Points accessible in average by rearrangement of sequences

TL;DR

This paper explores how rearranging a sequence affects the set of its average limit points, revealing the relationship between sequence properties and the structure of accessible average points.

Contribution

It characterizes the set of limit points of averages of rearranged sequences and links these sets to properties of the original sequence.

Findings

The structure of accessible points depends on sequence properties.

Certain sets of points can be realized as limit points of rearranged averages.

The study provides a framework for understanding how rearrangements influence average limits.

Abstract

We investigate the set of limit points of averages of rearrangements of a given sequence. We study how the properties of the sequence determine the structure of that set and what type of sets we can expect as the set of such accessible points.

Equations

$$\lim\limits_{n\to\infty}\mbox{${{{\cal A}}}$}(a_{1},\dots,a_{n})=\lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^{n}a_{i}}{n}=\alpha.$$

$$\frac{c_n}{\sum _{i=1}^{n-1}{c}_i}\to 0.$$

$$\frac{c_n}{\sum _{i=1}^{n-1}{c}_i}\to 0$$

$$\mbox{${{{\cal A}}}$}(d_{1},\dots,d_{m})=\frac{c+\sum\limits_{i=1}^{m-1}b_{i}}{m}=\frac{c}{m}+\mbox{${{{\cal A}}}$}(b_{1},\dots,b_{m-1})\frac{m-1}{m}$$

$$I_2=\{i\in \mathbb{N}:i>M,\exists l\in \mathbb{N}\text{ such that }i\in J_l\subset (M,m]\},$$

$$\mbox{${{{\cal A}}}$}(d_{1},\dots,d_{m})=\frac{\sum\limits_{i\in I_{1}}d_{i}}{m}+\frac{\sum\limits_{i\in I_{2}}d_{i}}{m}+\frac{\sum\limits_{i\in I_{3}}d_{i}}{m}.$$

$$\frac{k}{m}\frac{(a-\frac{ϵ}{4\alpha })r+(b-\frac{ϵ}{4\beta })(k-r)}{k}<\frac{k}{m}\frac{\sum _{i\in I_2}{d}_i}{k}<\frac{k}{m}\frac{(a+\frac{ϵ}{4\alpha })r+(b+\frac{ϵ}{4\beta })(k-r)}{k}$$

$$\frac{m-M-2\gamma }{\gamma (m-M)}\le \frac{r}{k}\le \frac{m-M}{\gamma (m-M-2\gamma )}.$$

$$\alpha a+\beta b-\frac{2ϵ}{3}<\frac{\sum _{i\in I_2}{d}_i}{m}<\alpha a+\beta b+\frac{2ϵ}{3}$$

$$\alpha a+\beta b-\epsilon<\mbox{${{{\cal A}}}$}(d_{1},\dots,d_{m})<\alpha a+\beta b+\epsilon$$

$$\mbox{${{{\cal A}}}$}(a_{1},\dots,a_{n})=\frac{\sum\limits_{i=1}^{N}a_{i}+\sum\limits_{i=N+1}^{n}a_{i}}{n}<\frac{\sum\limits_{i=1}^{N}a_{i}}{n}+\frac{M+c}{2}\cdot\frac{n-N}{n}.$$

$$\{a_{p_2(1)},\dots ,a_{p_2(N_2)}\}\subset \{a_{p_1(1)},\dots ,a_{p_1(n_1)}\}$$

$$|\mbox{${{{\cal A}}}$}\big{(}a_{p_{1}(i)}:1\leq i\leq n_{1},\ a_{p_{1}(i)}\notin\{a_{p_{2}(1)},\dots,a_{p_{2}(N_{2})}\}\big{)}-b_{1}|<\frac{2}{3}.$$

$$v_{1}=\mbox{${{{\cal A}}}$}(a_{p_{1}(i)}:1\leq i\leq n_{1}),$$

$$w_{2}=\mbox{${{{\cal A}}}$}(a_{p_{2}(i)}:1\leq i\leq N_{2}),$$

$$v^{\prime}_{1}=\mbox{${{{\cal A}}}$}\big{(}a_{p_{1}(i)}:1\leq i\leq n_{1},\ a_{p_{1}(i)}\notin\{a_{p_{2}(1)},\dots,a_{p_{2}(N_{2})}\}\big{)}.$$

$$v_1=\frac{(n_1-N_2)v_1^{\prime }+N_2{w}_2}{n_1}$$

$$v_1^{\prime }=\frac{n_1{v}_1-N_2{w}_2}{n_1-N_2}.$$

$$|v_1^{\prime }-v_1|=\frac{N_2|v_1-w_2|}{n_1-N_2}\le N_2\frac{|v_1-b_1|+|b_1-b_2|+|b_2-w_2|}{n_1-N_2}\le \frac{N_2}{n_1-N_2}$$

$$|v_1^{\prime }-b_1|\le |v_1^{\prime }-v_1|+|v_1-b_1|=\frac{N_2}{n_1-N_2}+\frac{1}{3}<\frac{2}{3}$$

$$\{a_{p_{k+1}(1)},\dots ,a_{p_{k+1}(N_{k+1})}\}\subset \{d_1\dots ,d_{m_{k-1}};a_{p_k(1)},\dots ,a_{p_k(n_k)}\}$$

$$|v_k-b_k|<\frac{1}{3k}.$$

$$|v_k^{\prime }-b_k|<\frac{2}{3k}.$$

$$v_{k}=\mbox{${{{\cal A}}}$}(d_{1}\dots,d_{m_{k-1}};\ a_{p_{k}(1)},\dots,a_{p_{k}(n_{k})}:a_{p_{k}(i)}\neq d_{l}\ (1\leq i\leq n_{k},\ 1\leq l\leq m_{k-1}),$$

$$v^{\prime}_{k}=\mbox{${{{\cal A}}}$}\big{(}d_{1}\dots,d_{m_{k-1}};\ a_{p_{k}(1)},\dots,a_{p_{k}(n_{k})}:d_{l}\neq a_{p_{k}(i)}\neq a_{p_{k+1}(j)}\neq d_{l}$$

$$(1\leq i\leq n_{k},\ 1\leq j\leq N_{k+1},\ 1\leq l\leq m_{k-1})\big{)}.$$

$$w_{k+1}=\mbox{${{{\cal A}}}$}(a_{p_{k+1}(i)}:1\leq i\leq N_{k+1}).$$

$$v_k=\frac{(n_k^{\prime }-N_{k+1})v_k^{\prime }+N_{k+1}{w}_{k+1}}{n_k^{\prime }}$$

$$v_k^{\prime }=\frac{n_k^{\prime }{v}_k-N_{k+1}{w}_{k+1}}{n_k^{\prime }-N_{k+1}}.$$

Full text

00footnotetext: AMS (2010) Subject Classifications: 40A05, 26E60

Key Words and Phrases: rearrangement of sequence, arithmetic mean

1 Introduction

When in [6] we started building the theory of means on infinite sets, at one point we faced the problem that how the average behaves for the rearrangements of an arbitrary bounded sequence. More precisely if a bounded sequence $(a_{n})$ is given, then we wanted to determine the set of points of the limit of averages of the rearranged sequences. I.e. take all rearrangements $(a^{\prime}_{n})$ of the sequence, choose those where the limit of the averages $\lim\limits_{n\to\infty}\frac{a^{\prime}_{1}+\dots+a^{\prime}_{n}}{n}$ exists, and examine the set of all such limit points.

Many authors studied the rearrangement of the underlying sequence of a series and investigated what effect it has for the sum of the series, see [1], [3], [4], [5], [8]. In their research the rearrangement was always associated to a series. In [9] Sarigöl investigates the permutations that preserves bounded variation of sequences.

It is well known that the accumulation points, hence limit point of a rearranged sequence are identical to such points of the original sequence. Hence it does not make sense to study. However if we take the average of the rearranged sequence, that is not so trivial.

In this paper our main aim will be to investigate which set of points can be accessed in average by rearrangement of sequences. How the properties of the sequence determine the structure of that set. What type of sets we can expect as the set of such accessible points.

In the first part of the paper we prove some generic results that will provide theorems for bounded sequences. Unbounded sequences behaves differently, their investigation is our goal in the remainder of the paper. For more details see subsection 1.2.

1.1 Basic notions and notations

Throughout this paper function \mbox{{{{\cal A}}}}() will denote the arithmetic mean of any number of variables. We will also use the notation \mbox{{{{\cal A}}}}(a_{i}:1\leq i\leq n) for \mbox{{{{\cal A}}}}(a_{1},\dots,a_{n}). If $H\subset\mathbb{R}$ is a finite set then \mbox{{{{\cal A}}}}(H) denotes the arithmetic mean of its distinct points.

Let us use the notation $\bar{\mathbb{R}}=\mathbb{R}\cup\{-\infty,+\infty\}$ and consider $\bar{\mathbb{R}}$ as a 2 point compactification of ${\mathbb{R}}$ i.e. a neighborhood base of $+\infty$ is $\{(c,+\infty]:c\in\mathbb{R}\}$.

**  Definition 1.1****.**

Let $(a_{n})$ be a sequence. We say that $a_{n}$ tends to $\alpha\in\bar{\mathbb{R}}$ in average if

[TABLE]

We denote it by a_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}\alpha. We also use the expression that $\alpha$ is the limit in average of $(a_{n})$.

With this notation if a series $\sum a_{n}$ is Cesaro summable with sum $c$ then we may say that s_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}c where $s_{n}=\sum\limits_{i=1}^{n}a_{i}$.

**  Definition 1.2****.**

Let $(a_{n})$ be a sequence, $\alpha\in\bar{\mathbb{R}}$. We say that $\alpha$ is accessible in average by rearrangement of $(a_{n})$ if there exists a rearrangement of $a_{n}$ i.e. a bijection $p:\mathbb{N}\to\mathbb{N}$ such that a_{p(n)}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}\alpha.

