Are there any tight bounds, or any nontrivial ones actually, known for, with the lack of a better name, the Alternating Mertens Function: $$ S(n) = \sum_{k=1}^{n} \left((-1)^k \mu\left(k \right)\right) $$


Thanks to i707107. As stated in the answer here, A convergence problem: splitting a double sum , this sum has a more general asymptotic as: $$ S_\theta(n) = \sum_{k=1}^{n} \left(e^\left(2 \pi i \theta k \right) \mu\left(k \right)\right) $$

For which, $ S(n) $ is just a special case where $ \theta = \frac{1}{2} $. It turns out due to a result by H. Davenport that $ S_\theta(n) $ is $ O(x/\ln^\mathcal{E}(x)) $ for all real $ \theta$ and all $ \mathcal{E} > 0 $.

My question is for a more improved bound though, as this bound is not too much greater thank that of $ O(x) $.

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    $\begingroup$ This doesn't have the answer to your question, but I thought it might interest you: oeis.org/A247418 $\endgroup$ – Robert Soupe Nov 25 '14 at 2:20
  • $\begingroup$ The sum I had was the negative of that, so it would imply for at least small values it tends to be somewhat regular. I really have no idea how it might diverge though. I assume it will, but as to what rate, I have no idea. $\endgroup$ – Paul LeVan Nov 25 '14 at 4:22
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    $\begingroup$ See this: math.stackexchange.com/questions/332832/… $\endgroup$ – i707107 Nov 26 '14 at 2:58
  • $\begingroup$ So, for any $K>0$, it is bounded above by $n/(\log n)^K$. $\endgroup$ – i707107 Nov 26 '14 at 7:58
  • $\begingroup$ the asymtotics of $\sum_{n < x} \mu(n) e^{2 i \pi n a}$ is connected to the equivalent of the prime number theorem for Dirichlet L-functions : $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s}$ and $\frac{1}{L(s,\chi)} = \sum_{n=1}^\infty \chi(n) \mu(n) n^{-s}$. As for $\zeta(s)$, you prove $L(s,\chi)$ has no zero on $Re(s)=1$, and then you write $\sum_{n < x} \chi(n) \mu(n)$ as a contour integral, and you show it is $o(x)$ ($\mathcal{O}(x^{1/2+\epsilon})$ if RH was true). Finally, because of the discrete Fourier transform : $e^{2 i \pi n a} = \sum_{\chi} c(\chi,a) \chi(n)$ and voila $\endgroup$ – reuns Jul 12 '16 at 23:18

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