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There is a result of Soundararajan on the upper bound of the partial sums of the Möbius function assuming GRH here. Suger and Halupczok find an analogous bound for $\displaystyle \sum_{\substack{n\leq x\\n \equiv a \bmod q}}\mu(n)$, also assuming GRH. However, without assuming GRH, what are some non-trivial upper bounds on $$\displaystyle \sum_{\substack{n\leq x\\n \equiv a \bmod q}}\mu(n)?$$ It does not have to be the most optimal one (although I would appreciate being linked to such a bound). In fact, I don't even know of a non-trivial bound not assuming GRH on $\displaystyle \sum_{n\leq x}\mu(n)$. Thanks!

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    $\begingroup$ For the sum over all $n\leq x$ see here $\endgroup$
    – Wojowu
    Apr 21, 2021 at 13:18
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    $\begingroup$ This question is also discussed here: mathoverflow.net/questions/381550/… $\endgroup$
    – Milo Moses
    Apr 21, 2021 at 15:12
  • $\begingroup$ Thanks for the links. $\endgroup$
    – user147650
    Apr 21, 2021 at 20:10

1 Answer 1

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A variant of the Siegel-Walfisz theorem states that there is a constant $c>0$ with the following property. For any $A>0$ and $q\leq(\log x)^A$, we have $$\displaystyle \sum_{\substack{n\leq x\\n \equiv a \bmod q}}\mu(n)\ll_A x\exp\left(-c\sqrt{\log x}\right).$$ See Exercise 13 for Section 11.3 of Montgomery-Vaughan: Multiplicative number theory I.

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  • $\begingroup$ Many thanks for your answer. Are there also some results when $q\ge (\log x)^A$? Actually, I am quite satisfied as long as I can be sure that $\displaystyle \sum_{\substack{n\leq x\\n \equiv a \bmod q}}\mu(n)=o(x)$ for all $a\in \mathbb{Z}$ and $q\in \mathbb{N}$. $\endgroup$
    – user147650
    Apr 21, 2021 at 20:10
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    $\begingroup$ @RemarkablyUnremarkable: One can go up to $q\leq\exp(2c\sqrt{\log x})$, with the same $c>0$ as before, but there will be a main term depending on a possible Siegel zero for quadratic Dirichlet characters modulo $q$. If the Siegel zero does not exist, then the same bound holds true as for $q\leq(\log x)^A$. See Exercises 12-13 for Section 11.3. $\endgroup$
    – GH from MO
    Apr 22, 2021 at 3:07
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    $\begingroup$ Thank you for the comment. $\endgroup$
    – user147650
    Apr 22, 2021 at 9:43

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