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Home , Limit comparison test

Section 8.3 and Comparison Tests

APEX C II Version 2.0

Authors Gregory Hartman, Ph.D. Department of Applied Mathemacs Virginia Military Instute

Brian Heinold, Ph.D. Department of Mathemacs and Computer Science Mount Saint Mary’s University

Troy Siemers, Ph.D. Department of Applied Mathemacs Virginia Military Instute

Dimplekumar Chalishajar, Ph.D. Department of Applied Mathemacs

Virginia Military Instute

Editor Jennifer Bowen, Ph.D. Department of Mathemacs and Computer Science The College of Wooster Copyright © 2014 Gregory Hartman Licensed to the public under Creave Commons Aribuon-Noncommercial 3.0 United States License Contents

Preface iii

Table of Contents v

5 Integraon 185 5.1 Anderivaves and Indefinite Integraon ...... 185 5.2 The Definite Integral ...... 194 5.3 Riemann Sums ...... 204 5.4 The Fundamental Theorem of ...... 221 5.5 Numerical Integraon ...... 233

6 Techniques of Andifferenaon 247 6.1 Substuon ...... 247 6.2 Integraon by Parts ...... 266 6.3 Trigonometric ...... 276 6.4 Trigonometric Substuon ...... 286 6.5 Paral Fracon Decomposion ...... 295 6.6 Hyperbolic Funcons ...... 303 6.7 L’Hôpital’s Rule ...... 313 6.8 Improper Integraon ...... 321

7 Applicaons of Integraon 333 7.1 Area Between Curves ...... 334 7.2 Volume by Cross-Seconal Area; Disk and Washer Methods ... 341 7.3 The Shell Method ...... 348 7.4 and Surface Area ...... 356 7.5 Work ...... 365 7.6 Fluid Forces ...... 375

8 Sequences and 383 8.1 Sequences ...... 383 8.2 Infinite Series ...... 395 8.3 Integral and Comparison Tests ...... 410 8.4 Rao and Root Tests ...... 419 8.5 Alternang Series and Absolute Convergence ...... 424 8.6 ...... 434 8.7 Taylor Polynomials ...... 446 8.8 ...... 457

A Soluons To Selected Problems A.1

Index A.11 Chapter 8 Sequences and Series

8.3 Integral and Comparison Tests

Knowing whether or not a series converges is very important, especially when we discuss Power Series in Secon 8.6. Theorems 60 and 61 give criteria for when Geometric and p- series converge, and Theorem 63 gives a quick test to determine if a series diverges. There are many important series whose conver- gence cannot be determined by these theorems, though, so we introduce a set of tests that allow us to handle a broad range of series. We start with the Inte- gral Test.

Integral Test

We stated in Secon 8.1 that a sequence {an} is a funcon a(n) whose do- main is N, the set of natural numbers. If we can extend a(n) to R, the real num- bers, and it is both posive and decreasing on [1, ∞), then the convergence of ∑∞ ∫ ∞ Note: Theorem 65 does not state that a is the same as a(x) dx. the integral and the summaon have the n n=1 1 same value.

. Theorem 65 Integral Test y Let a sequence {an} be defined by an = a(n), where a(n) is connuous, ∑∞ 2 posive and decreasing on [1, ∞). Then. an converges, if, and only if, y = a(x) = ∫ ∞ n 1 a(x) dx converges. 1 1

We can demonstrate the truth of the Integral Test with two simple graphs. . x 1 2 3 4 5 In Figure 8.15(a), the height of each rectangle is a(n) = an for n = 1, 2,..., (a) and clearly the rectangles enclose more area than the area under y = a(x). y Therefore we can conclude that

∫ ∞ ∑∞ 2 a(x) dx < an. (8.1) = ( ) y a x 1 n=1

In Figure 8.15(b), we draw rectangles under y = a(x) with the Right-Hand rule, 1 starng with n = 2. This me, the area of the rectangles is less than the area ∑∞ ∫ ∞ under y = a(x), so an < a(x) dx. Note how this summaon starts n=2 1 . x 1 2 3 4 5 with n = 2; adding a1 to both sides lets us rewrite the summaon starng with (b)

Figure 8.15: Illustrang the truth of the Integral Test. Notes:

410 8.3 Integral and Comparison Tests

n = 1: ∑∞ ∫ ∞ an < a1 + a(x) dx. (8.2) n=1 1 Combining Equaons (8.1) and (8.2), we have ∑∞ ∫ ∞ ∑∞ an < a1 + a(x) dx < a1 + an. (8.3) n=1 1 n=1 From Equaon (8.3) we can make the following two statements: ∑∞ ∫ ∞ ∑∞ ∫ ∞ 1. If an diverges, so does a(x) dx (because an < a1+ a(x) dx) n=1 1 n=1 1 ∑∞ ∫ ∞ ∫ ∞ ∑∞ 2. If an converges, so does a(x) dx (because a(x) dx < an.) n=1 1 1 n=1 Therefore the series and integral either both converge or both diverge. Theorem 64 allows us to extend this theorem to series where an is posive and decreasing on [b, ∞) for some b > 1.

