Does the function change the sign infinitely many times? Let $0<\epsilon<\frac{1}{2}$; $\mu$ denotes the Moebius function. Consider the sums
$$
\sum_{k=1}^n\frac{\mu(k)}{k^\epsilon}.
$$
Do they change the sign infinitely many times as $n\to\infty$? (like the Mertens function corresponding to $\epsilon=0$)
 A: One can prove that this changes sign infinitely often by the same method used to prove, say, that $\psi(x) - x$ changes sign infinitely often. A standard reference for this is chapter 15 of Montgomery and Vaughan.
A: Titchmarsh* at p. 370 shows that convergence of $W(x)= \sum_{n =1}^\infty\frac{\mu(n)}{n^s}$ for $\sigma > 1/2$ is a necessary and sufficient condition for the [truth of] the Riemann hypothesis. 
Also, assuming RH, the sum converges to $1/\zeta(s)$ for every $\sigma > 1/2.$ For values of $\sigma$ for which the sum converges it must be possible to bound the sum in some neighborhood for sufficiently high $n,$ and so the sum would not change sign. 
Also on RH, for $1/2<\sigma,$ Titchmarsh at 371-72 shows that
$M(x) = \sum_{n=1}^\infty\mu(x)=\Omega(x^{1/2})$ using exactly the same technique used by Schmidt to show that $\psi(x)-x=\Omega_\pm(x^{1/2}).$ This can't be extended to $W(x)$ (at least for this interval) if $W(x)$ doesn't change sign.
As for $0<\sigma<1/2,$ this MO question/answer seems to cover it. As the comments note, $W(x)$ diverges here. There is a comment on the MO question that I think is both correct and nice. Assume $W(x)$ is convergent for $\sigma_o <1/2$ on $(0,1/2).$ Then the series will converge in the half-plane $\sigma_o<\sigma.$ But there is at least one zero of $\zeta(s)$ for $\sigma = 1/2$ which is a singularity of $W(x),$ contradicting the assumption.    
*Titchmarsh, The Theory of the Riemann Zeta-function
