Is there a sequence whose arithmetic means lie dense in $[-1..1]$?

Question

Is there a sequence $(a_n)_{n \in \mathbb{N}}$ of real numbers in the range of $[-1..\,1]$ such that the sequence of their arithmetic means $(\alpha_n)_{n \in \mathbb{N}}$, given by $$\alpha_n = \frac{1}{n}\sum_{k=1}^n a_k,\quad n \in \mathbb{N}$$ has a dense image in $[-1..\, 1]$?

My thoughts: Yes, there is and I strongly suspect the sequence which alternates between $1$ and $-1$ such that it will be constantly $1$ for $2^k$ members and then constantly $-1$ for $2^{k+1}$ members and so on ... to do the trick. And if it doesn't something similiar will do.

Answer 1

The slabs you mention are not expanding quickly enough to guarantee full limit set (only the middle third interval, I believe) but the idea is good. Assume again that $a_k=1$ for every $k$ between $x_{2n}$ and $x_{2n+1}$ for some $n$ and that $a_k=-1$ for every $k$ between $x_{2n-1}$ and $x_{2n}$ for some $n$, for some increasing sequence $(x_n)_n$, but now, choose $x_n=2^{n^2}$ for every $n$. Then $x_n/x_{n+1}\to0$, and this is enough to guarantee that the whole interval $[-1,1]$ is the limit set of $(\alpha_k)_k$.

To prove the last assertion above, one might want to show the following:

• Every $\alpha_n$ is in $[-1,1]$.
• The sequence of general term $\alpha_{x_{2n}}$ converges to $-1$.
• The sequence of general term $\alpha_{x_{2n-1}}$ converges to $+1$.
• For every sequence $(a_k)_k$ such that $a_k\leqslant1$ for every $k$, $|\alpha_n-\alpha_{n-1}|\leqslant2/n$.
• Every sequence $(b_n)_n$ such that $|b_n|\leqslant1$ for every $n$, with a subsequence converging to $1$, with another subsequence converging to $-1$, and such that $|b_n-b_{n-1}|\to0$, has exactly $[-1,1]$ as limit set.

Answer 2

Given a sequence $b_n$ dense in $[0,1]$, consider a new sequence $$b_1 \text{ ($n_1$ times)}, b_2 \text{ ($n_2$ times)}, b_3 \text{ ($n_3$ times)}, \ldots$$ which has each $b_k$ repeated $n_k$ times, where $n_k \ge k (n_1 + \ldots + n_{k-1})$. Then the arithmetic mean after the $b_k$'s differs from $b_k$ by less than $1/k$. From this it is easy to show that the arithmetic means are dense in $[0,1]$.

Answer 3

Another alternative is:

Let $a_n$ be an enumeration of the rationals in [-1,1].

This set is dense in [-1,1]

Define:

• $b_1=a_1$ and
• $b_{n}=n.a_{n}-(n-1)a_{n-1}$ for $n>1$

The sequence of arithmetic means of this sequence is exactly $a_n$, since:

$\frac{1}{n}\sum\limits_{k=1}^{n}b_k=\frac{1}{n}(n.a_n)=a_n$