. Example 241 .Using the Integral Test y ∑∞ ln n 0.8 Determine the convergence of . (The terms of the sequence {a } = n2 n n=1 0.6 {ln n/n2} and the nth paral sums are given in Figure 8.16.)

∫ ∞ 0.4 ln x S Applying the Integral Test, we test the convergence of 2 dx. 1 x 0.2 Integrang this requires the use of Integraon by Parts, with u = ln x and dv = 1/x2 dx. . n ∫ ∞ ∫ 2 4 6 8 10 12 14 16 18 20 ln x b ln x dx = lim dx 2 →∞ 2 . an Sn 1 x b 1 x ∫ 1 b b 1 = lim − ln x + dx Figure 8.16: Plong the sequence and b→∞ x 1 x2 1 series in Example 241. 1 1 b = lim − ln x − b→∞ x x 1 1 ln b = lim 1 − − . Apply L’Hôpital’s Rule: b→∞ b b = 1. ∫ ∞ ∑∞ ln x ln n Since dx converges, so does . x2 n2 1 n=1

Notes:

411 Chapter 8 Sequences and Series

Note how the sequence {an} is not strictly decreasing; it increases from n = 1 to n = 2. However, this does not keep us from applying the Integral Test as the sequence in posive and decreasing on [2, ∞). .

Theorem 61 was given without jusficaon, stang that the general p-series ∑∞ 1 converges if, and only if, p > 1. In the following example, we (an + b)p n=1 prove this to be true by applying the Integral Test.

. Example 242 Using the Integral Test to establish Theorem 61. ∑∞ 1 Use the Integral Test to prove that converges if, and only if, p > 1. (an + b)p n=1

∫ ∞ 1 ≠ S Consider the integral p dx; assuming p 1, 1 (ax + b) ∫ ∞ ∫ 1 c 1 dx = lim dx (ax + b)p c→∞ (ax + b)p 1 1 1 − c = lim (ax + b)1 p c→∞ a(1 − p) 1 ( ) 1 − − = lim (ac + b)1 p − (a + b)1 p . c→∞ a(1 − p)

This converges if, and only if, p > 1. It is easy to show that the integral also diverges in the case of p = 1. (This result is similar to the work preceding Key Idea 21.) ∑∞ 1 Therefore converges if, and only if, p > 1. . (an + b)p n=1 We consider two more in this secon, both comparison tests. That is, we determine the convergence of one series by comparing it to another series with known convergence.

Notes:

412 8.3 Integral and Comparison Tests

Direct Comparison Test

. Theorem 66 Note: A sequence {an} is a posive sequence if an > 0 for all n. Let {an} and {bn} be posive sequences where an ≤ bn for all n ≥ N, ≥ for some N 1. Because of Theorem 64, any theorem that ∑∞ ∑∞ . relies on a posive sequence sll holds 1. If bn converges, then an converges. true when an > 0 for all but a finite num- n=1 n=1 ber of values of n. ∑∞ ∑∞ 2. If an diverges, then bn diverges. n=1 n=1

. Example 243 Applying the Direct Comparison Test ∑∞ 1 Determine the convergence of . 3n + n2 n=1

S This series is neither a geometric or p-series, but seems re- lated. We predict it will converge, so we look for a series with larger terms that converges. (Note too that the Integral Test seems difficult to apply here.) ∑∞ 1 1 1 Since 3n < 3n + n2, > for all n ≥ 1. The series is a 3n 3n + n2 3n n=1 ∑∞ 1 convergent ; by Theorem 66, converges. . 3n + n2 n=1 . Example 244 Applying the Direct Comparison Test ∑∞ 1 Determine the convergence of . n − ln n n=1 ∑∞ 1 S We know the Harmonic Series diverges, and it seems n n=1 that the given series is closely related to it, hence we predict it will diverge. 1 1 Since n ≥ n − ln n for all n ≥ 1, ≤ for all n ≥ 1. n n − ln n ∑∞ 1 The Harmonic Series diverges, so we conclude that diverges as n − ln n n=1 well. .

Notes:

413 Chapter 8 Sequences and Series

The concept of direct comparison is powerful and oen relavely easy to apply. Pracce helps one develop the necessary intuion to quickly pick a proper series with which to compare. However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test. ∑∞ 1 Consider . It is very similar to the divergent series given in Ex- n + ln n n=1 1 1 ample 244. We suspect that it also diverges, as ≈ for large n. How- n n + ln n ever, the inequality that we naturally want to use “goes the wrong way”: since 1 1 n ≤ n + ln n for all n ≥ 1, ≥ for all n ≥ 1. The given series has terms n n + ln n less than the terms of a divergent series, and we cannot conclude anything from this. Fortunately, we can apply another test to the given series to determine its convergence.

