Is $\{\{\ln(n)\} : n \in \mathbb{N)} \}$ dense in $[0,1]$? Is the sequence defined by $a_n=\{\ln(n)\}$ dense in $[0,1]$?
Note: $\{x\}$ denotes the fractional part of $x$
 A: Hint: $\ln (n+1) - \ln n \to 0.$
A: [Just to elaborate on zhw.'s hint.]
Suppose $(\{\ln n\})_n$ is not dense in $[0,1]$. Then there is some interval $(a,b)\subset [0,1]$ such that $\{\ln n\}\not\in(a,b)$ for every $n$. As $\ln n-\ln (n-1)\to0$ when $n\to\infty$ we can choose $N$ big enough so that $$\ln n-\ln (n-1)<b-a\,\,\,\text{ if }\,\,\,n> N.\hspace{1cm}(\star)$$ 
Since $\ln n\to+\infty$, there is $M>N$ so that $\ln N-\ln M\geq1$, which [along with $(\star)$] tells us that $P=\{\{\ln n\}\}_{n=N}^M$ is a partition of $[0,1]$ with diameter/norm $d_P<b-a$, implying the contradiction $P\cap(a,b)\neq\phi.$
A: Given an irrational number, $\alpha$, we know the set of $\{n\alpha\}$ is dense in $[0,1]$. 
If you know that $\ln 2$ is irrational, then $\{n\ln 2\}$ is dense in $[0,1]$ and since $n\ln 2 = \ln(2^n)$, this is a subsequence of your sequence, and hence your sequence is dense in $[0,1]$.
$\ln 2$ being irrational follows from $e$ being transcendental, since if $e^{p/q}=2$ then $e$ is a root of the rational polynomial $x^p-2^q$.  
This is all a "big gun" approach to this question.

You actually don't need that $e$ is transcendental to use this technique - you can show that at least one of $\ln 2$ and $\ln 3$ is irrational, because otherwise $2^p=3^q$ for some integers $p,q$. Then either $\{n\ln 2\}$ or $\{n\ln 3\}$ is dense in $[0,1]$. 
This works with any base -  $\{\log_b n\}$ is dense in $[0,1]$ - since at least one of $\log_b 2$ and $\log_b 3$ is irrational.
