# A truncated divisor sum

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). References are welcome.

• 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. – Greg Martin Jan 13 '19 at 20:47

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.
• 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 $d(n).$ – user164144 Jan 14 '19 at 16:38
• @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. – Aleksei Kulikov Jan 14 '19 at 23:26
• 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. – user164144 Jan 16 '19 at 0:51