The set of all such accessible points will be denoted by $AAR_{(a_{n})}$.

**  Definition 1.3****.**

If $(a_{n}),(b_{n})$ are two sequences then let $(c_{n})=(a_{n})||(b_{n})$ be the sequence defined by $c_{2n}=b_{n},c_{2n-1}=a_{n}\ (n\in\mathbb{N})$.

The following theorem is well know in the theory of Cesaro summation or can be proved easily.

**  Theorem 1.4****.**

If $a_{n}\to\alpha\ (\alpha\in\bar{\mathbb{R}})$ then a_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}\alpha. ∎

**  Corollary 1.5****.**

If $a_{n}\to\alpha\ (\alpha\in\bar{\mathbb{R}})$ then $AAR_{(a_{n})}=\{\alpha\}$.

Proof.

It is also well known that for every rearrangement $(a_{p_{n}})$ of $(a_{n})$, $a_{p_{n}}\to\alpha$. ∎

**  Proposition 1.6****.**

*If a_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}a\in\mathbb{R},\ b_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}b\in\mathbb{R},\ c\in\mathbb{R} then

a_{n}+c\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}a+c,\ ca_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}ca,\ a_{n}+b_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}a+b,\ a_{n}||b_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}\frac{a+b}{2}.∎*

1.2 Brief summary of the main results

We just enumerate some of the most interesting results to give a taste of the topic.

**  Proposition****.**

If $\alpha\in\bar{\mathbb{R}}$ is an accumulation point of $(a_{n})$ then $\alpha\in AAR_{(a_{n})}$.

**  Proposition****.**

If a_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}c then $c\in[\varliminf a_{n},\varlimsup a_{n}]$.

**  Theorem****.**

$AAR_{(a_{n})}$* is closed in $\bar{\mathbb{R}}$.*

**  Theorem****.**

Let $(a_{n})$ be a bounded sequence. Then $AAR_{(a_{n})}=[\varliminf a_{n},\varlimsup a_{n}]$.

**  Theorem****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b=\varlimsup b_{n}\in\mathbb{R},\ c_{n}\to+\infty$ and $(c_{n})$ is increasing. If $c>b$ is accessible in average by rearrangement of $(a_{n})$ then

[TABLE]

**  Theorem****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $\varlimsup b_{n}=b\in\mathbb{R},\ c_{n}\to+\infty$. If

[TABLE]

and $c>b$ then $c$ is accessible in average by rearrangement of $(a_{n})$.

**  Corollary****.**

Let $k\in\mathbb{N}$, $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,c_{n}=n^{k}$. Then $AAR_{(a_{n})}=[0,+\infty]$.

**  Corollary****.**

Let $d>1$, $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,c_{n}=d^{n}$. Then $AAR_{(a_{n})}=\{0,+\infty\}$.

**  Theorem****.**

Let $(a_{n})$ be a sequence bounded from below or above, $(a^{\prime}_{n})$ be one of its rearrangements. Let $c_{n}=\frac{\sum\limits_{i=1}^{n}a^{\prime}_{i}}{n}$. Then the set of accumulation points of $(c_{n})$ is $[\varliminf c_{n},\varlimsup c_{n}]$.

**  Theorem****.**

Let $(a_{n})$ be a sequence such that it has at least 4 accumulation points: $a,b,-\infty,+\infty\ (a,b\in\mathbb{R},a<b)$. Let $Z\subset[a,b]$ be a closed set. Then there is a rearrangement $(a^{\prime}_{n})$ of $(a_{n})$ such that the accumulation points of the sequence $(p_{n})$ is exactly $Z$ where p_{n}=\mbox{{{{\cal A}}}}(a^{\prime}_{1},\dots,a^{\prime}_{n}).

2 General results

First we need some preparation.

**  Lemma 2.1****.**

Let $(b_{n})$ be a sequence, $c\in\mathbb{R}$. Assume b_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}b. Then $\forall\epsilon>0$ we can merge $c$ into $(b_{n})$ i.e. create a new sequence $(d_{n})$ with $d_{i}=b_{i}\ (i<k),d_{k}=c,d_{i}=b_{i-1}\ (i>k)$ such that $n\geq k$ implies that b-\epsilon<\mbox{{{{\cal A}}}}(d_{1},\dots,d_{n})<b+\epsilon.

Proof.

Choose $k\in\mathbb{N}$ such that $n\geq k-1$ implies that

(1) b-\frac{\epsilon}{3}<\mbox{{{{\cal A}}}}(b_{1},\dots,b_{n})<b+\frac{\epsilon}{3} and

(2) $|\frac{c}{n}|<\frac{\epsilon}{3}$ and

(3) $b-\frac{2\epsilon}{3}<(b-\frac{\epsilon}{3})\frac{n-1}{n}$.

If $m\geq k$ then

[TABLE]

hence b-\epsilon<\mbox{{{{\cal A}}}}(d_{1},\dots,d_{m})<b+\frac{2\epsilon}{3}. ∎

**  Lemma 2.2****.**

Let $(b_{n})$ be a sequence, $c\in\mathbb{R}$. Assume b_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}+\infty. Then $\forall M>0$ we can create a new sequence $(d_{n})$ with $d_{i}=b_{i}\ (i<k),d_{k}=c,d_{i}=b_{i-1}\ (i>k)$ such that $n\geq k$ implies that M<\mbox{{{{\cal A}}}}(d_{1},\dots,d_{n}). Similar holds for $-\infty$.

Proof.

Choose $k\in\mathbb{N}$ such that $n\geq k-1$ implies that

(1) M+2<\mbox{{{{\cal A}}}}(b_{1},\dots,b_{n}) and

(2) $|\frac{c}{n}|<1$ and

(3) $M+1<(M+2)\frac{n-1}{n}$.

If $m\geq k$ then

[TABLE]

hence M<\mbox{{{{\cal A}}}}(d_{1},\dots,d_{m}). ∎

**  Lemma 2.3****.**

Let $(b_{n}),(c_{n})$ be two sequences. Assume b_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}b\in\bar{\mathbb{R}}. Then we can merge the two sequences into a new sequence $(d_{n})$ such that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}b.

Proof.

We define sequences $(b^{(l)}_{n})$ and associated constants $k_{l}$ recursively. Let $(b^{(0)}_{n})=(b_{n}),\ k_{0}=1$. Let first $b\in\mathbb{R}$. If $\epsilon=\frac{1}{2}$ then by Lemma 2.1 we can merge $c_{1}$ into $(b_{n})$ such that $d^{\prime}_{i}=b_{i}(i<k_{1}),d^{\prime}_{k_{1}}=c_{1},d^{\prime}_{i}=b_{i-1}(i>k_{1})$ and $n>k_{1}$ implies that b-\frac{1}{2}<\mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n})<b+\frac{1}{2}. Let $(b^{(1)}_{n})=(d^{\prime}_{n})$. If we have already defined $(b^{(i)}_{n})$ and $k_{i}$ for $i\leq l$ then apply 2.1 for $(b^{(l)}_{n}),c_{l},\epsilon=\frac{1}{2^{l+1}}$. Then we end up with sequence $(b^{(l+1)}_{n})$ and $k_{l+1}>k_{l}+1$ such that $n>k_{l+1}$ implies that b-\frac{1}{2^{l+1}}<\mbox{{{{\cal A}}}}(b^{(l+1)}_{1},\dots,b^{(l+1)}_{n})<b+\frac{1}{2^{l+1}}.

Then let us define $(d_{n})$ by $d_{j}=b^{(l)}_{j}$ where $k_{l}\leq j<k_{l+1}$. Obviousy $(d_{n})$ is a merge of the two original sequences and d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}b.

Now if $b=\pm\infty$ then replace $\epsilon=\frac{1}{2^{l+1}}$ by $M=2^{l+1}$ in the first part of the proof and apply 2.2 instead of 2.1. ∎

**  Corollary 2.4****.**

Let $(a_{n})$ be a sequence. If there is a subsequence $(a^{\prime}_{n})$ such that it can be rearranged to $(a^{\prime\prime}_{n})$ such that a^{\prime\prime}_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}\alpha then there is a rearrangement of $(a_{n})$ which tends to $\alpha$ in average.∎

**  Corollary 2.5****.**

If $\alpha\in\bar{\mathbb{R}}$ is an accumulation point of $(a_{n})$ then $\alpha\in AAR_{(a_{n})}$.

Proof.

Let $(b_{n})$ be a subsequence of $(a_{n})$ such that $b_{n}\to\alpha$ and $(c_{n})$ be the rest. Then apply 2.3. ∎

**  Proposition 2.6****.**

Let $(a_{n}),(b_{n})$ be two sequences with $a_{n}\to a,b_{n}\to b,\ a<b,\ a,b\in\mathbb{R}$. Then for $\forall\alpha>0,\forall\beta>0,\ \alpha+\beta=1$ the two sequences can be merged into a new sequence $(d_{n})$ such that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}\alpha a+\beta b.

Proof.

Let $\alpha\leq\beta$ (the opposite case can be handled similarly). Let $\gamma=1+\frac{\beta}{\alpha}=\frac{1}{\alpha}$. Obviously $\mathbb{N}\cap[1,\infty)=\bigcup\limits_{n=1}^{\infty}\mathbb{N}\cap[(n-1)\gamma,n\gamma)$. Because of $\gamma\geq 2$ the length of each such interval is at least 2 hence contains at least 2 integers. Set $J_{n}=\mathbb{N}\cap[(n-1)\gamma,n\gamma)$.

If $i\in\mathbb{N}$ is given then $i\in[(n-1)\gamma,n\gamma)$ for some $n\in\mathbb{N}$. For every first index $i$ of $[(n-1)\gamma,n\gamma)$ let $d_{i}$ come from the sequence $(a_{n})$, for all other indexes from $(b_{n})$ using the not-yet-used elements from the sequences and from the original order. In this way we have defined $(d_{n})$ as a merge of $(a_{n}),(b_{n})$.

