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What kind of upper bound can one get for $$ \sum_{d|n}\frac{1}{d^{\sigma}}, $$ where $0<\sigma<1$? The best I can do is $\ll n^{(1-\sigma)/2}$ by breaking the sum at $\sqrt{n}$, using the symmetry of the divisors, and replacing the sum on $d|n$, $d<\sqrt{n}$, with the sum over all $d<\sqrt{n}$. Since the hyperbola method gives that the average value is $O(1)$, I'm hoping for a tighter upper bound.

Update: Thanks to Lucia for reminding me of the better but still trivial bound $$ < d(n)\ll C(\epsilon)n^\epsilon. $$ The article Will Jagy cites below shows that on the Riemann Hypothesis the sum (in the $\sigma>1/2$ case I'm interested in) is $$ \exp\left(O\left(Li(\log(n)^{1-\sigma})\right)\right). $$

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    $\begingroup$ This is the topic in a Ramanujan article fragment math.univ-lyon1.fr/~nicolas/ramanujanNR.pdf The entirety of the original article was not published as there were shortages of many things, including paper. The recent movie gives a good feel for this. I should add that i wrote to Nicolas for confirmation of some asymptotics (at least under RH) and he confirmed that there was a sudden transition at one end of the $0 \leq \sigma \leq 1$ segment, cannot immediately recall. $\endgroup$ – Will Jagy Jul 14 '17 at 20:29
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    $\begingroup$ For example it is less than $d(n)$, which is less than $C(\epsilon)n^{\epsilon}$ for any $\epsilon >0$. $\endgroup$ – Lucia Jul 14 '17 at 20:49
  • $\begingroup$ Is this not a multiplicative function of n? If it is, solving it for prime powers should work. For prime powers $p^k$ I get something less than A/(A-1) where A is $p^{\sigma}$. Gerhard "Surely It Must Be Multiplicative" Paseman, 2017.07.14. $\endgroup$ – Gerhard Paseman Jul 14 '17 at 20:57
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    $\begingroup$ @WillJagy This is helpful; if you make it an answer I'll accept it. $\endgroup$ – Stopple Jul 14 '17 at 21:02
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This is the topic of an article fragment by Ramanujan. The story as I understood it was that there were wartime paper shortages, and the original article was shortened. Nicolas and Robin seem to be giving a different reason.

I also asked about aspects of this on MO some years ago but wound up mostly answering myself, Estimate term in Ramanujan Lost Notebook (classic analytic number theory)

I wrote to Nicolas about this a few years ago. He confirmed what my experiments at the time appeared to indicate, there is a sudden jump in the asymptotics at one end of the segment you are calling $0 \leq \sigma \leq 1,$ at least under RH.

I don't have the major portion of the original article. I learned the basics from Nicolas 1988

Now I remember what I was going to include. The original article by Alaoglu and Erdos on Colossally Abundant numbers is available. For those and for the original Superior Highly Composite numbers, we can solve both directions in closed form: given real $\delta > 0,$ and a prime $p,$ we can find the best exponent $k$ for Ramanujan's optimum construction. Also, given a prime $p$ and exponent $k,$ we can find the first (largest) $\delta > 0$ that causes $p$ to be given exponent $k.$ For the non-integer $\sigma$ under current discussion, the later direction cannot be solved in closed form. This makes the whole problem of programming these sums rather unwieldy. I did come up with something, but an awful lot of work.

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Combining both observations, one has the trivial bounds $$d(n)n^{-\sigma/2}\le S\le \frac{d(n)}{2}(1+n^{-\sigma})$$ The upper bound is optimal when $n$ is prime. Can one do better ?

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We have

$\displaystyle \sum_{d|n}\frac{1}{d^{\sigma }}\leq \sum_{d\leq d(n)}\frac{1}{d^{\sigma}}$, for $0< \sigma <1$

The smaller denominator means that the value of the fraction will be greater.

And

$\sum_{d=1}^{d(n)}\frac{1}{d^{\sigma}}=\zeta (\sigma) + \frac{1}{1-\sigma}d(n)^{1-\sigma} + \frac{1}{2}d(n)^{-\sigma} - \frac{\sigma}{12}d(n)^{-\sigma-1} + O(d(n)^{-\sigma-2})$

$\zeta (\sigma)=\frac{1}{1-\sigma} + \gamma + O(\sigma - 1)$

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  • $\begingroup$ This answer is helpful particularly if you're interested in explicit values for the implied constants. For example, with $\epsilon=1/4$, one can show $d(n)<4n^{1/4}$. So we improve on Lucia's answer to an estimate whose main term is (a known constant times) $n^{(1-\sigma)/4}$. (Also does not assume RH!) $\endgroup$ – Stopple Jul 15 '17 at 19:36
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For $k\ge 1$ Handbook of number theory I (Sándor, Jó.; Mitrinović, D. S. & Crstici, B.) gives a result of Sándor̈ (On Jordan’s arithmetical function. Math. Student 52 (1984), 91–96, 1988): $$\prod_{d\mid n}\frac{p^{2k}-1}{p^{k}-1}\left(\frac{1}{p^k}\right)\le\frac{\sigma_k(n)}{n^k}<\prod_{d\mid n}\frac{p^{k}}{p^{k}-1}.$$

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  • $\begingroup$ Thanks, but $k\ge 1$ is not the range I'm interested in. $\endgroup$ – Stopple Jul 17 '17 at 15:20

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