11 INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES 11.4 The Compariso Tests I this sectio, we will lear: How to fid the value of a series by comparig it with a kow series.
COMPARISON TESTS I the compariso tests, the idea is to compare a give series with oe that is kow to be coverget or diverget.
COMPARISON TESTS Series 1 Cosider the series 1 2 + 1 = 1 = This remids us of the series. 1 1/2 The latter is a geometric series with a = ½ ad r = ½ ad is therefore coverget.
COMPARISON TESTS As the series is similar to a coverget series, we have the feelig that it too must be coverget. Ideed, it is.
COMPARISON TESTS The iequality 1 1 < 2 + 1 2 shows that our give series has smaller terms tha those of the geometric series. Hece, all its partial sums are also smaller tha 1 (the sum of the geometric series).
COMPARISON TESTS Thus, Its partial sums form a bouded icreasig sequece, which is coverget. It also follows that the sum of the series is less tha the sum of the geometric series: = 1 1 2 + 1 < 1
COMPARISON TESTS Similar reasoig ca be used to prove the followig test which applies oly to series whose terms are positive.
COMPARISON TESTS The first part says that, if we have a series whose terms are smaller tha those of a kow coverget series, the our series is also coverget.
COMPARISON TESTS The secod part says that, if we start with a series whose terms are larger tha those of a kow diverget series, the it too is diverget.
THE COMPARISON TEST Suppose that Σ a ad Σ b are series with positive terms. i. If Σ b is coverget ad a b for all, the Σ a is also coverget. ii. If Σ b is diverget ad a b for all, the Σ a is also diverget.
THE COMPARISON TEST PROOF Part i Let s = a t = b t = b i i i= 1 i= 1 = 1 Sice both series have positive terms, the sequeces {s } ad {t } are icreasig (s +1 = s + a +1 s ). Also, t t; so t t for all.
THE COMPARISON TEST PROOF Part i Sice a i b i, we have s t. Hece, s t for all. This meas that {s } is icreasig ad bouded above. So, it coverges by the Mootoic Sequece Theorem. Thus, Σ a coverges.
THE COMPARISON TEST PROOF Part ii If Σ b is diverget, the t (sice {t } is icreasig). However, a i b i ; so s t. Thus, s ; so Σ a diverges.
SEQUENCE VS. SERIES It is importat to keep i mid the distictio betwee a sequece ad a series. A sequece is a list of umbers. A series is a sum.
SEQUENCE VS. SERIES With every series Σ a, there are associated two sequeces: 1. The sequece {a } of terms 2. The sequece {s } of partial sums
COMPARISON TEST I usig the Compariso Test, we must, of course, have some kow series Σ b for the purpose of compariso.
COMPARISON TEST Most of the time, we use oe of these: A p-series [Σ 1/ p coverges if p > 1 ad diverges if p 1] A geometric series [Σ ar 1 coverges if r < 1 ad diverges if r 1]
COMPARISON TEST Example 1 Determie whether the give series coverges or diverges: 5 2 + + = 1 2 4 3
COMPARISON TEST Example 1 For large, the domiat term i the deomiator is 2 2. So, we compare the give series with the series Σ 5/(2 2 ).
COMPARISON TEST Example 1 Observe that 5 5 < 2 + 4+ 3 2 2 2 sice the left side has a bigger deomiator. I the otatio of the Compariso Test, a is the left side ad b is the right side.
COMPARISON TEST Example 1 We kow that 5 5 1 = 2 2 2 2 = 1 = 1 is coverget because it s a costat times a p-series with p = 2 > 1.
COMPARISON TEST Example 1 Therefore, is coverget by part i of the Compariso Test. 5 2 + + = 1 2 4 3
NOTE 1 Although the coditio a b or a b i the Compariso Test is give for all, we eed verify oly that it holds for N, where N is some fixed iteger. This is because the covergece of a series is ot affected by a fiite umber of terms. This is illustrated i the ext example.
COMPARISON TEST Example 2 Test the give series for covergece or divergece: = 1 l
COMPARISON TEST Example 2 This series was tested (usig the Itegral Test) i Example 4 i Sectio 11.3 However, it is also possible to test it by comparig it with the harmoic series.
COMPARISON TEST Example 2 Observe that l > 1 for 3. So, l 1 > 3 We kow that Σ 1/ is diverget (p-series with p = 1). Thus, the series is diverget by the Compariso Test.
