Denote by $\mu$ the Möbius function and $\forall x\ge0$, $$M(x)=\sum_{n\le x}\mu(n)$$ the summatory Möbius function. It is know that $M(x)=o(x)$, which is a consequence of the prime number theorem. Are there any known real numbers $\alpha\in\left(\frac12,1\right)$ such that $M(x)=O(x^\alpha)$?
I'm guessing that there must necessarily be some known $\alpha$ for which this is true but that the proofs use powerful analytic number theory tools and extensive computing. I was most curious to know if a more elementary proof could be found for example in the case $\alpha=\frac34$.
The inverse zeta function has coefficients $\mu(n)$: $$ 1/\zeta(s) = \sum_{n \geq 1} \frac{\mu(n)}{n^s}. $$ Thus one way to understand partial sums of the Möbius function would be to use a Perron-type integral $$ \frac{1}{2\pi i} \int_{2 - i\infty}^{2 + i\infty} \frac{1}{\zeta(s)} \frac{X^s}{s} \; ds = \sum_{n \leq X} \mu(n), \tag{1}$$ valid for noninteger $X$. A zero of $\zeta(s)$ leads to a pole of $1/\zeta(s)$, and thus if $\beta = \limsup \sigma$ where $\sigma$ ranges over the real parts of zeros of $\zeta(s)$, then this Perron analysis implies that $$ \sum_{n \leq X} \mu(n) = \Omega(X^\beta). \tag{1}$$ On the other hand, if $$ \sum_{n \leq X} \mu(n) \ll X^\alpha, \tag{2} $$ then a straightforward partial summation argument shows that $1/\zeta(s)$ converges absolutely for $\mathrm{Re}(s) > \alpha$.
The obvious inequality $\beta \leq \alpha$ shows that any nontrivial $\alpha$ in $(2)$ would yield a zero-free region for $\zeta(s)$ of the form $\mathrm{Re}(s) > \alpha$. Any such result would be an enormous improvement over what's known.