LECTURE SERIES WITH NONNEGATIVE TERMS (II). SERIES WITH ARBITRARY TERMS

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1 LECTURE 4 SERIES WITH NONNEGATIVE TERMS II). SERIES WITH ARBITRARY TERMS Series with oegative terms II) Theorem 4.1 Kummer s Test) Let x be a series with positive terms. 1 If c ) N i 0, + ), r > 0 ad 0 N, such that c the the series x is coverget. x c +1 r, N, 0, If c ) N i 0, + ) ad 0 N, such that 1 the the series x is diverget. 1 c + ad c x c +1 0, N, 0, Proof. 1 Sice c x c +1 r, 0, it follows that for ay 0 + 1, 1 k 0 c k x k c k+1 x k+1 ) r 1 k 0 x k+1. Deotig s. x x, we deduce that c 0 x 0 c x r s s 0 ) ad therefore s s r c 0 x 0 c x ) s 0 + c 0 x 0. r Hece, the sequece of partial sums s ) is bouded, which meas that the series x is coverget by Lemma 3.13) 1

2 Sice c x c +1, 0, we have c 0 x 0 c x, 0. This yields 1 1 x, 0. c c 0 x 0 Sice the series 1 is diverget, we coclude that the series x is diverget as well, accordig c to the Compariso Test Theorem 3.18) Theorem 4. Raabe-Duhamel s Test) Let x be a series with positive terms. 1 x If q > 1, 0 N such that 1 q, 0, the x is coverget. x If 0 N such that 1 1, 0, the x is diverget. 3 If the followig limit exists x R. lim 1 R, the we have a) If R > 1, x is coverget. b) If R < 1, x is diverget. Proof. Follows from Kummer s Test Theorem 4.1)for c. for all N. Example 4.3 For ay a > 0 cosider the series! aa + 1)... a + ). This series is coverget for a > 1 ad diverget for a 0, 1].! Ideed, deotig x., we have aa + 1)... a + ) x Note that D. lim x x R. lim 1 + 1)! aa + 1)... a + ) aa + 1)... a + + 1)! 1, hece the Ratio Test is icoclusive. However, a lim 1 lim a a + 1 a, which allows us to coclude, by Raabe-Duhamel s Test, that the give series is coverget if a > 1 ad diverget if a 0, 1). Fially, for a 1 the give series becomes 1, which is diverget. + 1

3 Theorem 4.4 Bertrad s Test) Let x be a series with positive terms. If the followig limits exists [ ] x B. lim l ) 1 1 R, the we have a) If B > 1, the x is coverget. b) If B < 1, the x is diverget. Proof. Follows from Kummer s Test Theorem 4.1)for c. l, N,. [ ] 1)!! is diverget. Example 4.5 The series )!! [ ] [ ] 1)!! ) Ideed, deotig x. we have )!! 4... ) + 1 for all N. It is a simple exercise to check that x + D. lim R. lim 1; x x 1 ) lim [ ) + 1] lim , hece both the Ratio Test ad the Raabe-Duhamel s Test are icoclusive. O the other had, we have [ ] x 4 B. + 3 lim l ) 1 1 lim l ) < 1. We coclude by Bertrad s Test that the give series is diverget. Series with arbitrary terms Theorem 4.6 Abel-Dirichlet s Test) Let x be a series of real umbers. Assume that there exist two sequeces of real umbers, a ) N ad b ) N, satisfyig the followig three coditios: i) x a b, N. ii) M > 0 s.t. M A. a 1 + +a M, N, i.e., the sequece A ) N is bouded. iii) The sequece b ) N is mootoe ad coverget to 0. The the series x is coverget. Proof. Without loss of geerality we ca assume i iii) that b ) is decreasig. We will prove that x coverges by usig Cauchy s Criterio Theorem 3.11)To this aim, cosider a arbitrary ε > 0. O the oe had, by i), ii) ad the assumptio that b ) is decreasig, we have + x x +p a +1 b +1 + a + b a +p b +p A +1 A ) b +1 + A + A +1 ) b A +p A +p 1 ) b +p A b +1 + A +1 b +1 b + ) + + A +p 1 b +p 1 b +p ) + A +p b +p A b +1 + A +1 b +1 b A +p 1 b +p 1 b +p + A +p b +p A b +1 + A +1 b +1 b + ) + + A +p 1 b +p 1 b +p ) + A +p b +p M [b +1 + b +1 b + ) + b + b +3 ) b +p 1 b +p ) + b +p ] Mb +1,, p N. 3

