For a given sequence of real numbers $(x\_n)\_{n=1}^\infty \subset [0,1]$, let $A(a,b,N)$ to be the number of terms $x\_n$ of the sequence up to index $N$ such that $a \leq x\_n \leq b$. A sequence of real numbers $(x\_n)\_{n=1}^\infty$ is said to be uniformly distributed modulo 1 if $\displaystyle \frac{A(a,b,N)}{N}$ tends to $b-a$ as $N \rightarrow \infty$ for every $0 \leq a < b \leq 1$. 


My question concerns whether or not the following sequence is uniformly distributed. Define $x\_n = H\_n - \lfloor H\_n \rfloor$, where $H\_n = \displaystyle \sum\_{k=1}^n \frac{1}{k}$ is the $n$-th harmonic number. Then is $(x\_n)\_{n=1}^\infty$ uniformly distributed? 


What if we instead defined $x\_n$ to be a random element of the interval $[H\_{n-1}, H\_n]$ with respect to a uniform distribution, and ask if the random sequence $(x\_n - \lfloor x\_n \rfloor)\_{n=1}^\infty$ is uniformly distributed?



Is a sequence of the following type uniformly distributed modulo 1?
$H\_n$ is a asymptotic to $\log n+\gamma+O(1/n)$. This means that the values of $x\_n$ for $e^k \leq n\leq e^{k+1/2}$ all fall into the same interval of length about $1/2$. The sequence is not equidistributed (the proportion of $x\_n$ in that interval for $n\leq e^k$ and for $n\leq e^{k+1/2}$ differ much).


