Sequences $\{\frac{1}{2\pi}\sum_{k=1}^{n}\frac{1}{k}\}$ is is equidistributed modulo $1$.

Question

I want to prove that the sequences $\{\frac{1}{2\pi}\sum_{k=1}^{n}\frac{1}{k}\}$ is equidistributed modulo $1$.

This question is for the following question: Accumulation points form a circle. It need to show that $\{\sum_{k=1}^{n}\frac{1}{k}\}$ is equidistributed modulo $2\pi$. I think Weyl’s criterion will work! Weyl's equidistributed criterion. The following are equivalent: $$\{x_n\}\quad\text{is equidistributed modulo 1};$$

$$\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f ;$$

$$\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0.$$

Answer 1

Have a look at this general fact:

If $\{a_n\}$ is a sequence such that $a_n\ge0$ with $a_n\to0$ and $\sum a_n\to\infty$ then $\{\sum a_n\}$ is equidistributed modulo 1.

Proof:

Let $a,b\in [0,1]$ assume $a< b$ then there exist $n_0\in \mathbb{N}$ such that $0\le a_n< b-a$ for all $n\ge n_0$.

Now let us define partial sum $s_n=\sum_{i=1}^na_i$ then $s_n\to \infty$ (increasing sequence).

Choose $M$ such that it is the least positive integer greater or equal to $s_{n_0}$, i.e. $M\in \mathbb{N}$ such that $(M-1)<s_{n_0}\le M$.

We know there exist $k\in \mathbb{N}$ such that $s_k > (M+a)$.Hence the set $\{k\in\mathbb{N}|s_k > (M+a)\}$ is non empty subset of $\mathbb{N}$

Choose $n_1=min\{k\in\mathbb{N}|s_k > (M+a)\}$, (existance by well ordering principle).

Now observe that $(M+a)<s_{n_1}<(M+b)$ and hence $a<\{s_{n_1}\}<b$ . (here ${s_{n_1}}$ denotes fractional part of $s_{n_1}$ ).

So the fractional part of $s_n$ is dense in $[0,1]$

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Apply this to solve the above problem.