sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

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1 60. Ratio ad root tests Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece is stroger tha covergece, meaig that there are coverget series that do ot coverge absolutely. For example, the alteratig harmoic series a, a = ( 1) 1 /, is coverget, but ot absolutely coverget because the series of absolute values a = 1/ is othig but the harmoic series 1/ which is diverget (as a p series with p = 1). O the other had, the absolute covergece implies covergece. Theorem 38. (Covergece ad absolute covergece) Every absolutely coverget series is coverget Proof: For ay sequece {a }, the followig iequality holds 0 a + a 2 a because a is either a or a. It shows that the series b, where b = a + a, coverges by the compariso test because 2 a = 2 a coverges if a coverges absolutely. Hece, the series a = b a coverges as the differece of two coverget series. Example 91. Test the series [si() 2cos(2)]/ 3/2 for absolute covergece. Solutio: Makig use of the iequality A + B A + B ad the properties that si(x) 1 ad cos(x) 1, oe ifers a = si() 2cos(2) 3/2 si() + 2 cos(2) 3/2 3 3/2

2 60. RATIO AND ROOT TESTS 113 The series of absolute values a coverges by compariso with the coverget p series 3 3/2 (here p = 3/2 > 1). So the series i questio coverges absolutely. Defiitio 14. (Coditioal covergece) A series a is called coditioally coverget if it is coverget but ot absolutely coverget. Thus, all coverget series are separated ito two classes of coditioally ad absolutely coverget series. The key differece betwee properties of absolutely ad coditioally coverget series is studied i the ext sectio The ratio test. Theorem 39. (The ratio test) Give a series a, suppose the followig limit exists lim a +1 = c a where c 0 or c =. The If c < 1, a coverges absolutely If c > 1, a diverges If c = 1, the test gives o iformatio Proof: If c < 1, the the existece of the limit meas that for ay ε > 0 there is a iteger N such that ε < a +1 a +1 c < ε = < c + ε = q < 1 for all N a a Note that sice c is strictly less tha 1, oe ca always take ε > 0 small eough so that the umber q = c + ε < 1. I particular, put = N + k 1, where k 2. Applyig the iequality a +1 < q a cosequtively k times, a N+k < q a N+k 1 < q 2 a N+k 2 < < q k a N = a N q N q N+k This shows that (55) a < βq, β = a N q N, for all N The series a coverges by compariso with the coverget geometric series βq = β q because q < 1. So, a coverges absolutely. If c > 1, the there is a iteger N such that a +1 / a > 1 or a +1 > a 0 for all N. Hece, the ecessary coditio for a series to coverge, a 0 as does ot hold, that is, the series a diverges. If c = 1, it is sufficiet to give examples of a coverget

3 SEQUENCES AND SERIES ad diverget series for which c = 1. Cosider a p series p. Oe has a +1 p c = lim = lim a ( + 1) = lim 1 p (1 + 1/) = 1 p for ay p. But a p series coverges if p > 1 ad diverges otherwise. Example 92. Fid all values of p ad q for which the series =1 p q coverges absolutely. Solutio: Here a = p q. Oe has a +1 ( + 1) p q +1 c = lim = lim a p q (1 + 1/) p = q lim 1 = q So, for q < 1 ad ay p the series coverges absolutely by the ratio test. If q = ±1, the ratio test is icoclusive ad these cases have to be studied by differet meas. If q = 1, the a = p = 1/ p which is a p series that coverges if p > 1 or p < 1. Thus, the series coverges absolutely for all p if q < 1 ad for p < 1 if q = ±1. Note that for 1 p < 0 ad q = 1 the series coditioally coverges (i.e. it is coverget but ot absolutely coverget). I this case, it is a coverget alteratig p series ( 1) / p (see Exercise 14 i Sectio 1.6) The root test. Theorem 40. (The root test) Give a series a, suppose the followig limit exists a = c lim where c 0 or c =. The If c < 1, a coverges absolutely If c > 1, a diverges If c = 1, the test gives o iformatio Proof: If c < 1, the, as i the proof of the ratio test, the existece of the limit meas that for ay c < q < 1 there is a iteger N such a < q = a < q for all N This shows that the series a coverges by compariso with the coverget geometric series q, 0 < q < 1. So, a coverges absolutely. If c > 1, the there exists a iteger N such that a > 1 for all N ad, hece, the coditio a 0 as does ot hold. The series a diverges. If c = 1, cosider a p series: p = ( ) p 1 p = 1 by Theorem 1.6. But a p series coverges if p > 1 ad diverges if p < 1. So the root test is icoclusive.

