Density of $\{ \ \{\ln k\} \ \}_{k=1}^{\infty}$ Is this sequence dense in $(0,1)$? I want to say that it is, I think the transcendence of the logarithm function is leading me to believe that it is, but I don't know how to prove it.
 A: A start: As $n\to\infty$, $\ln(n+1)-\ln n\to 0$, but $\ln n\to\infty$.
More Detail: Let $a\in [0,1]$, and let $\epsilon \gt 0$. We show there is an $n$ such that the fractional part pf $\ln n$ differs from $a$ by less than $\epsilon$.
Let $K=K_{\epsilon}$ be a positive integer such that $\frac{1}{K}\lt \epsilon$. Divide the interval $[0,1]$ into $K$ subintervals of equal length. Then $a$ lies in one of these subintervals. We will show that for any subinterval of $[0,1]$ of length $\frac{1}{K}$, there is some $n$ such that the fractional part of $\ln n$ lies in that subinterval.
Note that $\ln(n+1)-\ln n$ approaches $0$ monotonically. So there is an $N$ such that for all $n\ge N$ we have $\ln(n+1)-\ln(n)\lt \frac{1}{K}$.
Let $N^\ast$ be the smallest integer $\gt eN$. We have $\ln N^\ast\gt 1+\ln N$. 
As $j$ travels through the integers from $N$ to $N^\ast$, $\ln j$ increases by at least $1$. But since $j\ge N$, we have $\ln(j+1)-\ln j\lt \frac{1}{K}$, so $\ln j$ increases  in steps of length $\lt \frac{1}{K}$. In particular, the fractional part of $\ln j$ enters at least once any subinterval of $[0,1]$ of length $\frac{1}{K}$.
Remark: The fact that the logarithm function is transcendental is not used in the argument. The same idea can be used to show that the set of fractional parts of $\sqrt{n}$ is dense in $(0,1)$. 