NOTE 2 The terms of the series beig tested must be smaller tha those of a coverget series or larger tha those of a diverget series. If the terms are larger tha the terms of a coverget series or smaller tha those of a diverget series, the Compariso Test does t apply.
NOTE 2 For istace, cosider 1 2 1 = 1 The iequality 1 1 > 2 1 2 is useless as far as the Compariso Test is cocered. This is because Σ b = Σ (½) is coverget ad a > b.
NOTE 2 Noetheless, we have the feelig that Σ1/(2-1) ought to be coverget because it is very similar to the coverget geometric series Σ (½). I such cases, the followig test ca be used.
LIMIT COMPARISON TEST Suppose that Σ a ad Σ b are series with positive terms. If a lim b = c where c is a fiite umber ad c > 0, either both series coverge or both diverge.
LIMIT COMPARISON TEST PROOF Let m ad M be positive umbers such that m < c < M. Sice a /b is close to c for large, there is a iteger N such that ad so a m< < M whe > N b mb < a < Mb whe > N
LIMIT COMPARISON TEST PROOF If Σ b coverges, so does Σ Mb. Thus, Σ a coverges by part i of the Compariso Test.
LIMIT COMPARISON TEST PROOF If Σ b diverges, so does Σ mb. Thus, Σ a diverges by part ii of the Compariso Test.
COMPARISON TESTS Example 3 Test the give series for covergece or divergece: 1 2 1 = 1
COMPARISON TESTS Example 3 We use the Limit Compariso Test with: a 1 1 = b = 2 1 2
COMPARISON TESTS Example 3 We obtai: ( ) a 1/ 2 1 lim = lim b 1/2 = lim 2 2 1 1 = lim = 1 > 0 1 1/2
COMPARISON TESTS Example 3 This limit exists ad Σ 1/2 is a coverget geometric series. Thus, the give series coverges by the Limit Compariso Test.
COMPARISON TESTS Example 4 Determie whether the give series coverges or diverges: = 1 2 2 + 3 5 + 5
COMPARISON TESTS Example 4 The domiat part of the umerator is 2 2. The domiat part of the deomiator is 5/2. This suggests takig: a 2 2 2 + 3 2 2 b 5 5/2 1/2 = = = 5 +
COMPARISON TESTS Example 4 We obtai: 2 1/2 a 2 + 3 lim = lim. b 5 + 5 2 = 5/2 3/2 2 + 3 lim 2 5 + 3 2 + 2 0 lim + = = = 1 5 2 0+ 1 2 + 1 5 5
COMPARISON TESTS Example 4 Σ b = 2 Σ 1/ 1/2 is diverget (p-series with p = ½ < 1). Thus, the give series diverges by the Limit Compariso Test.
COMPARISON TESTS Notice that, i testig may series, we fid a suitable compariso series Σ b by keepig oly the highest powers i the umerator ad deomiator.
ESTIMATING SUMS We have used the Compariso Test to show that a series Σ a coverges by compariso with a series Σ b. It follows that we may be able to estimate the sum Σ a by comparig remaiders.
ESTIMATING SUMS As i Sectio 11.3, we cosider the remaider R = s s = a +1 + a +2 + For the compariso series Σ b, we cosider the correspodig remaider T = t - t = b +1 + b +2 +
ESTIMATING SUMS As a b for all, we have R T. If Σ b is a p-series, we ca estimate its remaider T as i Sectio 11.3 If Σ b is a geometric series, the T is the sum of a geometric series ad we ca sum it exactly (Exercises 35 ad 36). I either case, we kow that R is smaller tha T
ESTIMATING SUMS Example 5 Use the sum of the first 100 terms to approximate the sum of the series Σ 1/( 3 +1). Estimate the error ivolved i this approximatio.
ESTIMATING SUMS Example 5 Sice 1 1 < + 1 3 3 the give series is coverget by the Compariso Test.
ESTIMATING SUMS Example 5 The remaider T for the compariso series Σ 1/ 3 was estimated i Example 5 i Sectio 11.3 (usig the Remaider Estimate for the Itegral Test). We foud that: T 1 1 dx = x 2 3 2
ESTIMATING SUMS Example 5 Therefore, the remaider for the give series satisfies: R T 1/2 2
ESTIMATING SUMS Example 5 With = 100, we have: 1 R = 100 2(100) 2 0.00005 With a programmable calculator or a computer, we fid that 100 1 1 + 1 + 1 3 3 = 1 = 1 with error less tha 0.00005 0.6864538