Let $\epsilon>0$ be given. Then there is $N\in\mathbb{N}$ such that $n>N$ implies that $a_{n}\in(a-\frac{\epsilon}{4\alpha},a+\frac{\epsilon}{4\alpha}),\ b_{n}\in(b-\frac{\epsilon}{4\beta},b+\frac{\epsilon}{4\beta})$. Let $M\in\mathbb{N}$ such that $\{a_{n}:n\leq N\}\cup\{b_{n}:n\leq N\}\subseteq\{d_{m}:m\leq M\}$. If $m>M$ then set $I_{1}=\{1,\dots,M\}$,

[TABLE]

$I_{3}=\{1,\dots,m\}-(I_{1}\cup I_{2})$. If $m>M$ then

[TABLE]

Clearly the first and third items can be arbitrarily small if $m\to\infty$ because the number of elements in the first sum is $M$, while it is at most $2\gamma$ in the third. Let us estimate the middle term now.

[TABLE]

where $k$ denotes the number of elements in the sum and $r$ is the number of $J_{l}$ intervals which are subset of $(M,m]$.

The obvious estimation gives that $m-M-2\gamma<k\leq m-M$ and $\frac{m-M-2\gamma}{\gamma}<r\leq\frac{m-M}{\gamma}$. Therefore

[TABLE]

When $m\to\infty$ then $\frac{k}{m}\to 1$ and $\frac{r}{k}\to\frac{1}{\gamma}=\alpha,\ \frac{k-r}{k}\to\beta$. Hence we get

[TABLE]

if $m$ is large enough. Finally

[TABLE]

if $m$ is large enough. ∎

**  Proposition 2.7****.**

Proposition 2.6 is valid too if $\alpha=0$ or $\beta=0$.

Proof.

Apply Lemma 2.3. ∎

**  Proposition 2.8****.**

If a_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}c then $c\in[\varliminf a_{n},\varlimsup a_{n}]$.

Proof.

Let $m=\varliminf a_{n},\ M=\varlimsup a_{n}$. If any of $m,M$ is infinite the we do not have to check that side. Hence assume that $M\in\mathbb{R}$ ($m$ can be handled similarly). First let $c\in\mathbb{R}$. Assume indirectly that $c>M$. Then there is $N$ such that $n>N$ implies that $a_{n}<\frac{M+c}{2}$. Then

[TABLE]

The latter can be smaller than $\frac{M+2c}{3}$ if $n$ is large enough which is a contradiction.

The case $c=+\infty$ can be handled similarly: just apply $M+1$ instead of $\frac{M+c}{2}$. ∎

**  Theorem 2.9****.**

$AAR_{(a_{n})}$* is closed in $\bar{\mathbb{R}}$.*

Proof.

Let us note first that if $\sup AAR_{(a_{n})}=+\infty$ then $+\infty\in AAR_{(a_{n})}$ by 2.8 and 2.5. And $-\infty$ can be handled similarly.

Let $(b_{n})$ be a sequence such that $\forall n\ b_{n}\in AAR_{(a_{n})}$ and $b_{n}\to b\in\mathbb{R}$. We have to show that $b\in AAR_{(a_{n})}$. For that it is enough to give a subsequence of $(a_{n})$ which tends to $b$ in average (see 2.3).

We can assume that $(b_{n})$ is increasing, moreover $b-b_{i}<\frac{1}{3i}$. The other case when $(b_{n})$ is decreasing is similar.

We know that for each $i\in\mathbb{N}$ there is a rearrangement $p_{i}:\mathbb{N}\to\mathbb{N}$ such that a_{p_{i}(n)}\mbox{ {{{\overset{\mbox{${{{\cal A}}}$}}{\longrightarrow}}}} }b_{i}. Let $N_{i}\in\mathbb{N}$ such that $n>N_{i}$ implies that

|\mbox{{{{\cal A}}}}(a_{p_{i}(1)},\dots,a_{p_{i}(n)})-b_{i}|<\frac{1}{3i}.

We define a new rearrangement $(d_{n})$ of $(a_{n})$ recursively. We will add some elements of $(a_{n})$ to $(d_{n})$ in each step. Without mentioning we will assume that we just add new elements i.e. that are not among the previously selected ones.

Step 1: Take $n_{1}\geq N_{1}$ elements from $(a_{p_{1}(n)})$ such that

[TABLE]

and

[TABLE]

This can be done. (1) is obvious because $p_{1}$ is a bijection. To show (2) let

[TABLE]

[TABLE]

[TABLE]

Then clearly

[TABLE]

which gives that

[TABLE]

From that we get that

[TABLE]

and

[TABLE]

if $n_{1}$ is chosen big enough.

Then add those elements $a_{p_{1}(1)},\dots,a_{p_{1}(n_{1})}$ to $(d_{n})$ as $(d_{1},\dots,d_{n_{1}})$.

Step k: Now $(d_{n})$ is already defined till index $m_{k-1}$ i.e. $(d_{1},\dots,d_{m_{k-1}})$.

Take $n_{k}\geq N_{k}$ elements from $(a_{p_{k}(n)})$ such that

[TABLE]

and

[TABLE]

and

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

This can be done. (3) is obvious because $p_{k}$ is a bijection and (4) is evident too. To show (5) let

[TABLE]

Let $n_{k}^{\prime}$ be the number of distinct elements in $(d_{1}\dots,d_{m_{k-1}};\ a_{p_{k}(1)},\dots,a_{p_{k}(n_{k})})$. Then clearly

[TABLE]

which gives that

[TABLE]

From that we get that

[TABLE]

and

[TABLE]

if $n_{k}$ is chosen big enough.

Then add those elements $a_{p_{k}(1)},\dots,a_{p_{k}(n_{k})}$ to $(d_{n})$.

In that way we have constructed $(d_{n})$. We show that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}b. Let $\epsilon=\frac{1}{k}$. First we show that b-\mbox{{{{\cal A}}}}(d_{1},\dots,d_{m_{k}})<\frac{1}{k}. It is clear that b-\mbox{{{{\cal A}}}}(a_{p_{k}(1)},\dots,a_{p_{k}(n_{k})})<\frac{1}{3k} and $(d_{1},\dots,d_{m_{k}})$ contains $(a_{p_{k}(1)},\dots,a_{p_{k}(n_{k})})$. But in step $k$ we had $|v_{k}-b_{k}|<\frac{1}{3k}$. We remark that v_{k}=\mbox{{{{\cal A}}}}(d_{1},\dots,d_{m_{k}}). Hence $|b-v_{k}|\leq|b-b_{k}|+|b_{k}-v_{k}|<\frac{1}{k}$.

Let $m_{k}<p\leq m_{k+1}$. By construction $d_{p}$ is elements from $(a_{p_{k+1}(n)})$. Let v=\mbox{{{{\cal A}}}}(d_{1},\dots,d_{p}) and let v^{\prime}=\mbox{{{{\cal A}}}}\big{(}elements of $(a_{p_{k+1}(n)})$ among d_{1},\dots,d_{p}\big{)}. Obviously

[TABLE]

i.e. $v$ is a weighted average of $v^{\prime}$ and $v^{\prime}_{k}$ therefore $v\in(v^{\prime},v^{\prime}_{k})$.

But

[TABLE]

and

[TABLE]

which gives that $|v-b|<\frac{1}{k}$. We got that if $m_{k}\leq p\leq m_{k+1}$ then

[TABLE]

which proves the claim. ∎

3 On bounded sequences

**  Theorem 3.1****.**

Let $(a_{n})$ be a bounded sequence. Then $AAR_{(a_{n})}=[\varliminf a_{n},\varlimsup a_{n}]$.

Proof.

Let $m=\varliminf a_{n},\ M=\varlimsup a_{n}$. Clearly if $(a^{\prime}_{n})$ is a rearrangement of $(a_{n})$ then $m=\varliminf a^{\prime}_{n},\ M=\varlimsup a^{\prime}_{n}$. Hence by 2.8 $AAR_{(a_{n})}\subset[m,M]$.

Now let $l$ be choosen such that $m\leq l\leq M$. We can devide $(a_{n})$ into three distinct sequences: $b_{n}\to m,c_{n}\to M$ and $(d_{n})$ is the rest i.e. $\{b_{n},c_{n},d_{n}:n\in\mathbb{N}\}=\{a_{n}:n\in\mathbb{N}\}$ and $b_{n}\neq c_{k}\neq d_{l}\neq b_{n}\ (\forall n,k,l)$. It can happen that either $(d_{n})$ or $(c_{n}),(d_{n})$ are empty. By Proposition 2.6 we can merge $(b_{n}),(c_{n})$ into a new sequence $(e_{n})$ such that e_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}l. By Lemma 2.3 we can add $(d_{n})$ as well in a way that the limit does not change. ∎

**  Theorem 3.2****.**

Let $m=\varliminf a_{n},\ M=\varlimsup a_{n}$. If $m<M,\ m,M\in\mathbb{R}$ then we can create a rearrangement $(e_{n})$ such that $\lim\limits_{n\to\infty}\frac{\sum_{i=1}^{n}e_{i}}{n}$ does not exist.

Proof.

Let us devide $H$ into three distinct sequences as in the proof of Theorem 3.1 (let us use the same notations). Let $p=m+\frac{M-m}{3},\ q=M-\frac{M-m}{3}$.

