small (and correspondingly many large) divisors.

Update: One thing we learn is that the relevant parameter is not $N$, but $u:=\log_n(N)$.