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Dirichlet’S Test 4. CONDITIONALLY CONVERGENT SERIES 23 4. Conditionally convergent series Here basic tests capable of detecting conditional convergence are studied. 4.1. Dirichlet’s test. Theorem 4.1. (Dirichlet’s test) n Let a complex sequence {An}, An = k=1 ak, be bounded, and the real sequence {bn} is decreasing monotonically, b1 ≥ b2 ≥ b3 ≥··· , so that P limn→∞ bn = 0. Then the series anbn converges. A proof is based on the identity:P m m m m−1 anbn = (An − An−1)bn = Anbn − Anbn+1 n=k n=k n=k n=k−1 X X X X m−1 = An(bn − bn+1)+ Akbk − Am−1bm n=k X Since {An} is bounded, there is a number M such that |An|≤ M for all n. Therefore it follows from the above identity and non-negativity and monotonicity of bn that m m−1 anbn ≤ An(bn − bn+1) + |Akbk| + |Am−1bm| n=k n=k X X m−1 ≤ M (bn − bn+1)+ bk + bm n=k ! X ≤ 2Mbk By the hypothesis, bk → 0 as k → ∞. Therefore the right side of the above inequality can be made arbitrary small for all sufficiently large k and m ≥ k. By the Cauchy criterion the series in question converges. This completes the proof. Example. By the root test, the series ∞ zn n n=1 X converges absolutely in the disk |z| < 1 in the complex plane. Let us investigate the convergence on the boundary of the disk. Put z = eiθ 24 1. THE THEORY OF CONVERGENCE ikθ so that ak = e and bn = 1/n → 0 monotonically as n →∞. Then 1 − einθ A = eiθ + e2iθ + ··· + einθ = eiθ n 1 − eiθ 1 − einθ 1 − cos(nθ) 1 1 |A | = = ≤ ≤ n 1 − eiθ 1 − cos θ 1 − cos(θ) | sin(θ/2)| r s Thus, the sequence {An} is bounded for all 0 <θ< 2π, and the series in question converges at all points of the closed disk |z| ≤ 1 except z = 1. The convergence on the boundary is not absolute. 4.2. Abel’s and Leibnitz tests. Dirichlet’s test provides a general test that allows one to detect conditional convergence. There are two con- sequences known as Abel’s and Leibnitz tests. They are also used to detect conditional convergence. Corollary 4.1. (Leibnitz criterion for alternating series) Suppose that |cn| is decreasing monotonically, |cn+1|≤|cn|, so that cn → 0 as n → ∞, and c2m−1 ≥ 0, c2m ≤ 0. Then the series cn converges. P n In Dirichlet’s test, put an = (−1) and bn = |cn|. Then |An| ≤ 1 and bn → 0 monotonically, Leibnitz criterion follows. Alternating harmonic series. By Leibnitz criterion, the alternating har- monic series ∞ (−1)n+1 n n=1 X converges, but not absolutely. Theorem 4.2. (Abel’s test) Suppose a complex series an converges and a real sequence {cn} is monotonic and bounded. Then the series anbn converges. P n A proof of Abel’s test is also based onP Dirichlet’s test. A bounded monotonic sequence (either increasing or decreasing) converges. Sup- pose cn ≤ cn+1 for all n. Then there exists c such that cn → c as n → ∞ monotonically. Therefore the sequence bn = c − cn ≥ 0 is non-negative and converges to 0. The convergence of an means that the sequence of partial sums An = a1 + a2 + ··· + an converges. But any convergent sequence is bounded. Therefore by DirichletP ’s test, the series n an(c − cn) converges. From the convergence of an and P P 4. CONDITIONALLY CONVERGENT SERIES 25 n an(c − cn) and the limit laws it follows that the series n ancn converges and P P ancn = c an − an(c − cn) n n X X X Similarly, if cn ≥ cn+1 for all n. Then put bn = cn − c, where c is the limit of a bounded monotonically decreasing sequence cn, in Dirichlet’s test, and the convergence of ancn follows from the limit laws. 4.3. Rearrangements. ConsiderP a one-to-one map of the set of all pos- itive integer onto itself so that {1, 2, 3,...} is mapped to {k1, k2, k3,...