Let $\mu(n)$ denote the Mobius function with the well-known Dirichlet series representation
$$
 \frac{1}{\zeta(s)} = \sum\_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}.
$$
Basic theorems about Dirichlet series imply that if the Dirichlet series on the right converges for some $s = \sigma + it$, then it converges for all $s$ with real part $> \sigma$. Hence, $\sum\_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ converging is a sufficient condition for the Riemann hypothesis.


One way to approach this question is to use partial summation. Let $M(x) = \sum\_{n \leq x} \mu(n)$. Then
$$
 \sum\_{n \leq x} \frac{\mu(n)}{\sqrt{n}} = \frac{M(x)}{\sqrt{x}} + \frac{1}{2} \int\_{1}^{x} \frac{M(t)}{t^{3/2}} \, dt.
$$
Odlyzko and te Riele proved that $\liminf\_{x \to \infty} \frac{M(x)}{\sqrt{x}} < -1.009$ and $\limsup\_{x \to \infty} \frac{M(x)}{\sqrt{x}} > 1.06$. Much earlier, Ingham had showed that $M(x)/\sqrt{x}$ was unbounded assuming the linear independence of the imaginary parts of the zeroes of $\zeta(s)$.


In addition, Gonek has an unpublished conjecture (mentioned in [Ng's paper "The distribution of the summatory function of the Mobius function"](http://arxiv.org/pdf/math/0310381)) that 
$$
-\infty < \liminf\_{x \to \infty} \frac{M(x)}{\sqrt{x} (\log \log \log x)^{5/4}} < 0 <\limsup\_{x \to \infty} \frac{M(x)}{\sqrt{x} (\log \log \log x)^{5/4}} < \infty.
$$


Using these results and conjectures to address the original question seems to be challenging, because of possible cancellation between $\frac{M(x)}{\sqrt{x}}$ and $\int\_{1}^{x} \frac{M(t)}{t^{3/2}} \, dt$. My questions are the following:


1. Are known results about $M(x)$ enough to determine if $\sum\_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?
2. If not, does Gonek's conjecture (or any other plausible conjectures) imply that $\sum\_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?



Is it possible to show that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?
One can show that $\sum\_{n=1}^{\infty} \mu(n)/\sqrt{n}$ diverges. Suppose to the contrary that it converges, which as you note implies RH. Put $M\_0(x)=\sum\_{n\le x} \mu(n)/\sqrt{n}$, and our assumption is that $M\_0(x)=C+o(1)$ as $x\to \infty$. 


Note that for any $s=\sigma+it$ with $\sigma>1/2$ we have 
$$ 
\int\_0^{\infty} sM\_0(e^x)e^{-sx} dx = \sum\_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}} \int\_{\log n}^{\infty} se^{-sx} dx = \frac{1}{\zeta(s+1/2)}. \tag{1}
$$ 
Since $1/\zeta(s+1/2)$ is analytic (by RH) in $\sigma >0$, the identity above also holds in this larger domain. But from our hypothesis we note that the LHS above is 
$$ 
\int\_0^{\infty} s(C+o(1)) e^{-sx} dx = C + o(|s|/\sigma).
$$ 
Now take $s=\sigma+i\gamma$, where $\gamma =14.1\ldots $ is the ordinate of the first zero of $\zeta(s)$. Then the RHS of (1) is $\sim C\_0/\sigma$ for a constant $C\_0 \neq 0$ (essentially $1/\zeta^{\prime}(1/2+i\gamma)$). Letting $\sigma \to 0$ from above, we get a contradiction. 


Note that the same heuristics underlying Gonek's conjecture should also suggest that $M\_0(x)$ grows like $(\log \log \log x)^{5/4}$. I'm sure all this is classical, but I don't know a reference offhand. 


