I am interested in an upper bound for


$$\sum\_{\substack{d\mid N\\ d>A}}\frac{1}{d^3},$$


in particular, I can show that above is


$$\ll\frac{\text{exp}\left(C\frac{\log(N)}{\log\log(N)}\right)}{A^3}$$


for some positive constant C. However I would like to do better. I think that the upper bound should be around
$$
\frac{\log^m(N)}{A^3}$$
for some other positive constant $m$.


We also have that $\log(N)<A<N^{1/2}$.
References are welcome.



A truncated divisor sum
I will prove below that your bound $\frac{\exp\left(C\frac{\log N}{\log \log N}\right)}{A^3}$ (which follows from $\sum\_{d\mid N, d > A} \frac{1}{d^3} \le \frac{d(N)}{A^3}$) is optimal at least in the regime $A = N^c$, where $0 < c < 1$ is fixed. (note that we can't have $A$ very small since $\sum\_{d\in \mathbb{N}} \frac{1}{d^3} < \infty$).


Put $N = p\_1p\_2\ldots p\_k$ where $k$ is some natural number and $p\_j$ are prime numbers. From, say, prime number theorem we have $k = \Theta \left(\frac{\log N}{\log \log N}\right)$. I will construct $\Theta\left(\exp\left(C\frac{\log N}{\log \log N}\right)\right)$ divisors of $N$ in the interval $(A, 2A]$ from what the desired estimate follows. Construction goes as follows:


we choose random subset of primes $p\_j$ with $j > [k\left(1 - \frac{1}{100}\min(c^5, (1-c)^5)\right)] = m$ and call their product $d\_1$(note that there are already required number of $d\_1$'s). It is also easy to see that $d\_1 < A$. Then we are doing the following greedy algorithm: initialize $d := d\_1$. Lets look at $p\_j$ starting with $p\_m$ in decreasing order and multiply $d$ by $p\_j$ until one more multiplication will make $d$ greater than $A$. Such a moment exists since $p\_1p\_2\ldots p\_m > A$. Call this moment $j$. We have now that $d \le A < p\_jd$. If $p\_j d \le 2A$ then let $d:=p\_j d$ and finish the algorithm. Otherwise by Bertrand's postulate there is some prime $p$ in the interval $(\frac{A}{d}, \frac{2A}{d}]$ and $p < p\_j$. Let $d := pd$ and finish the algorithm. 


In any case we will find divisor $d\in (A, 2A]$ of $N$ and all of them are obviously different. Thus our claim is proved.


