## Can the partial sums of a series be uniformly distributed modulo 1?

Earlier I asked in http://mathoverflow.net/questions/60203/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?

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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.

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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.

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 Can you tweak $b_n=\sqrt{n}$ to have $b_n'$ increasing with $\sqrt{n-1} There are many theorems in Kuipers and Niederreiter to the effect that if$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. - Here is how I interpret the question: Let$(b_n)$be an increasing sequence of reals,$s_n=b_n-\lfloor b_n\rfloor$be the fractional part of$b_n$and$a_n=b_{n+1}-b_n.$On the assumption that$(a_n)$is decreasing (or at least non-increasing), what conditions on the growth rate of$(b_n)$(or decay rate of$(a_n)$) allow us to conclude that$(s_n)$is uniformly distributed in$[0,1]?$Here I only consider the case that$\lim a_n=0$aside from a quick mention of the classic case$b_n=n\alpha$. Define$N_{b}$to be the last$n$with$b_n \lt b.$Conjecture: Given$\lim a_n=0$, the condition$\lim_{y \rightarrow \infty}\frac{N_{y}}{N_{y+1}}=1$is necessary and sufficient for$(s_n)$to be uniformly distributed. In case$b_n=\log{n}$it turns out that$b_n$grows too slowly (equivalently,$a_n$decays too rapidly). Here$\frac{N_{y}}{N_{y+1}}=\frac{e^{y}}{e^{y+1}}=\frac1e.$I note below that the distribution fluctuates with$\frac{\liminf r(\cdot)}{\limsup r(\cdot)}=\frac1e.$For this sequence$a_n=\log{\frac{n+1}{n}}=\frac{1}{n}+O(\frac{1}{n^2})$The previous question concerned$a_n=\frac{1}{n}.$Then the same proof works there but some minor details are slightly more involved. I claim that$b_n=n^p$for$0 \lt p \lt 1$does lead to a uniform distribution. For a given sequence of real numbers$(s_n)_{n=1}^\infty \subset [0,1]$, let$A(x,N)=\left|\lbrace n \le N: s_n \le x \rbrace \right|$and let$r(x,N)= \frac{A(x,N)}{N}.$The sequence is said to have a limiting distribution$r(x)$if$\displaystyle \lim_{N \rightarrow \infty}r_N(x)=r(x)$for every$0 \leq x \leq 1$. Let$0 \lt \epsilon$be small . Then for$K \in \mathbb{N},$we have that$A(\epsilon,N_{K+\epsilon})=A(\epsilon,N_{K+1}).$So$r(\epsilon,N)$fluctuates with a local maximum at each$r(\epsilon,N_{K+\epsilon})$and a local minimum at$r(\epsilon,N_{K+1}).$Furthermore $$\frac{r(\epsilon,N_{K+1})}{r(\epsilon,N_{K+\epsilon})}=\frac{N_{K+\epsilon}}{N_{K+1}}.$$ If there is any limiting distribution at all then the second ratio must converge to 1 in which case for any fixed$z$,$\lim_{y \rightarrow \infty}\frac{N_{y}}{N_{y+z}}=1.$This establishes that the condition is necessary. It helps to think of$\epsilon$as small, but all that is really used is that$0 \lt \epsilon \lt 1.$Assume for simplicity that$\epsilon$is not one of the countable number of values$s_n$, Then we can also express the common value above as$A(\epsilon,N_{K+\epsilon})=A(\epsilon,N_{K+1})=\sum_{J=0}^K(N_{J+\epsilon}-N_J)$With a little work this establishes the claim made above for$b_n=n^p$for$0 \lt p \lt 1$. It also establishes that in the case$b_n=\log n$we have$\frac{N_y}{N_{y+1}}=\frac1e.$$$\liminf r(x,N)=\frac{e^x-1}{e-1} \text{ and }\limsup r(x,N) =\frac{e^{1-x}(e^x-1)}{e-1}.$$ Recall that$\frac{N_{y}}{N_{y+1}}=\frac{e^{y}}{e^{y+1}}=\frac1e$in this case. There has been some mention of Weyl's criterion. It is very useful for the extreme case$b_n=n \alpha$in which$a_n$is identically$\alpha.$Then$(s_n)$is uniformly distributed if and only if$\alpha \gt 0$is irrational. The criterion is that $$\lim_{N \rightarrow \infty}\frac1N \sum_{j=1}^N e^{2\pi i \ell b_j}=0\text{ for all integers }\ell \gt 0.$$ I do not see that it helps for$b_n=\sqrt{n}$or$b_n=\sqrt{2n}$although it does apply to$b_n=1,\ 3/2,4/2,\ 7/3,8/3,9/3,\ 13/4,14/4,15/14,16/4,\ 21/5 \cdots$which has$b_n \approx \sqrt{2n}$. - I don't think the conjecture has any chance of being right. Define$N_b=b^3$say. Then between$N_b$and$N_{b+1}$, set the first$(b+1)^3-b^3-ba_n$terms to be zero and the remaining$ba_n$terms to be$1/b$. Clearly$a_n\to 0$. Then the partial sums have 0 fractional part for$n$between$N_b$and$N_{b+1}-b$(the large majority of the time), so there is no hope of uniform distribution. – Anthony Quas Apr 11 2011 at 7:31 But the paper that I mention above does prove that uniform distribution for$n^p$for any$p\not\in\mathbb Z$. – Anthony Quas Apr 11 2011 at 8:08 The conjecture does specify that$(a_n)$is non-increasing so it can be$1/b$for a while and then later$1/(b+1)$but it can't be$0$and then later$1/b$. It might be a pretty strong condition but I thought it managed to say something about growth rate and why$a_n=1/n$is too fast but$a_n=(\log n)^{\epsilon}/n$is probably ok. I wasn't totally sure about having$1/k$for$2^{2^k}$times and then$1/(k+1)$for$2^{2^{k+1}}$times. – Aaron Meyerowitz Apr 11 2011 at 8:12 Sorry - I didn't spot the fact that the$(a_n)$were decreasing as it was not stated inside the conjecture box!$1/k$for$2^{2^k}$should work just fine: imagine having an interval of length$\ell$. For any$k>100/\ell$the sums will land in the interval with the correct frequency up to a factor of$1\pm 1/100$. Now take the limit as$100\to\infty$. – Anthony Quas Apr 11 2011 at 19:26 Well, if$a_n = A \notin \mathbb{Q}\dots\$ you win. Otherwise, it seems implausible that any growth condition is sufficient.

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