This is quite an old question, and I guess OP already have an answer. For the future reference, however, I write down an answer.
The limit turns out to diverge to $+\infty$. Indeed, let $f : [0, 1] \to \Bbb{R}$ by
$ f(x) = (1 - \{1/x\})(\tfrac{1}{2} - x\{1/x\}) \quad \text{and} \quad f(0) = 0. $
It is easy to check that
- $0 \leq f(x) \leq \tfrac{1}{2}$,
- $f$ is Riemann-integrable, and hence
- $ n^{-1} \sum_{k=1}^{n} f(k/n) \to \int_{0}^{1} f(x) \, dx =: C > 0$ as $n \to \infty$.
Now fix any $m$. Then for all $n > m$, non-negativity of $f$ shows that
$ \sum_{k=1}^{n} \frac{\log k}{n} f\left(\frac{k}{n}\right) \geq \sum_{k=m}^{n} \frac{\log m}{n} f\left(\frac{k}{n}\right).$
Taking liminf as $n \to \infty$, we get
$ \liminf_{n\to\infty} \sum_{k=1}^{n} \frac{\log k}{n} f\left(\frac{k}{n}\right) \geq C \log m $
for any $m$ and therefore the limit diverges to $+\infty$. ////
In particular, it suggests that the original problem where $\log$ is replaced by von Mangoldt function $\Lambda$ requires some clever estimates on $\Lambda$ to produce a vanishing limit.