}. The series 0 0 0 0 akn = ak1 + ak2 + ak3 + ··· = a1 + a2 + a3 + ··· = an n n X X is called a rearrangement of the series n an = a1 + a2 + a3 + ··· . Theorem 4.3. (Rearrangement andP absolute convergence) A rearrangement of any absolutely convergent series converges abso- lutely to the same sum: 0 an = an . n n X X The convergence of a rearrangement of an absolutely convergent series follows from the Cauchy criterion. Indeed, given ε > 0, there is an integer N such that m |an| <ε, ∀m ≥ k ≥ N n=k 0 X Let Sn and Sn be partial sums of the series and its rearrangement, respectively. For any rearrangement there is an integer M ≥ N such that the first M terms of the rearrangement contains the terms a1, 0 a2,..., aN . Then the latter terms are canceled in the difference Sn − Sn 0 if n ≥ M ≥ N so that |Sn − Sn| does not exceed the sum of |aj| where j takes integer values no less than N. By the above inequality 0 0 |Sn − Sn| <ε for all n ≥ M. Therefore Sn converges to the same limit as Sn. In contrast, a rearrangement of a conditionally convergent series can converge to a different sum and even fail to converge. For example, consider the alternating harmonic series ∞ (−1)n+1 1 1 1 1 1 = 1 − + − + − + ··· = S n 2 3 4 5 6 n=1 X 26 1. THE THEORY OF CONVERGENCE and let Sn be the sequence of its partial sums so that Sn → S as n →∞. Evidently, 1 1 5 S<S3 = 1 − + = . 2 3 6 Consider the rearrangement of the series in which two positive terms are followed by one negative: ∞ 1 1 1 1 1 a0 =1+ − + + − + ··· n 3 2 5 7 6 n=1 0 X and let Sn be the sequence of its partial sums. It follows that 1 1 1 0 0 0 + − > 0 ⇒ S3 = S <S <S < ··· 4k − 3 4k − 1 2k 3 6 9 0 Therefore there are infinitely many terms of the sequence Sn greater than 5/6. This implies that 0 0 5 lim sup Sn >S3 = >S n→∞ 6 0 and, hence, the sequence Sn cannot converge to S. Theorem 4.4. (Properties of conditionally convergent series) + Let an be a real series that converges but not absolutely. If Sp is the − sum of the first p positive terms of the series and Sq is the sum of its firstPq negative terms, then + − lim Sp = ∞ , lim Sq = −∞ . p→∞ q→∞ For example, ∞ (−1)n+1 1 1 1 1 1 = 1 − + − + − + ··· n 2 3 4 5 6 n=1 X ∞ + 1 1 1 lim Sp = lim 1+ + ··· + = = ∞ p→∞ p→∞ 3 2p − 1 2p − 1 p=1 X∞ − 1 1 1 1 1 lim Sq = lim − − −···− = − = −∞ q→∞ q→∞ 2 4 2q 2 q q=1 X In general, the stated property follows from the following observa- tion. Consider two series n bn and n cn where 1 1 b = (a +P|a |) ,P c = (a −|a |) n 2 n n n 2 n n 4. CONDITIONALLY CONVERGENT SERIES 27 Clearly, bn = an if an > 0 and bn = 0 otherwise. Similarly, cn = an if an < 0 and cn = 0 otherwise. Let An, Bn, Cn, and Sn be the partial sums of an, bn, cn, and |an|, respectively. Then P P P1 P 1 B = (A + S ) , C = (A − S ) n 2 n n n 2 n n Since an converges but not absolutely, An converges to some number A but lim Sn = ∞. Therefore P lim Bn = ∞ , lim Cn = −∞ n→∞ n→∞ + − Evidently, Bn = Sp for any n and some p ≤ n, and Cn = Sq for any ± n and some q = n − p ≤ n (if S0 = 0 by definition). It is concluded that a real conditionally convergent series necessarily contains infinitely many negative terms and infinitely many positive terms (otherwise the corresponding partial sums Bn and Cn were finite). Therefore the limit n → ∞ implies that p → ∞ and q → ∞ and the conclusion of the theorem follows. Remark. As the series made of positive and negative terms of a condi- tionally convergent series diverge to infinity, the sum of the series looks like an indeterminate form “∞−∞” whose value depends on how the infinities in this form are “regularized”, or, more precisely, how negative and positive terms are ordered when computing the sequence of partial sums. This characteristic property of conditionally convergent series is quantified by the following theorem. Theorem 4.5. (Riemann) Let n an converges but not absolutely. For any two elements of the extended real number system −∞ ≤ α ≤ β ≤ ∞, there exists a re- P 0 0 arrangement n an with partial sums Sn such that P 0 0 lim inf Sn = α, lim sup Sn = β n→∞ n→∞ The Riemann theorem about rearrangements shows that a rearrange- ment of a conditionally convergent series may be constructed to con- verge to any desired number α = β, or diverge to ±∞, or to have a sequence of partial sums that does not converge and oscillates between any two numbers (α<β) (or has unbounded oscillations, α = −∞, β = ∞, or both). 28 1. THE THEORY OF CONVERGENCE 4.4. A rearrangement converging to any designated number. To illustrate the Riemann theorem3, let us construct a rearrangement of a condition- ally convergent series that converges to a given number β > 0 (the case 0 β ≤ 0 can be considered along a similar line of arguments). Let Sn denote the sequence of partial sums of the sought-after rearrangement.
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