# 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.

Come esempio, 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, eccetera.) to give a series that converges to a different sum: 1 + 1/2 − 1 + 1/3 + 1/4 − 1/2 + ⋯ = ln 2. Più generalmente, 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.

Contenuti 1 Definizioni 2 Enunciato del teorema 3 Alternating harmonic series 3.1 Changing the sum 3.2 Getting an arbitrary sum 4 Prova 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, finito o infinito 5 generalizzazioni 5.1 Sierpiński theorem 5.2 Il teorema di Steinitz 6 Guarda anche 7 References Definitions A series {textstyle sum _{n=1}^{infty }un_{n}} converges if there exists a value {stile di visualizzazione ell } such that the sequence of the partial sums {stile di visualizzazione (S_{1},S_{2},S_{3},ldot ),quad S_{n}=somma _{k=1}^{n}un_{K},} converge a {stile di visualizzazione ell } . Questo è, for any ε > 0, there exists an integer N such that if n ≥ N, poi {displaystyle leftvert S_{n}-ell rightvert leq epsilon .} A series converges conditionally if the series {textstyle sum _{n=1}^{infty }un_{n}} converges but the series {textstyle sum _{n=1}^{infty }leftvert a_{n}rightvert } diverges.

A permutation is simply a bijection from the set of positive integers to itself. This means that if {displaystyle sigma } is a permutation, then for any positive integer {stile di visualizzazione b,} there exists exactly one positive integer {stile di visualizzazione a} tale che {displaystyle sigma (un)=b.} In particolare, Se {displaystyle xneq y} , poi {displaystyle sigma (X)neq sigma (y)} .

Statement of the theorem Suppose that {stile di visualizzazione (un_{1},un_{2},un_{3},ldot )} is a sequence of real numbers, e quello {textstyle sum _{n=1}^{infty }un_{n}} is conditionally convergent. Permettere {stile di visualizzazione M} be a real number. Then there exists a permutation {displaystyle sigma } tale che {somma dello stile di visualizzazione _{n=1}^{infty }un_{sigma (n)}=M.} There also exists a permutation {displaystyle sigma } tale che {somma dello stile di visualizzazione _{n=1}^{infty }un_{sigma (n)}=infty .} The sum can also be rearranged to diverge to {displaystyle -infty } or to fail to approach any limit, finito o infinito.

Alternating harmonic series Changing the sum The alternating harmonic series is a classic example of a conditionally convergent series: {somma dello stile di visualizzazione _{n=1}^{infty }{frac {(-1)^{n+1}}{n}}} is convergent, whereas {somma dello stile di visualizzazione _{n=1}^{infty }sinistra|{frac {(-1)^{n+1}}{n}}Giusto|=somma _{n=1}^{infty }{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, {stile di visualizzazione 1-{frac {1}{2}}+{frac {1}{3}}-{frac {1}{4}}+cdot } and rearrange the terms: {stile di visualizzazione 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}}+cdot } 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. In generale, the sum is composed of blocks of three: {stile di visualizzazione {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). Da {stile di visualizzazione {frac {1}{2k-1}}-{frac {1}{2(2k-1)}}={frac {1}{2(2k-1)}},} this series can in fact be written: {stile di visualizzazione {inizio{allineato}&{frac {1}{2}}-{frac {1}{4}}+{frac {1}{6}}-{frac {1}{8}}+{frac {1}{10}}+cdot +{frac {1}{2(2k-1)}}-{frac {1}{2(2K)}}+cdots \={}&{frac {1}{2}}sinistra(1-{frac {1}{2}}+{frac {1}{3}}-cdot a destra)={frac {1}{2}}ln(2)fine{allineato}}} 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 {stile di visualizzazione 1+{1 Sopra 2}+{1 Sopra 3}+cdot +{1 over 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 (qui, 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 {stile di visualizzazione {1 Sopra 2}+{1 Sopra 4}+{1 Sopra 6}+cdot +{1 over 2q}={1 Sopra 2},gamma +{1 Sopra 2}ln q+o(1),} and by taking the difference, one sees that the sum of p odd terms satisfies {stile di visualizzazione {1}+{1 Sopra 3}+{1 Sopra 5}+cdot +{1 over 2p-1}={1 Sopra 2},gamma +{1 Sopra 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): {stile di visualizzazione {1}+{1 Sopra 3}+cdot +{1 over 2a-1}-{1 Sopra 2}-{1 Sopra 4}-cdot -{1 over 2b}+{1 over 2a+1}+cdot +{1 over 4a-1}-{1 over 2b+2}-cdot } 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, quindi {stile di visualizzazione S_{(a+b)n}={1 Sopra 2}ln p+ln 2-{1 Sopra 2}ln q+o(1)={1 Sopra 2}ln(a/b)+ln 2+o(1).} It follows that the sum of this rearranged series is {stile di visualizzazione {1 Sopra 2}ln(a/b)+ln 2=ln left(2{mq {a/b}}Giusto).} Suppose now that, più generalmente, 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. Quindi, the sum of such a rearrangement will be {stile di visualizzazione ln sinistra(2{mq {r}}Giusto),} 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.