$\sum_{d\leq x} (\mu(d)/d) \log(x/d)$: is (the analogue of) Mertens' conjecture still false? It is known to be false that $\sum_{m\leq x} \mu(m) \leq \sqrt{x}$ for all $x$ (Mertens' conjecture), and it is generally believed that $\lim \sup_{x\to \infty} |M(x)|/\sqrt{x} = \infty$. From the latter, it would follow that $$\lim \sup_{x\to\infty} \sqrt{x} \left|\sum_{m\leq x} \frac{\mu(m)}{m}\right| = \infty,$$
by partial summation. However, what about a smoothed sum, such as
$$\sum_{m\leq x} \frac{\mu(m)}{m} \log \frac{x}{m}?$$
Is it clear that 
$$\lim \sup_{x\to\infty} \sqrt{x} \left|\sum_{m\leq x} \frac{\mu(m)}{m} \log \frac{x}{m} - 1\right| = \infty?$$
If one can't deduce the truth of this easily from $\lim \sup_{x\to \infty} |M(x)|/\sqrt{x} = \infty$, I'd be interested in what standard random models imply on the matter. 
 A: We have that
\[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} = \frac{1}{2\pi i} \int_{\sigma_0 - i\infty}^{\sigma_0 + i\infty} \frac{1}{\zeta(s + 1)} \frac{x^s}{s^2} \, ds\]
for $\sigma_0$ sufficiently large; see the bit about Riesz typical means in Section 5.1 of Montgomery and Vaughan.
Now move the contour to the left (I'm ignoring the issue of the horizontal contours - these can be dealt with, though it takes a little effort). We pick up a pole at $s = 0$ with residue $1$. From here, the standard methods (Section 15.1 in Montgomery and Vaughan) show that
\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| > 0.\]
EDIT: Lucia is right that the Riemann hypothesis together with Linear Independence hypothesis do not imply that
\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \infty.\]
Rather, they imply that
\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}.\]
To see this, assume the Riemann hypothesis and the simplicity of the zeroes of $\zeta(s)$, and mimic the proof of Lemma 4 of Ng's paper on the summatory function of the Möbius function to get an explicit expression more or less of the form
\[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1 = \sum_{\rho} \frac{x^{\rho - 1}}{\zeta'(\rho) (\rho - 1)^2} + o\left(\frac{1}{\sqrt{x}}\right),\]
where the sum is over the nontrivial zeroes of $\zeta(s)$. This gives the bound
\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| \leq \sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}.\]
If one additionally assumes the Linear Independence hypothesis, then the Kronecker-Weyl equidistribution theorem implies that this is in fact an equality; this method goes back to the work of Ingham.
A conjecture due to Gonek and Hejhal states that
\[J_{-1/2}(T) := \sum_{0 < \gamma < T} \frac{1}{|\zeta'(\rho)|} \asymp T (\log T)^{1/4}.\]
Using methods from random matrix theory, Hughes, Keating, and O'Connell conjecture more generally that $J_k(T) := \sum_{0 < \gamma < T} |\zeta'(\rho)|^{2k} \sim c_k T (\log T)^{(k + 1)^2}$ whenever $\Re(k) > -3/2$, where
\[c_k = \frac{1}{2\pi} \frac{G(k + 2)^2}{G(2k + 3)} \prod_p \left(1 - \frac{1}{p}\right)^{k^2} \sum_{m = 0}^{\infty} \left(\frac{\Gamma(m + k)}{m! \Gamma(k)}\right)^2 \frac{1}{p^m},\]
with $G(z)$ being the Barnes $G$-function. So by partial summation, the Gonek-Hejhal conjecture implies that
\[\sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}\]
converges.
I'm not surprised that your numerical calculations show that this is very small; Kotnik and van de Lune do numerical calculations for a similar sum over zeroes (see Table 5 of their paper) and obtain a sequence of partial sums that seem to converge to something very small.
On the other hand, if the Riemann hypothesis is false or if $\zeta(s)$ has a zero of order greater than $1$, then
\[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \infty.\]
(These can both be proved using the methods in Section 15.1 of Montgomery and Vaughan.)