Now we define $(e_{n})$. Let the first element be $d_{1}$. Then take elements from $(b_{n})$ such that \mbox{{{{\cal A}}}}(d_{1},b_{1},\dots,b_{n_{1}})<p. Next element will be $d_{2}$. Then take elements from $(c_{n})$ such that \mbox{{{{\cal A}}}}(d_{1},b_{1},\dots,b_{n_{1}},d_{2},c_{1},\dots,c_{n_{2}})>q. Next element is $d_{3}$. Then take elements from $(b_{n})$ such that

[TABLE]

And so on. Obviously we exhaust all elements from $(a_{n})$ and $\frac{\sum_{i=1}^{n}e_{i}}{n}$ will not converge. ∎

4 On unbounded sequences

**  Lemma 4.1****.**

Let $(c_{n})$ be an increasing sequence such that $c_{n}\to+\infty,\ c_{n}>0$. Let $(c^{\prime}_{n})$ be any of its rearrangements. Then

[TABLE]

Proof.

Take a subsequence $(c_{n_{k}})$ of $(c_{n})$ such that

[TABLE]

We can assume that $\forall k$ if $m<n_{k}$ then $c_{m}<c_{n_{k}}$ because if there is $m<n_{k}$ such that $c_{m}=c_{n_{k}}$ then

[TABLE]

hence put $c_{m}$ into the subsequence instead of $c_{n_{k}}$.

Now find $c_{n_{k}}$ in $(c^{\prime}_{n})$, say $c_{n_{k}}=c^{\prime}_{m_{k}}$. Let

[TABLE]

Clearly if $n<l_{k}$ then $c^{\prime}_{n}<c^{\prime}_{m_{k}}=c_{n_{k}}$. This gives that

[TABLE]

because $c_{n_{k}}=c^{\prime}_{m_{k}}\leq c^{\prime}_{l_{k}}$ and $\{c^{\prime}_{i}:1\leq i\leq l_{k}-1\}\subset\{c_{i}:1\leq i\leq n_{k}-1\}$ since $\{c_{i}:1\leq i\leq n_{k}-1\}$ contains all elements that are strictly smaller than $c_{n_{k}}$. That yields (6). ∎

**  Corollary 4.2****.**

Let $(c_{n})$ be a sequence such that $c_{n}\to+\infty,\ c_{n}>0$ and

[TABLE]

If we rearrange it to an increasing sequence $(c^{\prime}_{n})$ then

[TABLE]

**  Theorem 4.3****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,\ c_{n}\to+\infty$. If

[TABLE]

then $1$ is accessible in average by rearrangement of $(a_{n})$.

Proof.

We can assume that $c_{n}$ is increasing (by 4.2) and $c_{n}>1$. Then let $d_{m_{n}}=c_{n}$ where $(m_{n})$ is a strictly increasing sequence determined by the followings: $d_{k}=0$ if $\forall n\ k\neq m_{n}$ and $|\{d_{i}:d_{i}=0,\ i<m_{n}\}|=\lfloor c_{1}+\dots+c_{n}\rfloor-n$. We show that d_{l}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}1. Obviously

[TABLE]

[TABLE]

With evident estimation

[TABLE]

If we take the reciprocal and apply the condition then we get that \lim\limits_{n\to\infty}\mbox{{{{\cal A}}}}(d_{1},\dots,d_{m_{n}-1})=1. To finish to proof we have to remark that if $m_{n-1}<l<m_{n}-1$ then

[TABLE]

**  Theorem 4.4****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,\ c_{n}\to+\infty$ and $(c_{n})$ is increasing. If $1$ is accessible in average by rearrangement of $(a_{n})$ then

[TABLE]

Proof.

Let $(d_{n})$ be a rearrangement such that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}1. This rearrangement defines a rearrangement of $(c_{n})$, namely take the elements from $(c_{n})$ exactly in the same order as they come in $(d_{n})$. Let us denote that rearranged sequence with $(c^{\prime}_{n})$ and $c^{\prime}_{n}=d_{m_{n}}$.

Let $\epsilon>0$. Then there is $N\in\mathbb{N}$ such that $m\geq N$ implies that

[TABLE]

Let $n$ be chosen such that $m_{n-1}>N$. Let $m=m_{n-1}$. We know that $\sum\limits_{i=1}^{m_{n-1}}d_{i}=\sum\limits_{i=1}^{n-1}c^{\prime}_{i}$ which gives that $1-\epsilon<\frac{s_{n}}{m}<1+\epsilon$ where $s_{n}=\sum\limits_{i=1}^{n-1}c^{\prime}_{i}$. Suppose that in $(d_{n})$ there are $k_{n}$ zeros between $c^{\prime}_{n-1}$ and $c^{\prime}_{n}$. It gives that

[TABLE]

[TABLE]

From (7) we get that

[TABLE]

By multiplying with the denominator and using (6) we get that

[TABLE]

and clearly both sides tend to $1$ when $\epsilon\to 0$. Which finally gives that $\frac{c^{\prime}_{n}}{s_{n}}\to 0$. Now 4.2 yields the statement. ∎

**  Theorem 4.5****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,\ c_{n}\to+\infty$. If $1\in AAR_{(a_{n})}$ then $AAR_{(a_{n})}=[0,+\infty]$.

Proof.

We have to verify that if $l\in\mathbb{R}^{+}$ then $l\in AAR_{(a_{n})}$.

Let $(d_{n})$ be a rearrangement such that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}1.

First we show that if $l\in\mathbb{N}$ then $l\in AAR_{(a_{n})}$.

Let $k_{n}$ denotes the number of zeros in the first $n$ terms of $(d_{n})$. We state that there is $N\in\mathbb{N}$ such that $n>N$ implies that $k_{n}>(1-\frac{1}{l})n$. Assume the contrary: $\forall N\ \exists n>N$ such that $k_{n}\leq(1-\frac{1}{l})n$ which gives that there are at least $n^{\prime}=\lceil\frac{1}{l}n\rceil$ elements (say $z_{1},\dots,z_{n^{\prime}}$) that are non zero. Then

[TABLE]

But the average of the non zero elements tends to infinite that gives a contradiction.

Now we construct a new rearrangement $(d_{n}^{\prime})$ of $(d_{n})$. Till index $N$ leave out the first $\lfloor(1-\frac{1}{l})N\rfloor$ many zeros. It can be done by the previous statement. Then we go on by recursion. Suppose we are done for $n>N$ and already left out $\lfloor(1-\frac{1}{l})n\rfloor$ many zeros. Now we are dealing with $n+1$. If $\lfloor(1-\frac{1}{l})n\rfloor=\lfloor(1-\frac{1}{l})(n+1)\rfloor$ then we do nothing. Otherwise leave out 1 more zeros. Again the previous statement guarantees that it is possible.

Let show that d^{\prime}_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}l. Let $n>N-\lfloor(1-\frac{1}{l})N\rfloor$. Set $k_{n}=n-\lfloor(1-\frac{1}{l})n\rfloor$ that is the number of remainder elements after managing $d_{n}$. Observe that $\sum\limits_{i=1}^{k_{n}}d^{\prime}_{i}=\sum\limits_{i=1}^{n}d_{i}$. By $n-(1-\frac{1}{l})n\leq k_{n}\leq n-(1-\frac{1}{l})n+1$ we get that

[TABLE]

and both sides tend to $l$ which proves the claim.

Now we show that if $l\in\mathbb{N},\ 0<l^{\prime}<l$ and $l\in AAR_{(a_{n})}$ then $l^{\prime}\in AAR_{(a_{n})}$. Let $(d_{n})$ be a rearrangement such that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}l. Let $L=\frac{l}{l^{\prime}}-1$. Now let us put $\lfloor 2L\rfloor$ many zeros between $d_{1}$ and $d_{2}\ (k\in\mathbb{N})$and put ($\lfloor k\cdot L\rfloor$-previously added number of zeros) many zeros between $d_{k-1}$ and $d_{k}\ (k\in\mathbb{N})$. Let us denote this new sequence by $(d^{\prime}_{k})$. Let us denote $d^{\prime}_{n_{k}}=d_{k}$. Observe that \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n_{k}})\geq\mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n})\geq\mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n_{k+1}-1}) if $n_{k}<n<n_{k+1}$. Clearly \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n_{k}})=\frac{\sum\limits_{i=1}^{k}d_{i}}{k+\lfloor k\cdot L\rfloor}. By obvious estimation

[TABLE]

Let us estimate \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n_{k}-1})=\frac{\sum\limits_{i=1}^{k}d_{i}}{k+\lfloor(k+1)\cdot L\rfloor}.

[TABLE]

Hence both \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n_{k}})\to l^{\prime} and \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n_{k}-1})\to l^{\prime} which give that \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n})\to l^{\prime}. ∎

**  Theorem 4.6****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\to 0,\ c_{n}\to+\infty$. If $1\in AAR_{(a_{n})}$ then $AAR_{(a_{n})}=[0,+\infty]$.

Proof.

We have to verify that if $l\in\mathbb{R}^{+}$ then $l\in AAR_{(a_{n})}$.

Let $(d_{n})$ be the rearranged sequence whose average tends to $1$. Let $k_{n}$ denote the number of elements from $(b_{n})$ among the first $n$ terms of $(d_{n})$. Then

[TABLE]

Clearly $\frac{k_{n}}{n}$ is bounded, $\frac{\sum\limits_{j=1}^{k_{n}}b_{i_{j}}}{k_{n}}\to 0$ therefore $\frac{\sum\limits_{j=1}^{n-k_{n}}c_{l_{j}}}{n}\to 1$. Let us replace all $b_{n}$ with [math] in $(d_{n})$ and denote the new sequence by $(d^{\prime}_{n})$. Then

[TABLE]

hence \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n})\to 1 i.e. $1\in AAR_{(d^{\prime}_{n})}$.

