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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.

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  • $\begingroup$ Speculative approach: split the sum into two ranges $A<d<B$ and $d\ge B$. Your argument for the latter range gives (better than) $N^\varepsilon/B^3$, which is tolerable if $B/A>N^\varepsilon$ say. The result would then follow if one could show that the number of divisors of $N$ between $A$ and $B$ is at most a power of $\log N$. I don't know if that's true or not in this situation. $\endgroup$
    – Greg Martin
    Commented Jan 13, 2019 at 20:47

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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.

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  • $\begingroup$ Aleksei, you have shown above that the number of divisors of $n$ in (A,2A] can be $\text{exp}(O(\log(n)/\log\log(n)))$. Why that implies what $\sum_{\substack{d|N \\ d>A}}d^{-3}$ has an optimal upper bound of $d(N)/A^3$? For instance, $\sum_{d|n}d^{-1}$ has an upper bound $\log\log(n)$ which is smaller than an upper bound for $d(n).$ $\endgroup$
    – user164144
    Commented Jan 14, 2019 at 16:38
  • $\begingroup$ @user164144 Sorry, I should have written it more clearly. Of course I didn't provide lower bound anywhere close to $d(N)/A^3$(and I doubt that this is possible), but I gave an example with $\exp(C\log(N)/\log\log(N))/A^3$ which differs from $d(N)/A^3$ only by constant $C$ in exponent. Yet it is more than enough to prove that nothing close to $\log(N)^m/A^3$ is possible. I also have similar example for small $A$ which gives something like $1/A^{2-o(1)}$, I can add it if it is of interest for you. $\endgroup$
    – Aleksei Kulikov
    Commented Jan 14, 2019 at 23:26
  • $\begingroup$ I get it now. he latter bound you mention of $A^{-2+o(1)}$ is one that I can show, but is not useful to me. Thanks though. $\endgroup$
    – user164144
    Commented Jan 16, 2019 at 0:51

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