Riemann Rearrangement Thoerem and Proof
Författare
David Klapheck
Last Updated
för 6 år sedan
Licens
LaTeX Project Public License 1.3c
Sammanfattning
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.
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\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}
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