By 4.5 $(d^{\prime}_{n})$ can be rearranged to $(d^{\prime\prime}_{n})$ such that \mbox{{{{\cal A}}}}(d^{\prime\prime}_{1},\dots,d^{\prime\prime}_{n})\to l. Let $k^{\prime}_{n}$ denote the number of zeros among the first $n$ terms of $(d^{\prime\prime}_{n})$. Let us replace all zeros with distinct elements from $(b_{n})$ in $(d^{\prime\prime}_{n})$ and denote the new sequence by $(d^{\prime\prime\prime}_{n})$. Then

[TABLE]

[TABLE]

But the first term $\frac{k^{\prime}_{n}}{n}\frac{\sum\limits_{j=1}^{k^{\prime}_{n}}b_{n_{j}}}{k^{\prime}_{n}}\to 0$ hence \mbox{{{{\cal A}}}}(d^{\prime\prime\prime}_{1},\dots,d^{\prime\prime\prime}_{n})\to l. ∎

**  Theorem 4.7****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\to a,\ c_{n}\to+\infty$. If there is $b\in\mathbb{R}$ such that $a<b,\ b\in AAR_{(a_{n})}$ then $AAR_{(a_{n})}=[a,+\infty]$.

Proof.

Let $l\geq a$. Let $b^{\prime}_{n}=\frac{b_{n}-a}{b-a},\ c^{\prime}_{n}=\frac{c_{n}-a}{b-a},\ (a^{\prime}_{n})=(b^{\prime}_{n})||(c^{\prime}_{n})$. Clearly $b_{n}^{\prime}\to 0,\ c^{\prime}_{n}\to+\infty$ hence by 4.6 $(a^{\prime}_{n})$ can be rearranged to $(d_{n})$ such that \mbox{{{{\cal A}}}}(d_{1},\dots,d_{n})\to\frac{l-a}{b-a}. Let $d^{\prime}_{n}=(b-a)d_{n}+a$. Then clearly \mbox{{{{\cal A}}}}(d^{\prime}_{1},\dots,d^{\prime}_{n})\to l and $(d^{\prime}_{n})$ is a rearrangement of $(a_{n})$. ∎

**  Proposition 4.8****.**

Let $k\in\mathbb{N}$, $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,c_{n}=n^{k}$. Then $AAR_{(a_{n})}=[0,+\infty]$.

Proof.

By 4.6 and 4.3 it is enough to show that

[TABLE]

It is known that $\sum\limits_{i=1}^{n-1}i^{k}=p(n-1)$ where $p(x)$ is a polynomial of degree $k+1$ i.e. $p(n-1)=d_{k+1}(n-1)^{k+1}+q(n-1)$ where $q(x)$ is a polynomial of degree $k$ and $d_{k+1}>0$. Hence

[TABLE]

∎

**  Proposition 4.9****.**

Let $d>1$, $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,c_{n}=d^{n}$. Then $AAR_{(a_{n})}=\{0,+\infty\}$.

Proof.

By 4.7 and 6.3 it is enough to show that $1\notin AAR_{(a_{n})}$ i.e.

[TABLE]

But

[TABLE]

**  Example 4.10****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,\ c_{n}\to+\infty$ and $1\in AAR_{(a_{n})}$. Let $(c^{\prime}_{n})$ be given such that $c^{\prime}_{n}<c_{n},\ c^{\prime}_{n}\to+\infty,\ (a^{\prime}_{n})=(b_{n})||(c^{\prime}_{n})$. These conditions do not imply that $1\in AAR_{(a^{\prime}_{n})}$.

Proof.

Let $c_{n}=n^{2}$. By 4.8 $1\in AAR_{(a_{n})}$.

We define $(c^{\prime}_{n})$ by recursion. Let $c^{\prime}_{1}=1$. If $(c^{\prime}_{n})$ is defined till $n$ then let

[TABLE]

Properties of $(c^{\prime}_{n})$:

$(c^{\prime}_{n})$ is increasing. 2. 2.

$c^{\prime}_{n}<c_{n}\ (n>1)$. It can be seen by induction starting from $n=2,\ c^{\prime}_{2}=1$. 3. 3.

$c^{\prime}_{n}\to+\infty$. Assume the contrary. Then there is $N\in\mathbb{N}$ such that $n\geq N$ implies that $c^{\prime}_{N}=c^{\prime}_{n}$. Then for such $n$ we get that $\sum\limits_{i=1}^{n}c^{\prime}_{i}\leq n\cdot c^{\prime}_{N}$ using the monotonicity of $(c^{\prime}_{n})$ too. But there is an $n$ such that $n\cdot c^{\prime}_{N}<(n+1)^{2}$ which is a contradiction. 4. 4.

$1\notin AAR_{(a^{\prime}_{n})}$. To show that it is enough to prove that

[TABLE]

by 6.3. There are infinitely many $n$ where $c^{\prime}_{n}\neq c^{\prime}_{n-1}$. For such $n$

[TABLE]

therefore

[TABLE]

**  Proposition 4.11****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b_{n}\equiv 0,\ c_{n}=n$. Let $(c^{\prime}_{n})$ be given such that $c^{\prime}_{n}<c_{n},\ c^{\prime}_{n}\to+\infty,\ (a^{\prime}_{n})=(b_{n})||(c^{\prime}_{n})$. Then $1\in AAR_{(a^{\prime}_{n})}$.

Proof.

We can assume that $c^{\prime}_{n}>0$.

Let $\epsilon>0$. Choose $k\in\mathbb{N}$ such that $\frac{1}{k}<\frac{\epsilon}{2}$. Then there is $N\in\mathbb{N}$ such that $n>N$ implies that $c^{\prime}_{n}\geq k$. Then

[TABLE]

if $n$ is big enough which gives the statement by 4.3. ∎

Now we give some equivalent forms of the condition in 4.3.

**  Proposition 4.12****.**

Let $(c_{n})$ be a sequence such that $\forall n\ c_{n}>0$. Then the followings hold.

[TABLE]

**  Proposition 4.13****.**

Let $f:\mathbb{R}^{+}\to\mathbb{R}^{+}$ be an increasing function that is integrable over each finite interval and $\lim\limits_{+\infty}f=+\infty$. Let $(c_{n})$ be defined by $c_{n}=f(n)$. Then

[TABLE]

Proof.

By obvious estimation we get that

[TABLE]

which gives the statement. ∎

**  Proposition 4.14****.**

Let $f:\mathbb{R}^{+}\to\mathbb{R}^{+}$ be an increasing function that has a primitive function $F$ and $\lim\limits_{+\infty}f=+\infty$. Let $(c_{n})$ be defined by $c_{n}=f(n)$. Then

[TABLE]

5 More on the basic concept

We introduce terminology for the most frequently used property of sequences.

**  Definition 5.1****.**

Let $(c_{n})$ be a sequence such that $c_{n}\to\infty$. We say that $(c_{n})$ is balanced if

[TABLE]

We present an equivalent form.

**  Proposition 5.2****.**

$(c_{n})$* is balanced if and only if*

[TABLE]

Let us use the notation $A^{(c_{k})}_{n}=\frac{c_{1}}{c_{n}}+\frac{c_{2}}{c_{n}}+\dots+\frac{c_{n-1}}{c_{n}}$.

**  Proposition 5.3****.**

Removing or adding finitely many elements to a sequence does not change the property of being balanced.∎

**  Proposition 5.4****.**

Let $(c_{n}),\ (d_{n})$ be balanced sequences, $K\in\mathbb{R}$. Then $(K\cdot c_{n}),\ (c_{n}+K),\ (c_{n}+d_{n})$ are all balanced as well.

Proof.

$(K\cdot c_{n})$ is balanced:

[TABLE]

$(c_{n}+K)$ is balanced:

[TABLE]

because $\sum\limits_{i=1}^{n-1}c_{i}\to\infty$ moreover $\frac{\sum\limits_{i=1}^{n-1}c_{i}}{n-1}\to\infty$.

$(c_{n}+d_{n})$ is balanced:

[TABLE]

since this is a weighted average of $\frac{\sum\limits_{i=1}^{n-1}c_{i}}{c_{n}}$ and $\frac{\sum\limits_{i=1}^{n-1}d_{i}}{d_{n}}$, of which both tend to infinity. ∎

**  Proposition 5.5****.**

Let $(c_{n})$ be a balanced sequence. Then $\varlimsup\frac{c_{n-1}}{c_{n}}=1$.

Proof.

Suppose indirectly that $\varlimsup\frac{c_{n-1}}{c_{n}}=q<1$. Let $q<p<1$. Then there is $N\in\mathbb{N}$ such that $n>N$ implies that $\frac{c_{n-1}}{c_{n}}<p$. Take such $n$. If $N+1\leq k\leq n-1$ then

[TABLE]

Then

[TABLE]

[TABLE]

which is a contradiction. ∎

**  Example 5.6****.**

$(c_{n})$* being balanced does not imply that $\frac{c_{n-1}}{c_{n}}\to 1$ i.e. $\varliminf\frac{c_{n-1}}{c_{n}}=1$.*

Proof.

Let $(c_{n})$ be the following sequence: $1,2,2,4,4,4,8,8,8,8,\dots$ i.e. we have $k+1$ pieces of $2^{k}$ in $(c_{n})$.

We show that $(c_{n})$ is balanced. Let $c_{n-1}\not=c_{n}=2^{k}$. Then

[TABLE]

If $c_{n-1}=c_{n}$ then clearly

[TABLE]

which together with the previous statement gives that $(c_{n})$ is balanced.

Obviously $\frac{c_{n-1}}{c_{n}}\not\to 1$. ∎

**  Proposition 5.7****.**

If $c_{n}\to+\infty$ and $\frac{c_{n-1}}{c_{n}}\to 1$ then $(c_{n})$ is balanced.

Proof.

We can assume that $c_{n}>0$.

Let $M\in\mathbb{N}$ be given. Choose $0<p<1$ such that $p^{2M}>\frac{1}{2}$. Find $N\in\mathbb{N}$ such that $n>N$ implies that $\frac{c_{n-1}}{c_{n}}>p$. Let $n>N+2M$. If $N+1\leq k\leq n-1$ then

[TABLE]

Then

[TABLE]

[TABLE]

**  Proposition 5.8****.**

If \big{(}A^{(c_{k})}_{n}\big{)} increasing and $\lim\limits_{n\to\infty}A^{(c_{k})}_{n}=+\infty$ then $\frac{c_{n-1}}{c_{n}}\to 1$.

