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.

As an example, 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. More generally, 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.

Contents 1 Definitions 2 Statement of the theorem 3 Alternating harmonic series 3.1 Changing the sum 3.2 Getting an arbitrary sum 4 Proof 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, finite or infinite 5 Generalizations 5.1 Sierpiński theorem 5.2 Steinitz's theorem 6 See also 7 References Definitions A series {textstyle sum _{n=1}^{infty }a_{n}} converges if there exists a value {displaystyle ell } such that the sequence of the partial sums {displaystyle (S_{1},S_{2},S_{3},ldots ),quad S_{n}=sum _{k=1}^{n}a_{k},} converges to {displaystyle ell } . That is, for any ε > 0, there exists an integer N such that if n ≥ N, then {displaystyle leftvert S_{n}-ell rightvert leq epsilon .} A series converges conditionally if the series {textstyle sum _{n=1}^{infty }a_{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 {displaystyle b,} there exists exactly one positive integer {displaystyle a} such that {displaystyle sigma (a)=b.} In particular, if {displaystyle xneq y} , then {displaystyle sigma (x)neq sigma (y)} .

Statement of the theorem Suppose that {displaystyle (a_{1},a_{2},a_{3},ldots )} is a sequence of real numbers, and that {textstyle sum _{n=1}^{infty }a_{n}} is conditionally convergent. Let {displaystyle M} be a real number. Then there exists a permutation {displaystyle sigma } such that {displaystyle sum _{n=1}^{infty }a_{sigma (n)}=M.} There also exists a permutation {displaystyle sigma } such that {displaystyle sum _{n=1}^{infty }a_{sigma (n)}=infty .} The sum can also be rearranged to diverge to {displaystyle -infty } or to fail to approach any limit, finite or infinite.

Alternating harmonic series Changing the sum The alternating harmonic series is a classic example of a conditionally convergent series: {displaystyle sum _{n=1}^{infty }{frac {(-1)^{n+1}}{n}}} is convergent, whereas {displaystyle sum _{n=1}^{infty }left|{frac {(-1)^{n+1}}{n}}right|=sum _{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, {displaystyle 1-{frac {1}{2}}+{frac {1}{3}}-{frac {1}{4}}+cdots } and rearrange the terms: {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 } 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 general, the sum is composed of blocks of three: {displaystyle {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). Since {displaystyle {frac {1}{2k-1}}-{frac {1}{2(2k-1)}}={frac {1}{2(2k-1)}},} this series can in fact be written: {displaystyle {begin{aligned}&{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}}left(1-{frac {1}{2}}+{frac {1}{3}}-cdots right)={frac {1}{2}}ln(2)end{aligned}}} 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 {displaystyle 1+{1 over 2}+{1 over 3}+cdots +{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 (here, 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 {displaystyle {1 over 2}+{1 over 4}+{1 over 6}+cdots +{1 over 2q}={1 over 2},gamma +{1 over 2}ln q+o(1),} and by taking the difference, one sees that the sum of p odd terms satisfies {displaystyle {1}+{1 over 3}+{1 over 5}+cdots +{1 over 2p-1}={1 over 2},gamma +{1 over 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): {displaystyle {1}+{1 over 3}+cdots +{1 over 2a-1}-{1 over 2}-{1 over 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, hence {displaystyle S_{(a+b)n}={1 over 2}ln p+ln 2-{1 over 2}ln q+o(1)={1 over 2}ln(a/b)+ln 2+o(1).} It follows that the sum of this rearranged series is {displaystyle {1 over 2}ln(a/b)+ln 2=ln left(2{sqrt {a/b}}right).} Suppose now that, more generally, 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. Then, the sum of such a rearrangement will be {displaystyle ln left(2{sqrt {r}}right),} 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.

