Mean of fractional part of $\log n$ Let $\{x\}$ be the fractional part of $x.$ Can we show that 
$$\lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N~\{\log n\}= _{\pm}\Omega(1/2)? $$
In case my usage is nonstandard, I mean that the l.h.s. is infinitely often less than and greater than 1/2. 
Caveat--I am not sure it's true because I haven't made any headway towards proving it.
(I see that someone has asked a similar question about log n. This isn't quite the same so am posting it.)
 A: For any real $x$, let $\lfloor x \rfloor \in \mathbb{Z}$ and $\{x\} = x - \lfloor x \rfloor \in [0,1)$ be the integer and fractional part of $x$.
For any $x > 1$, the function $\{\log x\}$ is discontinuous at $e, e^2, e^3, \ldots$. Since $e$ is transcendental, this set of numbers is disjoint with $\mathbb{Z}$, the set of discontinuities of the function $\lfloor x \rfloor$. 
Using this fact, the average at hand can be rewritten as a Riemann Stieltjes integral
and integrate by part without any issue.
For any $N > 2$, write $\log N$ as $n + s$ where $n = \lfloor\log N\rfloor$ and $s \in (0,1)$. We have
$$\begin{align}
\sum_{k=1}^N \{\log k\} &= \sum_{k=2}^N \{\log k\}
= \int_{1+}^{N+}\{\log x\} d\lfloor x \rfloor
= \bigg[\{\log x\}\lfloor x \rfloor\bigg]_{1+}^{N+}
- \int_{0+}^{n+s+} \lfloor e^y \rfloor d\{y\}\\
&= Ns -\underbrace{\int_{0+}^{n+s+} e^y d\{y\}}_I
+ \underbrace{\int_{0+}^{n+s+} \{ e^y \} d\{y\}}_J\tag{*1}
\end{align}
$$
For the $2^{nd}$ term, we have
$$\begin{align}I 
&= \int_{0+}^{n+s+} e^y d( y - \lfloor y \rfloor )
= \int_0^{n+s} e^y dy - \sum_{k=1}^n e^k\\
&= e^{n+s} - 1 - \frac{e}{e-1}(e^n-1)
= N\left( 1  - \frac{e^{1-s}}{e-1}\right) + \frac{1}{e-1}
\end{align}\tag{*2}$$
For the $3^{th}$ term, we have
$$J = \int_{n}^{n+s} \{e^y\} dy + \sum_{k=0}^{n-1}\int_{k+}^{k+1+} \{e^y\} d\{y\}
   = \int_{n}^{n+s} \{e^y\} dy + \sum_{k=0}^{n-1}\int_0^1 \left( \{ e^{k+y} \} - \{ e^{k+1} \}\right) dy$$
Notice the magnitude of all the $n+1$ integrands in above expression is bounded from above by $1$. We obtain following bound of $J$.
$$|J| \le \int_n^{n+s} dy + \sum_{k=0}^{n-1}\int_0^1 dy = n + s = \log N\tag{*3}$$ 
Substitute $(*2)$ and $(*3)$ into $(*1)$, we obtain
$$\frac{1}{N}\sum_{k=1}^N \{\log k\} = f(\{\log N\}) + \epsilon_N
\quad\text{ where }\quad f(s) = \frac{e^{1-s}}{e-1} + s - 1$$
and the error term $\epsilon_N$ satisfies
$$|\epsilon_N| \le \frac1N \left(\log N + \frac{1}{e-1}\right)$$
From this, we can conclude the average at hand oscillate like $f(\{\log N\})$ as $N \to \infty$.
Since
$$f([0,1]) = \left[\log\frac{e}{e-1}, \frac{1}{e-1}\right]
\approx [\;0.458675, 0.5819767\;]$$
the average will oscillate between $0.458675$ and $0.5819767$ and hence below/above $\frac12$ infinitely often. 
At the end is a picture plotting the average vs the approximation $f(\{\log N\})$. It clearly shows the average approximately follows $f(\{\log N\})$ even for $N$ as small as $20$.

A: Let $k$ be a nonnegative integer and let $0<\delta<1$. Note that $$\sum_{e^k\le n<e^{k+\delta}} \{\log n\} = \sum_{e^k\le n<e^{k+\delta}} (\log n - k)$$ is a Riemann sum for $$(e^{k+\delta}-e^k) \int_{e^k}^{e^{k+\delta}} (\log t - k)\,dt.$$
Indeed, since the integrand is increasing, the difference between the Riemann sum and the integral itself is bounded by the difference between consecutive sample points (which is $1$ here) and the change in the integrand from one endpoint to the other (which is $\delta<1$ here). Moreover, the value of the integral is
$$
\int_{e^k}^{e^{k+\delta}} (\log t - k)\,dt = \big( t\log t - (k+1)t \big) \bigg|_{e^k}^{e^{k+\delta}} = 
$$
argh ran out of time, somebody want to finish?
