Riemann series theorem

Riemann series theorem In mathematics, the Riemann series theorem (also called the Riemann rearrangement theorem), named after 19th-century German mathematician Bernhard Riemann, says that if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number, or diverges. This implies that a series of real numbers is absolutely convergent if and only if it is unconditionally convergent.

Par exemple, the series 1 − 1 + 1/2 − 1/2 + 1/3 − 1/3 + ⋯ converges to 0 (for a sufficiently large number of terms, the partial sum gets arbitrarily near to 0); but replacing all terms with their absolute values gives 1 + 1 + 1/2 + 1/2 + 1/3 + 1/3 + , which sums to infinity. Thus the original series is conditionally convergent, and can be rearranged (by taking the first two positive terms followed by the first negative term, followed by the next two positive terms and then the next negative term, etc.) to give a series that converges to a different sum: 1 + 1/2 − 1 + 1/3 + 1/4 − 1/2 + ⋯ = ln 2. Plus généralement, using this procedure with p positives followed by q negatives gives the sum ln(p/q). Other rearrangements give other finite sums or do not converge to any sum.

Contenu 1 Définitions 2 Énoncé du théorème 3 Alternating harmonic series 3.1 Changing the sum 3.2 Getting an arbitrary sum 4 Preuve 4.1 Existence of a rearrangement that sums to any positive real M 4.2 Existence of a rearrangement that diverges to infinity 4.3 Existence of a rearrangement that fails to approach any limit, fini ou infini 5 Généralisations 5.1 Sierpiński theorem 5.2 Théorème de Steinitz 6 Voir également 7 References Definitions A series {textstyle sum _{n=1}^{infime }un_{n}} converges if there exists a value {style d'affichage ell } such that the sequence of the partial sums {style d'affichage (S_{1},S_{2},S_{3},ldots ),quad S_{n}=somme _{k=1}^{n}un_{k},} converge vers {style d'affichage ell } . C'est-à-dire, for any ε > 0, there exists an integer N such that if n ≥ N, alors {displaystyle leftvert S_{n}-ell rightvert leq epsilon .} A series converges conditionally if the series {textstyle sum _{n=1}^{infime }un_{n}} converges but the series {textstyle sum _{n=1}^{infime }leftvert a_{n}rightvert } diverges.

A permutation is simply a bijection from the set of positive integers to itself. This means that if {style d'affichage sigma } is a permutation, then for any positive integer {style d'affichage b,} there exists exactly one positive integer {style d'affichage a} tel que {style d'affichage sigma (un)=b.} En particulier, si {style d'affichage xneq y} , alors {style d'affichage sigma (X)neq sigma (y)} .

Statement of the theorem Suppose that {style d'affichage (un_{1},un_{2},un_{3},ldots )} is a sequence of real numbers, et cela {textstyle sum _{n=1}^{infime }un_{n}} is conditionally convergent. Laisser {style d'affichage M} be a real number. Then there exists a permutation {style d'affichage sigma } tel que {somme de style d'affichage _{n=1}^{infime }un_{sigma (n)}=M.} There also exists a permutation {style d'affichage sigma } tel que {somme de style d'affichage _{n=1}^{infime }un_{sigma (n)}=infty .} The sum can also be rearranged to diverge to {style d'affichage -infty } or to fail to approach any limit, fini ou infini.

