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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.

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One option would be to consider an enumeration of the rationals in $[-1,1]$. –  Beni Bogosel Jan 21 '13 at 17:50
    
@BeniBogosel That's a good idea. It seems like every such enumeration must have dense arithmetic means. I would also like to see an example where the original sequence doesn't lie dense in $[-1..\,1]$, preferably with only two limit points. –  k.stm Jan 21 '13 at 17:55
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I don't agree that every such enumeration must have dense arithmetic means. For example, if $(r_k)_{k\geq 1}$ is an enumeration of the rational numbers in $(0,1]$, then $(0,r_1,-r_1,r_2,-r_2,\ldots)$ is an enumeration of the rational numbers in $[-1,1]$, but in this case the sequence of means converges to $0$, hence $0$ is the only limit point of the set of means. –  Jonas Meyer Jan 21 '13 at 18:01
    
@JonasMeyer Thanks for pointing that out. –  k.stm Jan 21 '13 at 18:01
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In fact @Jonas's example might even be typical in the sense that for "most" such enumerations, the sequence of means converges to zero. –  Did Jan 21 '13 at 18:27

3 Answers 3

up vote 5 down vote accepted

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.
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How can I see that $[-1..\,1]$ is really the limit set? –  k.stm Jan 21 '13 at 18:08
    
Which part of the checking eludes you? –  Did Jan 21 '13 at 18:10
    
Thanks. I think I can convince myself with the last two given hints. –  k.stm Jan 21 '13 at 18:23

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:

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

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Others have proposed this, but then quickly deleted it before I had the chance to upvote. I like the idea – so +1 – but I wanted the sequence to stay within $[-1..1]$, but for $a_1 = -1$ and $a_2 = 1$ you have $b_2 = 2 \cdot 1 - 1 \cdot (-1) = 3 \notin [-1..1]$. So you'd have to guarantee the existence of a enumeration $(a_n)_{n \in \mathbb{N}}$ such that $(b_n)_{\mathbb{N}}$ stays within $[-1..1]$, too. –  k.stm Jan 22 '13 at 18:28
    
Indeed! Thanks for the remark. –  Aloizio Macedo Jan 22 '13 at 18:35

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 $[0,1]$.

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That I find cool. –  k.stm Jan 21 '13 at 18:45

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