INFINITE SEQUENCES AND SERIES
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1 11 INFINITE SEQUENCES AND SERIES
2 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.
3 COMPARISON TESTS I the compariso tests, the idea is to compare a give series with oe that is kow to be coverget or diverget.
4 COMPARISON TESTS Series 1 Cosider the series = 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.
5 COMPARISON TESTS As the series is similar to a coverget series, we have the feelig that it too must be coverget. Ideed, it is.
6 COMPARISON TESTS The iequality 1 1 < 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).
7 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
8 COMPARISON TESTS Similar reasoig ca be used to prove the followig test which applies oly to series whose terms are positive.
9 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.
10 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.
11 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.
12 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.
13 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.
14 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.
15 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.
16 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
17 COMPARISON TEST I usig the Compariso Test, we must, of course, have some kow series Σ b for the purpose of compariso.
18 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]
19 COMPARISON TEST Example 1 Determie whether the give series coverges or diverges: =
20 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 ).
21 COMPARISON TEST Example 1 Observe that 5 5 < 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.
22 COMPARISON TEST Example 1 We kow that = = 1 = 1 is coverget because it s a costat times a p-series with p = 2 > 1.
23 COMPARISON TEST Example 1 Therefore, is coverget by part i of the Compariso Test =
24 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.
25 COMPARISON TEST Example 2 Test the give series for covergece or divergece: = 1 l
26 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.
27 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.
28 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.
29 NOTE 2 For istace, cosider = 1 The iequality 1 1 > is useless as far as the Compariso Test is cocered. This is because Σ b = Σ (½) is coverget ad a > b.
30 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.
31 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.
32 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
33 LIMIT COMPARISON TEST PROOF If Σ b coverges, so does Σ Mb. Thus, Σ a coverges by part i of the Compariso Test.
34 LIMIT COMPARISON TEST PROOF If Σ b diverges, so does Σ mb. Thus, Σ a diverges by part ii of the Compariso Test.
35 COMPARISON TESTS Example 3 Test the give series for covergece or divergece: = 1
36 COMPARISON TESTS Example 3 We use the Limit Compariso Test with: a 1 1 = b = 2 1 2
37 COMPARISON TESTS Example 3 We obtai: ( ) a 1/ 2 1 lim = lim b 1/2 = lim = lim = 1 > 0 1 1/2
38 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.
39 COMPARISON TESTS Example 4 Determie whether the give series coverges or diverges: =
40 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 b 5 5/2 1/2 = = = 5 +
41 COMPARISON TESTS Example 4 We obtai: 2 1/2 a lim = lim. b = 5/2 3/ lim lim + = = =
42 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.
43 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.
44 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.
45 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 +
46 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
47 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.
48 ESTIMATING SUMS Example 5 Sice 1 1 < the give series is coverget by the Compariso Test.
49 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
50 ESTIMATING SUMS Example 5 Therefore, the remaider for the give series satisfies: R T 1/2 2
51 ESTIMATING SUMS Example 5 With = 100, we have: 1 R = 100 2(100) With a programmable calculator or a computer, we fid that = 1 = 1 with error less tha
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