TITLE: Can the partial sums of a series be uniformly distributed modulo 1?
QUESTION [5 upvotes]: Earlier I asked in Is a sequence of the following type uniformly distributed modulo 1? whether the partial sums of the harmonic series is uniformly distributed modulo 1. Here I ask for necessary and sufficient conditions for the partial sums of a series to be uniformly distributed modulo 1 (look at the original question for the definition of uniformly distributed modulo 1). 
Suppose $(a_n)_{n=1}^\infty$ is a sequence of positive numbers such that the sum of $a_n$'s diverges. Define $b_N = \displaystyle \sum_{n=1}^N a_n$. Under what conditions imposed on $a_n$ (such as growth rate) can we say that $(b_N)_{N=1}^\infty$ is uniformly distributed modulo 1?

REPLY [5 votes]: Every sequence of reals that is uniformly distributed modulo 1 arises as the partial sums of a divergent series of positive terms, so you are really asking for characterizations of sequences that are uniformly distributed modulo 1. Weyl’s criterion seems to fit the bill.

REPLY [4 votes]: I don't think any kind of growth rate condition will do it for you by itself. (You could just tweak the $b_n$'s without messing up the growth rate to avoid some set).
You might look at a very interesting paper of Boshernitzan ("Uniform distribution and Hardy fields). He gives beautiful necessary and sufficient conditions on the $(a_n)$ for uniform distribution mod 1 provided that the $(a_n)$ look like $f(n)$ for polynomially-growing functions in a reasonable class (a "Hardy field"). 
One such Hardy field is the collection of all functions that can be obtained starting from $x$, allowing yourself all real constants, allowing addition, multiplication, exponentiation, logarithms etc. (so that for example it contains the function $(x^{2.76}+15x\log x)/(x+e^{-1.4\sqrt x})$). The theorem in that paper shows that the sequence $f(n)$ is uniformly distributed modulo 1. 
ADDED: (in response to Aaron Meyerovitch's question). He asks Can you tweak $b_n=\sqrt n$ to have $b_n′$ increasing with $\sqrt{n−1} < b_n′ < \sqrt{n+1}$ and avoid a uniform distribution? Answer: no. More generally if $(b_n)$ is uniformly distributed and $b_n'=b_n+o(1)$, then $b_n'$ is also uniformly distributed. For a proof, use Weyl's criterion and look at the difference of the expressions you obtain using the $b_n$'s and the $b_n'$s. I don't really see this suggested restriction as being in the spirit of an honest growth condition on the $b_n$'s FWIW.

REPLY [2 votes]: There are many theorems in Kuipers and Niederreiter to the effect that if $b_n$ has a certain growth rate then it is uniformly distributed modulo one. Presumably you could turn those into results about $a_n$. 
EDIT: Let me elaborate on this. Corollary 2.1 is Fejer's Theorem: Let $f(x)$ be defined for $x\ge1$ and differentiable for $x\ge x_0$. If $f'(x)$ tends monotonically to $0$ as $x\to\infty$ and if $\lim_{x\to\infty}x|f'(x)|=\infty$ then the sequence $f(1),f(2),\dots$ is uniformly distributed modulo one. 
The authors point out that this implies the following sequences are u.d. mod 1; 
$\alpha n^{\sigma}(\log n)^{\tau}$ with $\alpha\ne0$, $0\lt\sigma\lt1$, $\tau$ arbitrary; 
$\alpha(\log n)^{\tau}$ with $\alpha\ne0$ and $\tau\gt1$; 
$\alpha n(\log n)^{\tau}$ with $\alpha\ne0$ and $\tau<0$. 
Later in the book we get other general theorems which imply u.d. mod 1 for
$\alpha n(\log n)^{\tau}$, $\alpha\ne0$, $\tau\gt0$; 
$n\log\log n$; 
$\alpha n^{\sigma}$, $\alpha\ne0$, $\sigma\gt0$, $\sigma$ not an integer; 
$\alpha n^k(\log n)^{\tau}$ with $\alpha\ne0$, $k$ a positive integer, $\tau\lt0$; same is true for $\tau\gt1$; 
and there is more. Combining these with the fact that if $u_n$ is u.d. mod 1 and $\lim_{N\to\infty}N^{-1}\sum_{n=1}^Nx_n=0$ then $u_n+x_n$ is u.d. mod 1 should give you the wiggle room to get some results on growth rates of $a_n$ implying u.d. mod 1.