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 nontrivial 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 nontrivial bound not assuming GRH on $\displaystyle \sum_{n\leq x}\mu(n)$. Thanks!

3$\begingroup$ For the sum over all $n\leq x$ see here $\endgroup$ – Wojowu Apr 21 at 13:18

2$\begingroup$ This question is also discussed here: mathoverflow.net/questions/381550/… $\endgroup$ – Milo Moses Apr 21 at 15:12

$\begingroup$ Thanks for the links. $\endgroup$ – Remarkably Unremarkable Apr 21 at 20:10
A variant of the SiegelWalfisz 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 MontgomeryVaughan: Multiplicative number theory I.

$\begingroup$ 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 $q\in \mathbb{N}$. $\endgroup$ – Remarkably Unremarkable Apr 21 at 20:10

2$\begingroup$ @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 1213 for Section 11.3. $\endgroup$ – GH from MO Apr 22 at 3:07

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