61. REARRANGEMENTS 119 61. Rearrangements Here the difference between conditionally and absolutely convergent series is further refined through the concept of rearrangement. Definition 15. (Rearrangement) Let k , n = 1, 2,..., be an integer-valued positive sequence in which { n} every positive integer appears only once (i.e. kn = kn0 if and only if n = n0). Given a series an, put an0 = akn . The series an0 is called a rearrangement of an. P P For a finite sum, aP rearrangement of its terms does not change the value of the sum. This is not generally so for convergent series. Consider an alternating p series − n 1 ∞ ∞ ( 1) − 1 1 1 1 1 (57) a = − = 1 + + + n n − 2 3 − 4 5 − 6 ··· n=1 n=1 X X The series is convergent but not absolutely convergent (its sum is s = ln 2; see Problem 18 in Section 1.6). One of its rearrangements reads ∞ 1 1 1 1 1 1 1 1 (58) a0 =1+ + + + + + n 3 − 2 5 7 − 4 9 11 − 6 ··· n=1 X in which two positive terms are always followed by one negative. Let sn and sn0 be partial sums of (57) and (58), respectively. Put hn = 1 + 1/2 + + 1/n (a partial sum of the harmonic series). Then s = h ···h . Furthermore, 2n 2n − n 1 1 1 1 1 1 1 s0 = 1+ + + + + 3n 3 5 7 ··· 4n 1 − 2 − 4 −···− 2n 1 1 1 −1 1 1 = h h = h h h 4n − 2 − 4 −···− 4n − 2 n 4n − 2 2n − 2 n 1 1 = (h h )+ (h h )= s + s 4n − 2n 2 2n − n 4n 2 2n Taking the limit n in this equality, one finds s0 = s + s/2 = 3s/2 →∞ where s and s0 are the sums of (57) and (58), respectively. Thus, a rearrangement of the series has changed its sum! This fact is not specific for the example considered but inherent to all conditionally convergent series. Terms of a conditionally convergent series occur with different signs (positive and negative). By regrouping positive and negative terms, it will be proved that the sum of a conditionally convergent series can be made any number or . The analysis begins with studying the properties of sums of positive±∞ and negative terms of a conditionally convergent series. 120 9. SEQUENCES AND SERIES + Given a number x, put x± = (x x )/2). The number x = x if + ±| | x > 0 and x = 0 otherwise. Similarly, x− = x if x < 0 and x− = 0 otherwise. Lemma . + 1 Given a series an, consider two series an and an− where an± = (an an )/2 (the series of positive and negative terms). Then ±| | P P P + (i) if an converges absolutely, then an and an− converge. + (ii) if an is conditionally convergent, then a and a− diverge. P P Pn n Proof: Let a = s < and a = t where t < if a P n ∞ | n| P P ∞ n converges absolutely and t = if it is conditionally convergent. Let s± P ∞ P P n be partial sums of an±, sn be partial sums of an and tn be partial sums of a . Since s s and t t as n , one infers that | n| n → n → →∞ + P + P + an an− = an sn sn− = sn s s− = s + −P = + − = + − a + a− = an ⇒ s + s− = tn ⇒ s + s− = t n n | | n n where s± are the limits of s±. If a converges absolutely, then t< n n ∞ and, hence, s± = (t s)/2, that is, both the series a± converge. If ± P n an is conditionally convergent, then t = , and both the sequences ∞ P s± diverge. Pn Theorem 43. (Riemann’s rearrangement theorem) Let an be a series which converges, but not absolutely. Then for any c which is a real number or there exists a rearrangement a0 P ±∞ n whose sequence of partial sums s0 converges to c. { n} P Proof: Let p1, p2,.... denote non-negative terms of an in the order in which they occur, and let q1,q2,... denote negative terms of P an in the order in which they occur. In notations of Lemma 1, the + series pn and a as well as qn and a− may only differ by P n n zero terms (if some an = 0). So, the series pn and qn diverge. ConsiderP the followingP rearrangement.P GivenP a number c, take first P P k1 terms pn so that the number c lies between the partial sums sk1 = p1 + p2 + + pk1 and sk1 1, that is, k1 is defined by the condition s p <c<s··· or c s− < p . If c< 0, then skip this first step. k1 − k1 k1 | − k1 | k1 Next, take first m1 terms qn where m1 is the smallest integer such that sk1+m1 = sk1 + q1 + + qm1 <c, that is, m1 is defined by the conition s + q >c>s··· or c s < q . This can always be k1+m1 m1 k1+m1 | − k1+m1 | | m1 | done because of partial sums of pn can be larger than any number, while partial sums of qn can be smaller than any number thanks to the divergence of these series. So,P P s s s , 1 n k where c s < p 1 ≤ n ≤ k1 ≤ ≤ 1 | − k1 | k1 s s s , k n k + m where c s < q k1 ≥ n ≥ k1+m1 1 ≤ ≤ 1 1 | − k1+m1 | | k1 | 61. REARRANGEMENTS 121 Next, take k2 next terms of pn where k2 is the smallest integer such that sk1+m1+k2 > c, and take m2 next term of qn where m2 is the smallest integer for which sk1+m1+k1+m2 < c and so on. At the nth step of the procedure, let n1 be the integer for which the last term in sn1 is pkn and let n2 be the integer for which the last term in sn2 is qmn , i.e. n2 = n1 + mn. The partial sums of the constructed rearrangement oscillate about c reaching local minima sn1 and local maxima sn1 : s s s , n n n n1 ≤ n ≤ n2 1 ≤ ≤ 2 (59) c s < p , c s < q | − n1 | kn | − n2 | | mn | By convergence of the series a , a 0 as n 0. Hence, p and q n n → → n n also converge to zero and so do the subsequences pkn 0 and qmn 0 Thus, all local maxima and minimaP of the sequence of→ partial sum →s { n} converge to c by (59) which shows that sn c. Finally, if c = , one can take any divergent sequence c → (or ) and construct±∞ n → ∞ −∞ a rearrangement such that sk1 overshoots c1, sk1+m1 undershoots c1, then sk1+m1+k2 overshoots c2 and sk1+m1+k2+m2 undershoots c2 and so on. Obviously, this sequence of partial sum diverges. Absolutely convergent series have a drastically different property. Theorem 44. (Rearrangement and absolute convergence) If a series an converges absolutely, then every rearrangement of an converges, and they all converge to the same sum P P Proof: Let t = a + a + + a is a partial sum of the n | 1| | 2| ··· | n| series of absolute values. The sequence tn converges to a number t by the hypothesis, that is, for any ε > {0 there} is an integer N such that t t <ε for all n > N. Therefore | − n| n a = t t = t t + t t | k| | n − N+1| | n − − N+1| k=XN+1 t t + t t < 2ε ≤ | n − | | − N+1| So, by taking N large enough the sum of any number of terms ak , k > N, can be made smaller than any preassigned positive number.| | Let sn and sn0 be partial sums of an and its rearrangement an0 . One can take n > N large enough so that s0 contains a1, a2,...,aN (i.e. the P n P integers 1, 2,...,N are in the set of integers k1, k2,...,kn in notations of Definition 15). Then the difference sn0 sn contains only terms a with k > N (the terms a , a ,..., |a −are cancelled).| Therefore | k| 1 2 N s0 s < 2ε for all n > N. If s0 s0 and s s, then s0 s < 2ε | n − n| n → n → | − | which shows that s0 = s because ε> 0 is arbitrary. 122 9. SEQUENCES AND SERIES Thus, an absolutely convergent series is much like a finite sum. The sum does not depend on the order in which the summation is carried out. In contrast, the sum of a conditionally convergent series depends on the summation order. This is the characteristic difference between these two classes of convergent series. 61.1. Strategy for testing series. It would not be wise to apply test for convergence in a specific order to find one that finally works. Instead, a proper strategy, as with integration, is to classify the series according to its form. One should also keep in mind that a conclusion about the convergence of a series can be reached in different ways. 1. Special series. A series an coincide with (or is a combination of or is equivalent to) special series such as a p series, alternating p series, geometric series, telescopicP series, etc.− Their convergence properties− are known. 2. Series similar to special ones. A series an has a form that is similar to one of the special series, then one of the comparison P tests should be considered. For example, if an is a rational or alge- braic function (contains roots of polynomials), then the series should be compared with a p series. − 3. Necessary condition for convergence. The condition an 0 as n is always easier to check than to investigate the series→ → ∞ an for convergence.