Proof.

Obvious calculation shows that

[TABLE]

Then $A^{(c_{k})}_{n}\leq A^{(c_{k})}_{n+1}$ gives that

[TABLE]

[TABLE]

[TABLE]

Which implies that

[TABLE]

hence $\frac{c_{n+1}}{c_{n}}-1\to 0$ and $\frac{c_{n+1}}{c_{n}}\to 1$ and finally $\frac{c_{n-1}}{c_{n}}\to 1$. ∎

**  Example 5.9****.**

$\lim\limits_{n\to\infty}A^{(c_{k})}_{n}=+\infty$* and $\frac{c_{n-1}}{c_{n}}\to 1$ do not imply that \big{(}A^{(c_{k})}_{n}\big{)} is increasing.*

Proof.

Let $(c_{n})$ be the following sequence: $1,1,2,2,2,3,3,3,3,...$ where there are $n+1$ pieces from $n$ in the sequence.

Obviously $c_{n}\to+\infty,\ (c_{n})$ is increasing and $\frac{c_{n-1}}{c_{n}}\to 1$.

We show that $\lim\limits_{n\to\infty}A^{(c_{k})}_{n}=+\infty$. If $c_{n}=c_{n+1}$ then clearly $A^{(c_{k})}_{n}<A^{(c_{k})}_{n+1}$. If $c_{n}\not=c_{n+1},\ c_{n+1}=k+1$ then $A^{(c_{k})}_{n+1}>(k+1)\cdot\frac{k}{k+1}=k$. These two statements gives the claim.

We finally show that \big{(}A^{(c_{k})}_{n}\big{)} is not increasing. Let $c_{n}=k,\ c_{n+1}=k+1$. Then $A^{(c_{k})}_{n+1}<A^{(c_{k})}_{n}$ because

[TABLE]

[TABLE]

and all terms in the latter are smaller than in the former, except the last terms which are equal. ∎

**  Example 5.10****.**

$(c_{n})$* being balanced does not imply that $(c^{2}_{n})$ is balanced as well.*

Proof.

Let $(c_{n})$ be the following sequence: 4 pieces of 1 then 9 pieces of 2 then 16 pieces of 6 and so on; generally there are $(n+1)^{2}$ pieces of $n!$ in the sequence.

We show that $(c_{n})$ is balanced. If $c_{n}=c_{n+1}$ then clearly $A^{(c_{k})}_{n}<A^{(c_{k})}_{n+1}$. If $c_{n}\not=c_{n+1},\ c_{n+1}=k!$ then $A^{(c_{k})}_{n+1}>k^{2}\cdot\frac{(k-1)!}{k!}=k$. These two statements gives the claim.

We show that $(c^{2}_{n})$ is not balanced. Let $c_{n}\not=c_{n+1},\ c_{n+1}=k!$. Then

[TABLE]

[TABLE]

showing that $(c^{2}_{n})$ is not balanced. ∎

6 Generalizing results

First let us present a generalized form of 3.1.

**  Theorem 6.1****.**

Let $(a_{n})$ be a sequence and $a\leq b\ (a,b\in\mathbb{R})$ are two of its accumulation points. Then $[a,b]\subset AAR_{(a_{n})}$.

Proof.

The proof can copy the proof of 3.1 remarking that we have not used that $(d_{n})$ is bounded. ∎

**  Theorem 6.2****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $b=\varlimsup b_{n}\in\mathbb{R},\ c_{n}\to+\infty$ and $(c_{n})$ is increasing. If $c>b$ is accessible in average by rearrangement of $(a_{n})$ then

[TABLE]

Proof.

We can assume that $c>0$ (i.e. $b\geq 0$) because otherwise consider sequences $(b_{n}-b),(c_{n}-b)$. If the statement is true for these sequences then by 5.4 it is true for the original sequences as well.

Let $(d_{n})$ be a rearrangement such that d_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}c. This rearrangement defines a rearrangement of $(c_{n})$, namely take the elements from $(c_{n})$ exactly in the same order as they come in $(d_{n})$. Let us denote that rearranged sequence with $(c^{\prime}_{n})$ and $c^{\prime}_{n}=d_{m_{n}}$.

Let $\epsilon>0$. Then there is $N\in\mathbb{N}$ such that $m\geq N$ implies that

[TABLE]

Let $n$ be chosen such that $m_{n-1}>N$. Set $s_{n}=\sum\limits_{i=1}^{n-1}c^{\prime}_{i}$ and

[TABLE]

i.e. $w_{n-1}$ is the sum of elements form $(b_{i})$ which are among $d_{1},\dots,d_{m_{n-1}}$. We get that

[TABLE]

[TABLE]

From (8) we get that

[TABLE]

By multiplying with the denominator and using (7) we get that

[TABLE]

[TABLE]

and clearly both sides tend to $1$ when $\epsilon\to 0$.

It gives that $\frac{c^{\prime}_{n}}{s_{n}+w_{n}}\to 0$.

We show that it implies that $\frac{c^{\prime}_{n}}{s_{n}}\to 0$. Let the number of terms in $w_{n}$ be $k_{n}=m_{n}-n$, the number of terms in $s_{n}$ be $l_{n}=n-1$. Clearly

[TABLE]

Therefore it is enough to prove that $\frac{w_{n}}{s_{n}}$ is bounded. Assume the contrary and assume first that $\varlimsup\frac{w_{n}}{s_{n}}=+\infty$. Let $N\in\mathbb{N}$ such that $n>N$ implies that

[TABLE]

Let $K\in\mathbb{R}$ such that

[TABLE]

Now choose $n>N$ such that $\frac{w_{n}}{s_{n}}>K$. Then

[TABLE]

[TABLE]

[TABLE]

which is a contradiction.

Assume now that $\varliminf\frac{w_{n}}{s_{n}}=-\infty$. Let $N\in\mathbb{N}$ such that $n>N$ implies that

[TABLE]

For $K=-2$ choose $n>N$ such that

[TABLE]

[TABLE]

[TABLE]

– a contradiction again.

Finally we got that $\frac{c^{\prime}_{n}}{s_{n}}\to 0$. Now Corollary 4.2 yields the statement. ∎

**  Theorem 6.3****.**

Let $(a_{n})=(b_{n})||(c_{n})$ where $\varlimsup b_{n}=b\in\mathbb{R},\ c_{n}\to+\infty$. If

[TABLE]

and $c>b$ then $c$ is accessible in average by rearrangement of $(a_{n})$.

Proof.

Set $v=\frac{1}{c-b}$. Set $s_{n}=\sum\limits_{i=1}^{n}c_{i}$. Note that $s_{n}\to+\infty$ moreover $\frac{s_{n}}{n}\to+\infty$. With that notation the assumption gets the form: $\frac{c_{n}}{s_{n-1}}\to 0$.

Take a subsequence $(b^{\prime}_{n})$ from $(b_{n})$ such that $b^{\prime}_{n}\to b$.

We can assume that $c_{n}$ is increasing (by Corollary 4.2) and $c_{n}>\max\{1,2(c-b)\}$. Then let m_{n}=\Big{\lfloor}v\cdot\sum\limits_{i=1}^{n}c_{i}\Big{\rfloor}. The previous assumption on $c_{n}$ gives that $(m_{n})$ is a strictly increasing sequence of integers.

If $k\in\mathbb{N}$ then let

[TABLE]

i.e. on the $m_{n}$ position put $c_{n}$, on the other positions put elements from $(b^{\prime}_{n})$.

We show that d_{k}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}c. Let $m_{n-1}\leq k<m_{n}$. Obviously

[TABLE]

Let us investigate the two terms. For the first one we get that

[TABLE]

Clearly

[TABLE]

because $b^{\prime}_{i}\to b$. Regarding the second factor we have

[TABLE]

Hence

[TABLE]

Now let us analyze the second term.

[TABLE]

[TABLE]

Evidently both sides tends to $\frac{1}{v}=c-b$ using that $\frac{c_{n}}{s_{n-1}}\to 0$ by assumption. Therefore

[TABLE]

which gives that \lim\limits_{k\to\infty}\mbox{{{{\cal A}}}}(d_{1},\dots,d_{k})=c

By [lapaar, Corollary 2.4] we can add the remaining elements from $(a_{n})$ (that are not in $(b^{\prime}_{n})||(c_{n})$) to $(d_{n})$ such that the limit in average does not change. ∎

7 Sequences composed of two sequences, one tending to $+\infty$, the other tending to $-\infty$

We are going to investigate the case when a sequence does not have finite accumulation point but it has both $\pm\infty$ as accumulation points.

**  Theorem 7.1****.**

Let $(a_{n})=(b_{n})||(c_{n}),\ b_{n}\to-\infty,\ c_{n}\to+\infty$. If there is $a\in AAR_{(a_{n})},\ a\in\mathbb{R}$ then $\varliminf\frac{b_{n}}{n}=0$ and $\varliminf\frac{c_{n}}{n}=0$.

Proof.

Suppose indirectly that $\varliminf\frac{c_{n}}{n}=2p>0$ ($(b_{n})$ can be handled similarly). Then there is $N\in\mathbb{N}$ such that $n>N$ implies that $\frac{c_{n}}{n}>p$.

Let $\frac{p}{9}>\epsilon>0$. If $a\in AAR_{(a_{n})}$ then there is a rearrangement of $(a_{n})$ say $(a_{n_{k}})$ such that a_{n_{k}}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}a. Which yields that there is $M\in\mathbb{N}$ such that $v>M$ implies that

[TABLE]

Let $M$ be chosen such that $\{c_{n}:n\leq N\}\subset\{a_{n_{k}}:k\leq M\}$ moreover $\frac{a+1}{M+1}<\epsilon$. Find a $c_{n}$ such that $n>N$ and $c_{n}=a_{n_{v+1}}$ and $v>M$ and $v+1<2n+1$. It can be done because let

[TABLE]

and let $v+1=\min\{w:a_{n_{w}}=c_{m},\ m>l\}$.