Alternating harmonic series Changing the sum The alternating harmonic series is a classic example of a conditionally convergent series: {somme de style d'affichage _{n=1}^{infime }{frac {(-1)^{n+1}}{n}}} is convergent, whereas {somme de style d'affichage _{n=1}^{infime }la gauche|{frac {(-1)^{n+1}}{n}}droit|=somme _{n=1}^{infime }{frac {1}{n}}} is the ordinary harmonic series, which diverges. Although in standard presentation the alternating harmonic series converges to ln(2), its terms can be arranged to converge to any number, or even to diverge. One instance of this is as follows. Begin with the series written in the usual order, {style d'affichage 1-{frac {1}{2}}+{frac {1}{3}}-{frac {1}{4}}+cdots } and rearrange the terms: {style d'affichage 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 } where the pattern is: the first two terms are 1 and −1/2, whose sum is 1/2. The next term is −1/4. The next two terms are 1/3 and −1/6, whose sum is 1/6. The next term is −1/8. The next two terms are 1/5 and −1/10, whose sum is 1/10. En général, the sum is composed of blocks of three: {style d'affichage {frac {1}{2k-1}}-{frac {1}{2(2k-1)}}-{frac {1}{4k}},quad k=1,2,dots .} This is indeed a rearrangement of the alternating harmonic series: every odd integer occurs once positively, and the even integers occur once each, negatively (half of them as multiples of 4, the other half as twice odd integers). Depuis {style d'affichage {frac {1}{2k-1}}-{frac {1}{2(2k-1)}}={frac {1}{2(2k-1)}},} this series can in fact be written: {style d'affichage {commencer{aligné}&{frac {1}{2}}-{frac {1}{4}}+{frac {1}{6}}-{frac {1}{8}}+{frac {1}{10}}+cdots +{frac {1}{2(2k-1)}}-{frac {1}{2(2k)}}+cdots \={}&{frac {1}{2}}la gauche(1-{frac {1}{2}}+{frac {1}{3}}-cdots à droite)={frac {1}{2}}dans(2)fin{aligné}}} which is half the usual sum.

Getting an arbitrary sum An efficient way to recover and generalize the result of the previous section is to use the fact that {style d'affichage 1+{1 plus de 2}+{1 plus de 3}+cdots +{1 sur n}=gamma +ln n+o(1),} where γ is the Euler–Mascheroni constant, and where the notation o(1) denotes a quantity that depends upon the current variable (ici, the variable is n) in such a way that this quantity goes to 0 when the variable tends to infinity.

It follows that the sum of q even terms satisfies {style d'affichage {1 plus de 2}+{1 plus de 4}+{1 plus de 6}+cdots +{1 over 2q}={1 plus de 2},gamma +{1 plus de 2}ln q+o(1),} and by taking the difference, one sees that the sum of p odd terms satisfies {style d'affichage {1}+{1 plus de 3}+{1 plus de 5}+cdots +{1 over 2p-1}={1 plus de 2},gamma +{1 plus de 2}ln p+ln 2+o(1).} Suppose that two positive integers a and b are given, and that a rearrangement of the alternating harmonic series is formed by taking, in order, a positive terms from the alternating harmonic series, followed by b negative terms, and repeating this pattern at infinity (the alternating series itself corresponds to a = b = 1, the example in the preceding section corresponds to a = 1, b = 2): {style d'affichage {1}+{1 plus de 3}+cdots +{1 over 2a-1}-{1 plus de 2}-{1 plus de 4}-cdots -{1 over 2b}+{1 over 2a+1}+cdots +{1 over 4a-1}-{1 over 2b+2}-cdots } Then the partial sum of order (a+b)n of this rearranged series contains p = an positive odd terms and q = bn negative even terms, Par conséquent {style d'affichage S_{(a+b)n}={1 plus de 2}ln p+ln 2-{1 plus de 2}ln q+o(1)={1 plus de 2}dans(a/b)+ln 2+o(1).} It follows that the sum of this rearranged series is {style d'affichage {1 plus de 2}dans(a/b)+ln 2=ln left(2{sqrt {a/b}}droit).} Suppose now that, plus généralement, a rearranged series of the alternating harmonic series is organized in such a way that the ratio pn/qn between the number of positive and negative terms in the partial sum of order n tends to a positive limit r. Alors, the sum of such a rearrangement will be {style d'affichage à gauche(2{sqrt {r}}droit),} and this explains that any real number x can be obtained as sum of a rearranged series of the alternating harmonic series: it suffices to form a rearrangement for which the limit r is equal to e2x/ 4.