Let $f:\mathbb{R}^+\to \mathbb{R}^+$ be given by

$f(t) = 1$ if $0\leq t\leq 1$

$f(t) = 1/t^2$ if $t>1$.


What is the sum $\sum_{n\; \text{squarefree}} f(n/x)$, for $x$ large?

Just by the simplest inclusion-exclusion, one can get an estimate of the form $x/\zeta(2) + C \sqrt{x}$, where $C$ is a largish constant, in part by using highly optimized estimates for $\sum_{n\leq x: n\; \text{squarefree}} 1$ in the literature (e.g. Cohen and Dress, MR0952866). However, I'd imagine the smoothing inherent in the problem allows one to do better than that. How much better? (Experiments suggest a very small error term.)

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    $\begingroup$ There must be some typo since the sum in the question is $+ \infty$. $\endgroup$ – user25235 Jan 28 '13 at 15:24
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    $\begingroup$ Since the generating function for square-free numbers is $\zeta(s)/\zeta(2s)$, I believe you can get an error of $o(\sqrt{x})$ by using the classical zero-free region for the zeta-function (even without smoothing the sum). To get an error like $O(x^{1/2-\delta})$ for some $\delta>0$ seems more or less equivalent to a quasi-Riemann hypothesis. $\endgroup$ – Micah Milinovich Jan 28 '13 at 15:25
  • $\begingroup$ I'm with km. Indeed, for any set of integers n of positive density, I see the desired sum over that set as infinite. Gerhard "Ask Me About Unbounded Confusion" Paseman, 2013.01.28 $\endgroup$ – Gerhard Paseman Jan 28 '13 at 16:48
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    $\begingroup$ FWIW, there's an update to the Cohen-Dress paper (by Cohen, Dress, and El Marraki) available through ams.org/mathscinet-getitem?mr=2357309 $\endgroup$ – Barry Cipra Jan 28 '13 at 19:34
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    $\begingroup$ When I do the inclusion-exclusion, I get a main term of $2x/\zeta(2)$ rather than $x/\zeta(2)$ - each of the ranges $n\le x$ and $n>x$ seems to contribute an $x/\zeta(2)$. Do you think that's right? $\endgroup$ – Greg Martin Jan 29 '13 at 2:02

There are two approaches I can think of.

(a) Analytic. We need only information about $\zeta(s)$, not about other $L$-functions. Hence we can use the fact that RH has been verified up to a very large height. This should imply a result of the form

$\sum_{n\; \text{squarefree}} f(n/x) = (1+\epsilon) 2 \zeta(2) + O^*(C_{\epsilon'} x^{1/4+\epsilon'})$,

where $O^*(K)$ means " a quantity of absolute value at most $K$", $\epsilon$ is a very tiny constant, $\epsilon'>0$ is arbitrarily small and $C_{\epsilon'}$ depends only on $\epsilon'$.

Difficulty: giving an explicit, and preferably small, value for $C_{\epsilon'}$ does not seem very easy. We cannot shift the line of integration all the way to $\Re(s)=1/2$ without having a bound on the residues of $1/\zeta(s)$; if we shift the line of integration only to $\Re(s)=1/2+\epsilon'$, we still need upper bounds for $1/|\zeta(s)|$ (i.e., lower bounds for $\zeta(s)$). There is some non-explicit work in this direction, but I do not know of anything explicit with reasonable constants. (I would be very glad to be surprised.)

(b) "Elementary". A fully elementary bound would presumably follow Cohen and Dress, MR0952866, and give bounds of about the same quality (error term = $O(x^{1/2})$, where the implied constant is not large but also not very small). A more mixed approach would proceed as in the paper by Cohen, Dress and El Marraki cited by Barry Cipra above. Half of the work (sieving out factors m^2 with $m\leq c\cdot \sqrt{x}$) would indeed be improved by the smoothing. The other half would rely on estimates on $M(N) = \sum_{n\leq N} \mu(n)$. Here estimates of the form $|M(N)|\leq c N$, $c$ a small constant (for $N$ larger than a constant) are known, and lead to estimates of the form

$\sum_{n\; \text{squarefree}} f(n/x) = 2 \zeta(2) + O^*(c x^{1/2})$,

with $c$ rather small (but not "very tiny"; we are speaking about $10^{-3}$ rather than $10^{-20}$ , say) provided that $x$ is larger than a constant. Getting useful error-term estimates better than $O^*(c x^{1/2})$ is hard, however, in that it involves estimating $M(x)$: this seems hard for the same reason given above, namely, estimates for the residues of $1/\zeta(s)$ or upper bounds for $1/|\zeta(s)|$ become necessary.

Question: is there a third way? Or is there a way to give good upper bounds for $1/|\zeta(s)|$ for $|\Im(s)|\leq H_0$, assuming that the Riemann hypothesis holds for $|\Im(s)|\leq H_0$ (or a little further)?

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