Let us estimate how much the new element $a_{n_{v+1}}$ modifies the average.

[TABLE]

Clearly

[TABLE]

and

[TABLE]

which gives that

[TABLE]

which is a contradiction. ∎

**  Theorem 7.2****.**

Let $(a_{n})=(b_{n})||(c_{n}),\ b_{n}\to-\infty,\ c_{n}\to+\infty$. If $\varliminf\frac{b_{n}}{n}=0$ and $\varliminf\frac{c_{n}}{n}=0$ then $\mathbb{R}\subset AAR_{(a_{n})}$.

Proof.

If enough to prove that $0\in AAR_{(a_{n})}$ because $(b_{n}-a)||(c_{n}-a)$ satisfies the conditions of the theorem and if a_{n}-a\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}0 then a_{n}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}a.

Let $\epsilon>0$.

First assume that $\frac{b_{n}}{n}\to 0$ and $\frac{c_{n}}{n}\to 0$. We define a rearrangement of $(a_{n})$. First take elements from $(c_{n})$ in the order as they are in $(c_{n})$ such that the average would be $\geq 0$. It can be done as $c_{n}\to+\infty$. Then add elements from $(b_{n})$ in the order as they are in $(b_{n})$ such that the average of all selected elements (the previous ones and the just added ones) would be $\leq 0$. It can be done as $b_{n}\to-\infty$. Then add not used elements from $(c_{n})$ in the order as they are in $(c_{n})$ such that the average of all selected elements (the previous ones and the just added ones) would be $\geq 0$. And so on. Let us denote the constructed sequence by $(a_{n_{k}})$.

We show that the rearranged sequence tends to [math] in average. Let $N\in\mathbb{N}$ such that $n>N$ implies that $\frac{|b_{n}|}{n}<\frac{\epsilon}{2},\ \frac{c_{n}}{n}<\frac{\epsilon}{2}$.

Choose $V\in\mathbb{N}$ such that $\{b_{n},c_{n}:n\leq N\}\subset\{a_{n_{k}}:k\leq V\}$. Let

[TABLE]

Suppose that $s\geq 0$ (the other case is similar).

First we show that there is $v\geq V$ such that

[TABLE]

If $s<\epsilon$ then we are done. If not then suppose $\frac{\sum\limits_{k=1}^{v}a_{n_{k}}}{v}\geq\epsilon$. Then clearly

[TABLE]

for $a_{n_{v+1}}=b_{n}$. But

[TABLE]

hence $\frac{\sum\limits_{k=1}^{v+1}a_{n_{k}}}{v+1}>-\frac{\epsilon}{2}$ i.e. sooner or later we will step into $(-\epsilon,\epsilon)$ referring also to the definition of $(a_{n_{k}})$.

Now suppose that $\frac{\sum\limits_{k=1}^{v}a_{n_{k}}}{v}\in(-\epsilon,\epsilon)$ and let us examine how much the average changes if we add a new element. If $v>V,\ a_{n_{k+1}}=c_{n}$ ($b_{n}$ is similar) then

[TABLE]

Which means that by the definition of $(a_{n_{k}})$ we will always remain in the interval $(-\epsilon,\epsilon)$. I.e. we showed that a_{n_{k}}\overset{\mbox{{{{\cal A}}}}}{\longrightarrow}0.

Let us turn to the general case and let $(b_{n_{k}}),\ (c_{m_{k}})$ subsequences of $(b_{n}),\ (c_{n})$ respectively such that $b_{n_{k}}\to 0,\ c_{m_{k}}\to 0$. Apply the previous assertion for sequence $(d_{n})=(b_{n_{k}})||(c_{m_{k}})$ which gives that $\mathbb{R}\subset AAR_{(d_{n})}$. Now Corollary 2.4 gives the statement. ∎

We can summarize the previous two theorems in the following way.

**  Theorem 7.3****.**

Let $(a_{n})=(b_{n})||(c_{n}),\ b_{n}\to-\infty,\ c_{n}\to+\infty$. Then $a\in AAR_{(a_{n})},\ a\in\mathbb{R}$ if and only if $\varliminf\frac{b_{n}}{n}=0$ and $\varliminf\frac{c_{n}}{n}=0$. ∎

8 Summary of all cases

We are now in the position that we can describe all possible cases regarding $AAR_{(a_{n})}$. Let us denote the accumulation points of $(a_{n})$ by $(a_{n})^{\prime}$.

There is no finite accumulation point of $(a_{n})$

1. (a)

If $(a_{n})^{\prime}=\{-\infty\}$ then $AAR_{(a_{n})}=\{-\infty\}$. 2. (b)

If $(a_{n})^{\prime}=\{+\infty\}$ then $AAR_{(a_{n})}=\{+\infty\}$. 3. (c)

If $(a_{n})^{\prime}=\{\pm\infty\}$ let $(a_{n})=(b_{n})||(c_{n}),\ b_{n}\to-\infty,\ c_{n}\to+\infty$. Then $AAR_{(a_{n})}=\bar{\mathbb{R}}$ if and only if $\varliminf\frac{b_{n}}{n}=0$ and $\varliminf\frac{c_{n}}{n}=0$. Otherwise $AAR_{(a_{n})}=\{\pm\infty\}$. 2. 2.

If $\pm\infty\not\in(a_{n})^{\prime}$ then let $\varliminf a_{n}=a,\ \varlimsup a_{n}=b\ (a,b\in\mathbb{R})$. In this case $AAR_{(a_{n})}=[a,b]$. 3. 3.

There are also finite and infinite accumulation points of $(a_{n})$. Let $\varliminf a_{n}=\alpha\in\bar{\mathbb{R}},\ \varlimsup a_{n}=\beta\in\bar{\mathbb{R}},\ \min(a_{n})^{\prime}-\{-\infty\}=a\in\bar{\mathbb{R}},\ \max(a_{n})^{\prime}-\{+\infty\}=b\in\bar{\mathbb{R}}$. Obviously either $\alpha=-\infty$ or $\beta=+\infty$.

Then we can decompose $(a_{n})$ in the following way: $(a_{n})=(b_{n})\cup(c_{n})\cup(d_{n}),\ b_{n}\to\alpha,\ c_{n}\to\beta,\ (d_{n})$ is the rest of the sequence (if needed). This decomposition is not unique in general. However if $\alpha=-\infty,\ a>-\infty$ then $(b_{n})$ is unique up to finitely many elements and if $\beta=+\infty,\ b<+\infty$ then $(c_{n})$ is unique up to finitely many elements.

1. (a)

If $a=-\infty$ or $b=+\infty$ then $AAR_{(a_{n})}=[a,b]\cup\{\alpha,\beta\}$. 2. (b)

If $-\infty<a\leq b<+\infty$ then we have the following cases:

1. i.

Let $\alpha=-\infty,\ \beta<+\infty$. If $(b_{n})$ is balanced then $AAR_{(a_{n})}=[-\infty,b]$. If $(b_{n})$ is not balanced then $AAR_{(a_{n})}=[a,b]\cup\{-\infty\}$. 2. ii.

Let $\alpha>-\infty,\ \beta=+\infty$. If $(c_{n})$ is balanced then $AAR_{(a_{n})}=[a,+\infty]$. If $(c_{n})$ is not balanced then $AAR_{(a_{n})}=[a,b]\cup\{+\infty\}$. 3. iii.

Let $\alpha=-\infty,\ \beta=+\infty$. If $(b_{n}),(c_{n})$ are both balanced then $AAR_{(a_{n})}=\bar{\mathbb{R}}$. If $(b_{n})$ is balanced, $(c_{n})$ is not balanced then $AAR_{(a_{n})}=[-\infty,b]\cup\{+\infty\}$. If $(b_{n})$ is not balanced, $(c_{n})$ is balanced then $AAR_{(a_{n})}=[a,+\infty]\cup\{-\infty\}$. If none of $(b_{n})$, $(c_{n})$ is balanced then $AAR_{(a_{n})}=[a,b]\cup\{-\infty,+\infty\}$.

9 On accumulation points

**  Theorem 9.1****.**

Let $(a_{n})$ be a sequence bounded from below or above, $(a^{\prime}_{n})$ be one of its rearrangements. Let $c_{n}=\frac{\sum\limits_{i=1}^{n}a^{\prime}_{i}}{n}$. Then the set of accumulation points of $(c_{n})$ is $[\varliminf c_{n},\varlimsup c_{n}]$.

Proof.

If $\varliminf c_{n}=\varlimsup c_{n}$ then there is nothing to prove. Assume that $\varliminf c_{n}<\varlimsup c_{n}$. Assume that $(a_{n})$ is bounded from below and $a_{n}>K$. The other case is similar.

Let $\varliminf c_{n}<p<\varlimsup c_{n}$. Obviously it true for infinitely many $n$ that $c_{n}\leq p<c_{n-1}$. Then

[TABLE]

which gives that

[TABLE]

Then we get that

[TABLE]

[TABLE]

if $n$ is big enough. ∎

**  Lemma 9.2****.**

Let $(b_{n})$ be a sequence, $c\in\mathbb{R}$. Assume that \mbox{{{{\cal A}}}}(b_{1},\dots,b_{n})\in(a,b) if $n>N$ ($a<b,a,b\in\mathbb{R}$). Then $\forall\epsilon>0$ we can create a new sequence $(d_{n})$ with $d_{i}=b_{i}\ (i<k),d_{k}=c,d_{i}=b_{i-1}\ (i>k>N)$ such that $n\geq k$ implies that \mbox{{{{\cal A}}}}(d_{1},\dots,d_{n})\in(a-\epsilon,b+\epsilon).