4 O the other had, sice lim b 0 by iii), there exists ε N such that b < ε M, N, ε. We coclude that + x x +p < ε, N, ε, p N. Defiitio 4.7 A series x is called alteratig if either x 1 0, x 0, x 3 0,... i.e., x 1) +1 x for all N) or x 1 0, x 0, x 3 0,... i.e., x 1) x for all N). Theorem 4.8 Leibiz s Criterio for Alteratig Series) Cosider a alteratig series x. If the sequece x ) N is decreasig, the the followig assertios are equivalet: 1 The series x is coverget. The sequece x ) N coverges to 0. Proof. Assume that x 1) +1 x for all N. The the coclusio follows by Abel-Dirichlet s Test for a. 1) +1 ad b. x. Defiitio 4.9 A series of real umbers is coverget. x is called absolutely coverget if the series Theorem 4.10 If a series of real umbers x is absolutely coverget, the it is also coverget. x Proof. Let ε > 0. Sice x is coverget, there exists i view of the Cauchy s Criterio Theorem 3.11) a umber ε N such that + + x +p < ε, N, ε, p N. Notig that + + x +p + + x +p + + x +p, we ifer + + x +p < ε, N, ε, p N. By Cauchy s Criterio Theorem 3.11)we coclude that x is coverget. Defiitio 4.11 A series of real umbers x is called semi-coverget or coditioally coverget) if it is coverget but ot absolutely coverget. Remark 4.1 A series x with oegative terms is absolutely coverget if ad oly if it is coverget. 4

5 Example 4.13 The alteratig geeralized harmoic series) Let p R. The so-called alteratig geeralized harmoic series 1) +1 is diverget for p, 0], semi-coverget for p 0, 1] ad absolutely coverget for p 1, ). I particular, for p 1 we get the alteratig harmoic series, whose sum is Example 4.14 The series 1) +1 p 1) +1 l. is absolutely coverget. Example 4.15 The series 1) +1 si 1 is semi-coverget. Example 4.16 The series 1) is diverget. Example 4.17 The series cosπ) is diverget. Theorem 4.18 Cauchy) If a series x is absolutely coverget, the for ay bijectio permutatio) σ : N N the series x σ) is absolutely coverget ad its sum coicides with the sum of the iitial series, i.e., x σ) x. 1 1 Theorem 4.19 Riema) If a series x is semi-coverget, the for every s R there exists a bijectio σ : N N such that x σ) s. 1 Example 4.0 Cosider the alteratig harmoic series see Example 4.13), whose sum is ) l. If we permute its terms by alteratig p. positive terms followed by q. 3 egative terms we obtai p l. q Ideed, cosider the Euler s costat γ. lim γ see Exercise of Semiar 3), where γ l, for all N. 5

6 Deote by s ) N the sequece of partial sums of the permuted series. The, for ay k N, we have s 5k ) ) k k 1 1 6k 4 1 6k 1 ) 6k hece k l 3k 3k γ 4k 1 γ k 1 γ 3k + l 4k 6k, O the other had, we also have ) l 4k ) l k k ) + l 4k 1 l k 1 l 3k lim s 5k γ 1 k γ 1 γ + l 4 l. 6 3 s 5k+1 s 5k + 1 4k + 1, s 5k+ s 5k + 1 4k k + 3, s 5k+3 s 5k + 1 4k k k +, s 5k+4 s 5k + 1 4k k k + 1 6k + 4, which show that lim s 5k lim s 5k+1 lim s 5k+ lim s 5k+3 lim s 5k+4. k k k k k We coclude that lim s l. 3 6