Riemann rearrangement theorem

The Riemann rearrangement theorem (after Bernhard Riemann ) is a mathematical theorem about conditionally convergent series .

formulation

If there is a conditionally convergent series of real numbers, then there is a rearrangement of the series members for any given real number , so that the rearranged series converges against . To there is a rearrangement , so that the rearranged row diverges against certain. ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{n}}$ ${\displaystyle S}$ ${\displaystyle \sigma \,}$ ${\displaystyle a_{n}}$ ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{\sigma (n)}}$ ${\displaystyle S}$ ${\displaystyle S\in \{-\infty ,+\infty \}}$ ${\displaystyle \sigma }$ ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{\sigma (n)}}$ ${\displaystyle S}$

The rearrangement is understood as a bijective mapping of the set of natural numbers onto itself (a permutation ). ${\displaystyle \sigma }$ ${\displaystyle \sigma \colon \mathbb {N} \to \mathbb {N} }$

Reason

The sequence is divided into two subsequences and , which contain only the non-negative and the negative sequence members of . For example: ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (p_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (q_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$

${\displaystyle {\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline (a_{n})_{n}&a_{0}&a_{1}&a_{2}&a_{3}&a_{4}&a_{5}&a_{6}&a_{7}&a_{8}&a_{9}&a_{10}&a_{11}&a_{12}&a_{13}&a_{14}&a_{15}&a_{16}&\cdots \\\hline \operatorname {sgn}(a_{n})&-&+&+&+&0&-&-&-&+&0&+&-&+&-&+&+&-&\cdots \\\hline (p_{n})_{n}&&p_{0}&p_{1}&p_{2}&p_{3}&&&&p_{4}&p_{5}&p_{6}&&p_{7}&&p_{8}&p_{9}&&\cdots \\\hline (q_{n})_{n}&q_{0}&&&&&q_{1}&q_{2}&q_{3}&&&&q_{4}&&q_{5}&&&q_{6}&\cdots \\\hline \end{array}}}$

The ranks and are both definitely divergent. If one of the two series were convergent, then the other would also converge, since it could be written as the difference between the original series and the first series (with inserted zeros). But that would also be absolutely convergent, contrary to the prerequisite. ${\displaystyle \textstyle \sum _{n=0}^{\infty }p_{n}}$ ${\displaystyle \textstyle \sum _{n=0}^{\infty }q_{n}}$ ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{n}}$ ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{n}}$

In particular, it follows from this that there are an infinite number of terms with a positive sign and an infinite number of terms with a negative sign.

Construction of the rearrangement

A series that converges to the real number can be constructed as follows: One adds up non-negative sequential terms until one crosses the target for the first time (in this case this is the empty sum). ${\displaystyle S}$ ${\displaystyle p_{0}+p_{1}+\dotsb +p_{k_{1}}}$ ${\displaystyle S}$ ${\displaystyle S<0}$

Then one adds up the negative sequence elements until the partial sum falls below the value . ${\displaystyle q_{0}+q_{1}+\dotsb +q_{l_{1}}}$ ${\displaystyle S}$

Then one continues alternately with non-negative and negative sequence terms. This consideration results in a rearrangement of the original series.

Since there is a zero sequence, there is an index for every stripe , no matter how small , from which all partial sums are included. The series rearranged in this way converges to . ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ ${\displaystyle \varepsilon }$ ${\displaystyle S}$ ${\displaystyle S}$

If , then one chooses the -th partial series of non-negative sequence members in the above construction so that the number is exceeded. Then one chooses the index smallest, not yet used, negative sequence term. The resulting rearrangement diverges against . The case can be handled accordingly. ${\displaystyle S=+\infty }$ ${\displaystyle n}$ ${\displaystyle n}$ ${\displaystyle +\infty }$ ${\displaystyle S=-\infty }$

example

Using the example of the alternating harmonic series , the effect of a rearrangement should be shown. This series is convergent, but not absolutely convergent: the series

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}}$

converges while the harmonic series

${\displaystyle \sum _{n=1}^{\infty }\left|{\frac {(-1)^{n+1}}{n}}\right|=\sum _{n=1}^{\infty }{\frac {1}{n}}}$

diverges. Although the alternating harmonic series converges to ln (2) in the normal representation, it can be rearranged according to the Riemann rearrangement theorem so that it converges to any other number, or even diverges. In the example it will only reach the limit value ln (2) / 2 through rearrangement.

The usual notation of this series is:

${\displaystyle 1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots =\ln 2}$

If you rearrange the summands, you get:

${\displaystyle 1-{\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{3}}-{\frac {1}{6}}-{\frac {1}{8}}+{\frac {1}{5}}-{\frac {1}{10}}-{\frac {1}{12}}+\cdots }$

In general, this sum is made up of blocks of three:

${\displaystyle {\frac {1}{2k-1}}-{\frac {1}{2(2k-1)}}-{\frac {1}{4k}},\quad k=1,2,\dots .}$

Such a block can be transformed into:

${\displaystyle {\frac {1}{2k-1}}-{\frac {1}{2(2k-1)}}-{\frac {1}{4k}}={\frac {2-1}{2(2k-1)}}-{\frac {1}{2\cdot 2k}}={\frac {1}{2}}\left({\frac {1}{2k-1}}-{\frac {1}{2k}}\right)}$

The total sum is exactly half of the alternating harmonic series:

${\displaystyle {\frac {1}{2}}\left({\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots \right)={\frac {1}{2}}\ln(2)}$

Steinitz rearrangement theorem

→ Main article : Steinitz's rearrangement theorem

Steinitz's Reordering Theorem is a generalization of Riemann's Reordering Theorem. If there is a convergent series with , then is the set of limits of all convergent rearranged series ${\displaystyle \textstyle \sum _{k=0}^{\infty }a_{k}}$ ${\displaystyle a_{k}\in \mathbb {R} ^{n}}$

${\displaystyle U=\left\{\sum _{k=0}^{\infty }a_{\sigma (k)}\ {\text{konvergent}}{\Bigg |}\sigma \ {\text{Umordnung}}\right\}}$

an affine subspace of the . If in particular , then in the complex plane there is either a point, a straight line or whole . The series is absolutely convergent if and only contains a single point. ${\displaystyle \mathbb {R} ^{n}}$ ${\displaystyle a_{k}\in \mathbb {C} \cong \mathbb {R} ^{2}}$ ${\displaystyle U}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle \textstyle \sum _{k=0}^{\infty }a_{k}}$ ${\displaystyle U}$

swell

• Harro Heuser : Textbook of Analysis. Part I, Teubner-Verlag.
• Otto Forster : Analysis 1. Vieweg Mathematics.