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

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?

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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 $\sqrt{n-1}<b_n'<\sqrt{n+1}$ and avoid a uniform distribution? –  Aaron Meyerowitz Apr 11 '11 at 5:33
Thanks. I just wondered about "If $1/n$ fails, is it the growth rate?" and the answer is, "yes, kind of, it is going to zero too quickly." –  Aaron Meyerowitz Apr 11 '11 at 8:16

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.

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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 $b_n \approx \sqrt{2n}$.

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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 $(b+1)^3-b^3-b$ $a_n$ terms to be zero and the remaining $b$ $a_n$ 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 '11 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 '11 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 '11 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 '11 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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