Proof.

Choose $k\in\mathbb{N}$ such that $n\geq k-1$ implies that

(1) $|\frac{c}{n}|<\frac{\epsilon}{2}$ and

(2) $a-\frac{\epsilon}{2}<a\cdot\frac{n-1}{n}$.

If $m\geq k$ then

[TABLE]

hence a-\epsilon<\mbox{{{{\cal A}}}}(d_{1},\dots,d_{m})<b+\epsilon. ∎

**  Lemma 9.3****.**

Let $(b_{n})$ be a bounded sequence, $a=\varliminf b_{n},b=\varlimsup b_{n},a<b,\ (a,b\in\mathbb{R})$. Let $c,d\in\mathbb{R}$ such that $a<c<d<b$. Then there is a rearrangement $(b^{\prime}_{n})$ of $(b_{n})$ and there is $N\in\mathbb{N}$ such that $n>N$ implies that \mbox{{{{\cal A}}}}(b^{\prime}_{1},\dots,b^{\prime}_{n})\in(c,d).

Proof.

It is an obvious consequence of 2.6. ∎

**  Theorem 9.4****.**

Let $(a_{n})$ be a sequence such that it has at least 4 accumulation points: $a,b,-\infty,+\infty\ (a,b\in\mathbb{R},a<b)$. Let $Z\subset[a,b]$ be a closed set. Then there is a rearrangement $(a^{\prime}_{n})$ of $(a_{n})$ such that the accumulation points of the sequence $(p_{n})$ is exactly $Z$ where p_{n}=\mbox{{{{\cal A}}}}(a^{\prime}_{1},\dots,a^{\prime}_{n}).

Proof.

Let $(a_{n})=(b_{n})||(c_{n})||(d_{n})||(e_{n})||(f_{n})$ where $b_{n}\to a,c_{n}\to b,d_{n}\to-\infty,e_{n}\to+\infty$ and $(f_{n})$ is the rest of the elements in $(a_{n})$. Assume that $\forall n\ a-1<b_{n},c_{n}<b+1$. Let $K=\max\{|a-1|,|b+1|\}$.

First let us note that there is a sequence $(t_{n})$ such that the accumulation points of $(t_{n})$ is exactly $Z$. We know that $Z$ is separable because $\mathbb{R}$ is hereditary separable, hence let $Z_{0}\subset Z$ such that $cl(Z_{0})=Z$ and $|Z_{0}|\leq\aleph_{0}$. Let $Z_{0}=\{w_{n}:n\in\mathbb{N}\}$ and then the sequence $(w_{1},w_{2},w_{1},w_{2},w_{3},w_{1},w_{2},w_{3},w_{4},\dots)$ satisfies the requirement.

We are going to create a rearrangement $(a^{\prime}_{n})$ of $(a_{n})$ such that there is a sequence $(n_{k})$ such that $n_{k}\in\mathbb{N},(n_{k})$ is increasing and $n_{k}\leq n<n_{k+1}$ implies that $a_{n}\in(t_{k}-\frac{1}{k},t_{k}+\frac{1}{k})$. It is easy to see that $(a^{\prime}_{n})$ fulfills the requirements.

Let us generalize the problem slightly. We have a sequence of disjoint finite open intervals $(I_{i})$ with $I_{i}\subset(a,b)$ and want to create a rearrangement $(a^{\prime}_{n})$ of $(a_{n})$ such that there is a sequence $(n_{k})$ such that $n_{k}\leq n<n_{k+1}$ implies that $a_{n}\in I_{k}$.

We define sequences by recursion. By 9.3 we can find a rearrangement $(r^{1}_{n})$ of $(b_{n})||(c_{n})$ and $N_{1}\in\mathbb{N}$ such that $n\geq N_{1}$ implies that \mbox{{{{\cal A}}}}(r^{1}_{1},\dots,r^{1}_{n})\in I_{1}. We can also assume that $r^{1}_{1}=a_{1}$.

Assume that for $k\in\mathbb{N},k>1$ there are a sequence $(r^{k}_{n})$ and positive integers $n_{1}<n_{2}<\dots<n_{k-1}<n_{k}$ such that

(0) if $n>n_{k}$ then $r^{k}_{n}$ is an element of $(b_{n})||(c_{n})$,

(1) and if $n\leq n_{k-1}$ then $r^{k-1}_{n}=r^{k}_{n}$,

(2) and if $1\leq i\leq k-1$ and $n_{i}\leq n<n_{i+1}$ then \mbox{{{{\cal A}}}}(r^{i}_{1},\dots,r^{i}_{n})\in I_{i},

(3) and if $n_{k}\leq n$ then \mbox{{{{\cal A}}}}(r^{k}_{1},\dots,r^{k}_{n})\in I_{k},

(4) and there is $i\leq n_{k}$ such that $r^{k}_{i}=a_{k}$.

We will see that $n_{1},\dots,n_{k}$ can be chosen independently of $k$.

Assume that $\sup I_{k}<\inf I_{k+1}$. (The other inequality can be handled similarly.)

If $a_{k+1}$ is already in $(r^{k}_{n})$, say $r^{k}_{i}=a_{k+1}$, then let $N_{k}=\max\{i,n_{k}\}$. If not, then let $I_{k}=(a^{\prime},b^{\prime}),\ \epsilon=\frac{b^{\prime}-a^{\prime}}{4},\ I^{\prime}=(a^{\prime}+\epsilon,b-\epsilon)$ and apply 9.2 for $(r^{k}_{n}),\ I^{\prime},\ N=n_{k},\ \epsilon,\ c=a_{k+1}$. Hence in that way we can merge $a_{k+1}$ into $(r^{k}_{n})$ in a way that all conditions (0),(1),(2),(3),(4) remain valid (maybe we have to modify $n_{k}$). Let us keep the notation $(r^{k}_{n})$ for this new sequence too. Let $N_{k}$ be the index of $a_{k+1}$ in $(r^{k}_{n})$.

Let $h$ be the length of $I_{k+1}$. Let $M>\max\{N_{k},\frac{9K}{h},\frac{18(b-a+1)}{h}\}$. If $m\geq M$ then $\frac{9(b-a+1)}{h}<\frac{m}{2}<\frac{m^{2}}{m+1}$ which gives that $(b-a+1)(m+1)<\frac{h}{9}m^{2}$. I.e. if we take intervals \big{(}(b-a+1)m,\frac{h}{9}m^{2}\big{)} for all such $m$, then those intervals will overlap. Hence there are a $P>M$ and an unused element $e$ from $(e_{n})$ such that $(b-a+1)P<e<\frac{h}{9}P^{2}$ and $\frac{P-1}{P}a>a-1$.

First note that b<\mbox{{{{\cal A}}}}(r^{k}_{1},\dots,r^{k}_{P-1},e) because

[TABLE]

If $m>P$ then let $s=\sum\limits_{i=1}^{m-1}r^{k}_{i},\ z=r^{k}_{m}$. Then

[TABLE]

[TABLE]

Now there is a sufficiently large $m$ such that $m>P$ and

[TABLE]

because \mbox{{{{\cal A}}}}(r^{k}_{1},\dots,r^{k}_{P},e)>b and

[TABLE]

and by |\mbox{{{{\cal A}}}}(r^{k}_{1},\dots,r^{k}_{m-1},e)-\mbox{{{{\cal A}}}}(r^{k}_{1},\dots,r^{k}_{m},e)|<\frac{h}{3}. Fix such $m$.

Let

[TABLE]

If $n>m$ then choose not-used elements from $(b_{n})||(c_{n})$ such that

[TABLE]

We show that this can be done (we apply a similar argument as above). We go by induction. It is true for $n=m$. Assume that we are done till $n$ and looking for $r^{k+1}_{n+1}$. Let $s=\sum\limits_{i=1}^{n-1}r^{k+1}_{i},\ z\in(a-1,b+1)$.

[TABLE]

[TABLE]

If \mbox{{{{\cal A}}}}(r^{k+1}_{1},\dots,r^{k+1}_{n})\in I_{k+1}, say \mbox{{{{\cal A}}}}(r^{k+1}_{1},\dots,r^{k+1}_{n})\leq\inf I_{k+1}+\frac{h}{2}, then we can choose $r^{k+1}_{n+1}$ from $(c_{n})$ such that r^{k+1}_{n+1}>\mbox{{{{\cal A}}}}(r^{k+1}_{1},\dots,r^{k+1}_{n}) and that will satisfy $A(r^{k+1}_{1},\dots,r^{k+1}_{n})\in I_{k+1}$. The other case is similar.

Let now $n_{k+1}=m$. We show that all conditions (0),(1),(2),(3),(4) will be valid for $(r^{k+1}_{n})$. (0) and (1) are obvious by definition. (2) is valid because we keep all previous $n_{i}\ (1\leq i\leq k)$ and by (3) for $(r^{k}_{n})$. (3) holds by definition. The first step was to include $a_{k+1}$ hence (4) is valid too.

Let $a^{\prime}_{n}=r^{k}_{n}$ if $n_{k}\leq n<n_{k+1}$. Clearly the accumulation points of p_{n}=\mbox{{{{\cal A}}}}(a^{\prime}_{1},\dots,a^{\prime}_{n}) is exactly $Z$. ∎

**  Remark 9.5****.**

In 9.4 if we want to weaken the condition then by 9.1 we cannot abandon neither $-\infty$ nor $+\infty.$

**  Remark 9.6****.**

In 9.4 we cannot say more in general in a way that any closed sets $Z$ outside $[a,b]$ could be the accumulation points of the averages of a rearranged sequence.

Proof.

Let $(a_{n})=(-2^{n})||(0)||(2^{n})$. By 6.2 $AAR_{(a_{n})}=\{-\infty,0,+\infty\}$. ∎