Proof Existence of a rearrangement that sums to any positive real M For simplicity, this proof assumes first that an ≠ 0 for every n. The general case requires a simple modification, given below. Recall that a conditionally convergent series of real terms has both infinitely many negative terms and infinitely many positive terms. First, define two quantities, {displaystyle a_{n}^{+}} and {displaystyle a_{n}^{-}} by: {displaystyle a_{n}^{+}={frac {a_{n}+|a_{n}|}{2}},quad a_{n}^{-}={frac {a_{n}-|a_{n}|}{2}}.} That is, the series {textstyle sum _{n=1}^{infty }a_{n}^{+}} includes all an positive, with all negative terms replaced by zeroes, and the series {textstyle sum _{n=1}^{infty }a_{n}^{-}} includes all an negative, with all positive terms replaced by zeroes. Since {textstyle sum _{n=1}^{infty }a_{n}} is conditionally convergent, both the positive and the negative series diverge. Let M be a positive real number. Take, in order, just enough positive terms {displaystyle a_{n}^{+}} so that their sum exceeds M. Suppose we require p terms – then the following statement is true: {displaystyle sum _{n=1}^{p-1}a_{n}^{+}leq M 0 because the partial sums of {displaystyle a_{n}^{+}} tend to {displaystyle +infty } . Discarding the zero terms one may write {displaystyle sum _{n=1}^{p}a_{n}^{+}=a_{sigma (1)}+cdots +a_{sigma (m_{1})},quad a_{sigma (j)}>0, sigma (1) 0), or as image of m1 + 1 (if a1 < 0). Now repeat the process of adding just enough positive terms to exceed M, starting with n = p + 1, and then adding just enough negative terms to be less than M, starting with n = q + 1. Extend σ in an injective manner, in order to cover all terms selected so far, and observe that a2 must have been selected now or before, thus 2 belongs to the range of this extension. The process will have infinitely many such "changes of direction". One eventually obtains a rearrangement {textstyle sum {a_{sigma (n)}}} . After the first change of direction, each partial sum of {textstyle sum {a_{sigma (n)}}} differs from M by at most the absolute value {displaystyle a_{p_{j}}^{+}} or {displaystyle |a_{q_{j}}^{-}|} of the term that appeared at the latest change of direction. But {textstyle sum {a_{n}}} converges, so as n tends to infinity, each of an, {displaystyle a_{p_{j}}^{+}} and {displaystyle a_{q_{j}}^{-}} go to 0. Thus, the partial sums of {textstyle sum {a_{sigma (n)}}} tend to M, so the following is true: {displaystyle sum _{n=1}^{infty }a_{sigma (n)}=M.} The same method can be used to show convergence to M negative or zero. One can now give a formal inductive definition of the rearrangement σ, that works in general. For every integer k ≥ 0, a finite set Ak of integers and a real number Sk are defined. For every k > 0, the induction defines the value {textstyle sigma (k)} , the set Ak consists of the values {textstyle sigma (j)} for j ≤ k and Sk is the partial sum of the rearranged series. The definition is as follows: For k = 0, the induction starts with A0 empty and S0 = 0. For every k ≥ 0, there are two cases: if Sk ≤ M, then {textstyle sigma (k+1)} is the smallest integer n ≥ 1 such that n is not in Ak and an ≥ 0; if Sk > M, then {textstyle sigma (k+1)} is the smallest integer n ≥ 1 such that n is not in Ak and an < 0. In both cases one sets {displaystyle A_{k+1}=A_{k}cup {sigma (k+1)},;quad S_{k+1}=S_{k}+a_{sigma (k+1)}.} It can be proved, using the reasonings above, that σ is a permutation of the integers and that the permuted series converges to the given real number M. Existence of a rearrangement that diverges to infinity Let {textstyle sum _{i=1}^{infty }a_{i}} be a conditionally convergent series. The following is a proof that there exists a rearrangement of this series that tends to {displaystyle infty } (a similar argument can be used to show that {displaystyle -infty } can also be attained). Let {displaystyle p_{1}

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