Where does this series converge $\sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}$, being $\mu(n)$ the Möbius function?

Question

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.

Answer 1

That is not a Dirichlet series strictly speaking, since $(-1)^{\mu(n)}$ is not a multiplicative function.
We have $$(-1)^{\mu(n)}=1-2\mu(n)^2$$ and square-free numbers have a positive density among integers, hence $$ \sum_{n\geq 1}\frac{(-1)^{\mu(n)}}{n^s} = \zeta(s)-2\sum_{n\geq 1}\frac{\mu(n)^2}{n^s} = \zeta(s)-\frac{2\,\zeta(s)}{\zeta(2s)}$$ and the abscissa of convergence is one.

Answer 2

The density of squarefree integers is $6/\pi^2$, so about 60% of the integers up to $X$ are squarefree (and this is very accurate for large $X$). So 60% of the time, your summands are $-1/n^s$, while 40% of the time your summands are $1/n^s$.

At $s = 1$, the expected value of your sum up to $X$ is $-\frac{1}{5}X$, which clearly diverges. Therefore the abscissa of convergence is $1$.