Let $\mu(n)$ the Möbius function and $s=\sigma+it$ the complex variable, then I've defined the Dirichlet series $$\epsilon(s):=\sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}.$$ And now I know that using absolute convergence it converges for $\Re s>1$, since $$\left|\sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}\right|\leq\sum_{n=1}^\infty\frac{1}{n^{\Re s}}.$$
Question. Can you improve this abscissa where the series is convergent? If it is not feasible explain why, if your explanation is possible. Thanks in advance.
The motivation of this series is to learn more about convergence and calculations with series.