Riemann Rearrangement Thoerem and Proof
Author
David Klapheck
Last Updated
7 years ago
License
LaTeX Project Public License 1.3c
Abstract
A simple proof of Riemann's Rearrangement Theorem. Also called Riemann's series theorem.
A simple proof of Riemann's Rearrangement Theorem. Also called Riemann's series theorem.
\documentclass[11pt,letterpaper]{article}
\usepackage{amssymb}
\usepackage{fullpage}
\usepackage{latexsym,amsmath,amssymb,amscd,bm,amsfonts}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{color}
\usepackage{upgreek,enumitem}
\newcommand{\com}{\color{red}}
\newcommand{\modulo}{\text{ mod }}
\newcommand{\wn}[2]{\ensuremath{\gamma ({#1},{#2})}}
\newcommand{\Cal}{\mathcal}
\newcommand{\tsize}{\textstyle}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\ba}{\begin{align}}
\newcommand{\ea}{\end{align}}
\newcommand{\sumstar}[2]{\sideset{}{^*}\sum_{#1}^{#2}}
\newcommand{\tmod}{\text{ mod }}
\newcommand{\summ}[3]{\sum_{#1=#2}^{#3}}
\newcommand{\phipm}{\phi(p^m)}
\newcommand{\chip}{\chi'}
\newcommand{\ent}{\mathbb Z}
\newcommand{\lsym}[2]{\left( \frac{#1}{#2}\right)}
\usepackage[top=1in, bottom=1in, left=1in, right=1in, paperwidth=8.5in, paperheight=11in]{geometry}
\newcommand{\workingDate}{\textsc{2013 $|$ January $|$ 01}}
\newcommand{\userName}{Your Name}
\newcommand{\institution}{Your University}
\newcommand{\Cbb}{\mathbb{C}}
\newcommand{\Rbb}{\mathbb{R}}
\newcommand{\Nbb}{\mathbb{N}}
\newcommand{\Qbb}{\mathbb{Q}}
\newcommand{\Zbb}{\mathbb{Z}}
\newcommand{\Wbb}{\mathbb{W}}
\newcommand{\Js}{\mathcal{J}}
\newcommand{\calh}{\mathcal{H}}
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\abs}[1]{\left|#1\right|}
\newcommand{\bbar}[1]{\mkern 1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern 1.5mu}
%\makeatletter
%\newcommand*{\ov}[1]{$\m@th\overline{\mbox{#1}}$}
%\newcommand*{\ovC}[1]{$\m@th\overline{\mbox{#1\strut}}$}
\newcommand\textbox[1]{\parbox{.333\textwidth}{#1}}
\let\mplies\Rightarrow
\let\then\Rightarrow
\let\ff\Leftrightarrow
\newcommand{\ifa}{\Longleftarrow}
\newcommand{\st}{\textnormal{ such that }}
\newcommand{\ep}{\upvarepsilon}
\newcommand{\fa}{\forall}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{conjecture}{Conjecture}[section]
\begin{document}
\title{Riemann Rearrangement Thoerem and Proof}
\author{David Klapheck}
\maketitle
\noindent {\bf Riemann Rearrangement Theorem:}\\
Given a conditionally convergent real series $$\sum_{n=1}^\infty a_n$$ and a value $M \in\Rbb$, there exists a rearrangement of the series such that $\sum a_{\sigma (n)} = M$.
\begin{proof}
Given $\sum a_n$ is conditionally convergent, $\sum \abs{a_n}=\infty$.
Define subsequences\footnote{Is there a less cumbersome way to define these subsequences?} $\left( a_{n_j}\right)_{n_j \in A}$ and $\left( a_{n_k}\right)_{n_k \in B}$ of $a_n$ by $i\in A \ff a_i < 0$ and $i\in B \ff a_i \geq 0$.\\
Claim: $\sum_{j=1}^\infty a_{n_j}=-\infty$ and $\sum_{k=1}^\infty a_{n_k}=\infty$. Suppose both series converge. Then by series addition $\sum \abs{a_n} = \sum_{k=1}^\infty a_{n_k} - \sum_{j=1}^\infty a_{n_j}$ converges. A contradiction. Suppose one series converges and the other series diverges. Then $\sum_{k=1}^\infty a_{n_k} + \sum_{j=1}^\infty a_{n_j}=\sum {a_n} $ diverges. Another contradiction.\\
Now for the construction of permutation $\sigma$ of $\Nbb$. Let $j_1$ be the smallest $\Nbb$ such that $$\sum_{j=1}^{j_1} a_{n_j}<M.$$ Define $\sigma(j)=n_j \in A$, $\forall j\in [1..j_1]$.\footnote{Here the notation [a..b] refers to all the integers from a through b. Also (a..b) is the set of all integers between a and b.} Let $k_1$ be the smallest $\Nbb$ such that $$\sum_{j=1}^{j_1} a_{n_j}+\sum_{k=1}^{k_1} a_{n_k}>M.$$ Define $\sigma(j_1 +k)=n_k \in B$, $\forall k\in [1..k_1]$.
Step 2: Let $j_2$ be the smallest $\Nbb$ such that $$\sum_{j=1}^{j_2} a_{n_j}+\sum_{k=1}^{k_1} a_{n_k}<M.$$ Define $\sigma(j+k_1)=n_j \in A$, $\forall j\in (j_1..j_2]$. Let $k_2$ be the smallest $\Nbb$ such that $$\sum_{j=1}^{j_2} a_{n_j}+\sum_{k=1}^{k_2} a_{n_k}>M.$$ Define $\sigma(j_2+k)=n_k \in B$, $\forall k\in (k_1..k_2]$.
Continue defining $\sigma$ as above and it will be a permutation of $\Nbb$ such that the series rearrangement $\sum a_{\sigma (n)}$ will continue to oscillate around $M$. First by summing, in order, the negative terms from the sequence $(a_n)$ until the last negative term drops it below $M$. Then by adding to the sum, in order, from the non-negative terms of sequence $(a_n)$ until the last term pushes is over $M$.
Let $\ep>0$. By the divergence test $\abs{ a_n } \to 0$. Thus $\exists N \in \Nbb \st \forall n\geq N$ $ \abs{a_n} <\ep$. Now $\exists i \in \Nbb \st j_i+k_i>N$. Then since $$\sum_{j=1}^{j_i} a_{n_j}+\sum_{k=1}^{k_i} a_{n_k}>M\geq \sum_{j=1}^{j_i} a_{n_j}+\sum_{k=1}^{k_i-1} a_{n_k},$$ we have $\forall p\geq j_i+k_i,$ $\abs{M-\sum _{n=1}^{p} a_{\sigma (n)} }<\ep$. Therefore $\sum a_{\sigma (n)} = M$. \end{proof}
\end{document}