4 60. RATIO AND ROOT TESTS 115 Example 93. Test the covergece of the series a where a = [( )/( )] Solutio: Here a = a ad the absolute covergece is equivalet to the covergece. Oe has lim So the series coverges a = lim = lim 2 + 5/ / = < Oscillatory behavior of sequeces i the root ad ratio tests. Cosider a sequece defied recursively by a 1 = 1 ad a +1 = 1 si()a 2. A attempt to test the covergece of a by the ratio test leads to the sequece c = a +1 / a = 1 si() that does 2 ot coverge as it oscillates betwee 0 ad 1/2. Similarly, the sequece used i the root test may also have a oscillatory behavior ad be ocoverget, e.g., a = ( 1 2 si()) so that c = a = 1 si(). The 2 ratio ad root tests, as stated i Theorems ad 40, assume the existece of the limit (c c). What ca be said about the covergece of a series whe this limit does ot exist? To aswer this questio, recall that i the proof of the ratio or root test the existece of lim c = c < 1 has bee used oly to establish the boudedess that c < 1 for all N which is sufficiet for the series a to coverge. But the boudedess property does ot imply the covergece! Evidetly, the boudedess coditio holds i the above examples, c = 1 si() 1 < 1 for all. Similarly, 2 2 the existece of the limit value c > 1 has oly bee used to show that a 1 or a 1 for ifiitely may to coclude that the sequece {a } caot coverge to 0 ad, hece, a diverges. If a +1 / a 1 for all N, the agai {a } caot coverge to 0 (by the proof of the ratio test). Thus, the covergece of {c } i the root or ratio test is ot really ecessary. Theorem 41. (Ratio ad root tests refied) Give a series a, put c = a +1 / a or c = a. The { c < 1 for all N = a coverges a 1 for ifiitely may a +1 = a a 1 for all N diverges for some iteger N.

5 SEQUENCES AND SERIES Wider scope of the root test. If the limit of a +1 / a exists, the so does the limit of a ad (56) lim a = lim a +1 a The coverse is ot true, that is, the existece of the limit of a does ot geerally imply the existece of the limit of a +1 / a (the latter may or may ot exist). Furthermore, if the sequece a does ot coverge, either does a +1 / a. A proof of these assertios is give i more advaced calculus courses. Thus, the ratio test have the same predictig power as the root test oly if a +1 / a coverges. I geeral, the root test (as i Theorem 41) has wider scope, meaig that wheever the ratio test shows covergece, the root test does too, ad wheever the root test is icoclusive, the ratio test is too. The subtlety to ote here is that the coverse of the latter statemet is ot geerally true, that is, the icoclusiveess of the ratio test does ot imply the icoclusiveess of the root test. The assertio ca be illustrated with the followig example. Cosider a coverget series obtaied from the sum of two geometric series i which the order of summatio is chaged: a = = =1 k=0 1 2 k =0 1 3 k = 3 2 where the sum of a geometric series has bee used (Theorem 27). Now ote that if = 2k is eve, the a 2k = (1/2) k ad a 2k 1 = (1/3) k if = 2k 1 is odd. Take the subsequece of ratios for eve = 2k, c 2k = a 2k+1 /a 2k = (2/3) k /9. It coverges to 0 as k. O the other had, the subsequece of ratios for odd = 2k 1 diverges: c 2k 1 = (3/2) k as k. So, the limit of c does ot exist ad, moreover, the ratio test (as i Theorem 41) fails miserably because c is ot eve bouded. The series coverges by the root test. Ideed, c 2k = 2k a 2k = 1/ 2 < 1 ad c 2k 1 = 2k 1 a 2k 1 = 1/ 3 < 1. Although the sequece c does ot coverge (it oscillates betwee 1/ 3 ad 1/ 2), it is bouded c < 1 for all ad, hece, the series coverges by Theorem 41. A similar example is give i Exercise 19. Thus, the ratio test is sesitive to the order of summatio, while this is ot so for the root test Whe the ratio test is icoclusive. Theorem 42. (De Morga s ratio test) Let a be a series i which a +1 / a 1 as. The series

6 coverges absolutely if 60. RATIO AND ROOT TESTS 117 ( lim a ) +1 1 = b < 1 a A proof of this theorem is left to the reader as a exercise (see Exercise 18). Cosider the asymptotic behavior of the ratio c = a +1 / a as. The theorem asserts that if c behaves as c 1 + b/ for large (i.e. eglectig terms of order 1/ 2 ad higher), the the series a coverges if b < 1. For a p series the ratio test is icoclusive (see the proof of the ratio test). However, De Mogra s test resolves the icoclusiveess. Ideed, for large c = p ( + 1) = 1 p (1 + 1/) 1 p p where the asymptotic behavior has bee foud from the liearizatio f(x) = (1 + x) p f(0) + f (0)x = 1 px for small x = 1/. So, here b = p ad the series coverges if b < 1 or p > 1. This illustrates a basic techical trick to apply De Morga s test. Suppose that there is a fuctio f(x) such that a +1 / a = f(1/). If f is differetiable at x = 0, i.e., f(x) f(0) + f (0)x as x 0, the a +1 / a = f(1/) f(0) + f (0)(1/) = 1 + f (0)/ ad the series a coverges absolutely if f (0) < 1. Note that the property f(0) = 1 follows from the icoclusiveess of the ratio test.