Limit Comparison Test

. Theorem 67 Limit Comparison Test

Let {an} and {bn} be posive sequences. ∑∞ an 1. If lim = L, where L is a posive real number, then an and n→∞ b n n=1 ∑∞ bn either both converge or. both diverge. n=1 ∑∞ ∑∞ an 2. If lim = 0, then if bn converges, then so does an. n→∞ b n n=1 n=1 ∑∞ ∑∞ an 3. If lim = ∞, then if bn diverges, then so does an. n→∞ b n n=1 n=1

It is helpful to remember that when using Theorem 67, the terms of the series with known convergence go in the denominator of the fracon. We use the Limit Comparison Test in the next example to examine the series ∑∞ 1 which movated this new test. n + ln n n=1

Notes:

414 8.3 Integral and Comparison Tests

. Example 245 Applying the Limit Comparison Test ∑∞ 1 Determine the convergence of using the Limit Comparison Test. n + ln n n=1 ∑∞ 1 S We compare the terms of to the terms of the n + ln n n=1 ∑∞ 1 Harmonic Sequence : n n=1 1/(n + ln n) n lim = lim n→∞ 1/n n→∞ n + ln n = 1 (aer applying L’Hôpital’s Rule). ∑∞ 1 Since the Harmonic Series diverges, we conclude that diverges as n + ln n n=1 well. .

. Example 246 Applying the Limit Comparison Test ∑∞ 1 Determine the convergence of 3n − n2 n=1

S This series is similar to the one in Example 243, but now we are considering “3n − n2” instead of “3n + n2.” This difference makes applying the Direct Comparison Test difficult. ∑∞ 1 Instead, we use the Limit Comparison Test and compare with the series : 3n n=1 1/(3n − n2) 3n lim = lim n→∞ 1/3n n→∞ 3n − n2 = 1 (aer applying L’Hôpital’s Rule twice). ∑∞ ∑∞ 1 1 We know is a convergent geometric series, hence converges 3n 3n − n2 n=1 n=1 as well. .

As menoned before, pracce helps one develop the intuion to quickly choose a series with which to compare. A general rule of thumb is to pick a series based on the dominant term in the expression of {an}. It is also helpful to note that dominate exponenals, which dominate algebraic func- ons (e.g., polynomials), which dominate logarithms. In the previous example,

Notes:

415 Chapter 8 Sequences and Series

∑∞ 1 1 the dominant term of was 3n, so we compared the series to . It is 3n − n2 3n n=1 hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L’Hôpital’s Rule to n!.

. Example 247 Applying the Limit Comparison Test ∑∞ √ x + 3 Determine the convergence of . x2 − x + 1 n=1

S We naïvely aempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is 1/x2. Knowing ∑∞ 1 that converges, we aempt to apply the Limit Comparison Test: n2 n=1 √ √ ( x + 3)/(x2 − x + 1) x2( x + 3) lim = lim n→∞ 1/x2 n→∞ x2 − x + 1 = ∞ (Apply L’Hôpital’s Rule). ∑∞ Theorem 67 part (3) only applies when bn diverges; in our case, it con- n=1 verges. Ulmately, our test has not revealed anything about the convergence of our series. The problem is that we chose a poor series with which to compare. Since the numerator and denominator of the terms of the series are both algebraic funcons, we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator. The dominant term of the numerator is x1/2 and the dominant term of the denominator is x2. Thus we should compare the terms of the given series to x1/2/x2 = 1/x3/2: √ √ ( x + 3)/(x2 − x + 1) x3/2( x + 3) lim = lim n→∞ 1/x3/2 n→∞ x2 − x + 1 = 1 (Applying L’Hôpital’s Rule). ∑∞ ∑∞ √ 1 x + 3 Since the p-series converges, we conclude that con- x3/2 x2 − x + 1 n=1 n=1 verges as well. .

Notes:

416 Exercises 8.3 ∑∞ 1 Terms and Concepts 16. n! + n n=1 1. In order to apply the Integral Test to a sequence {an}, the ∑∞ funcon a(n) = an must be , and . 1 17. √ 2 − 2. T/F: The Integral Test can be used to determine the sum of n=2 n 1 a . ∑∞ 1 18. √ 3. What test(s) in this secon do not work well with factori- n − 2 als? n=5 ∞ ∑∞ ∑ n2 + n + 1 19. 4. Suppose an is convergent, and there are sequences n3 − 5 n=0 n=1 { } { } ≤ ≤ bn and cn such that bn an cn for all n. What ∑∞ ∑∞ ∑∞ 2n 20. can be said about the series bn and cn? 5n + 10 n=1 n=0 n=0 ∑∞ n 21. n2 − 1 Problems n=2 ∑∞ 1 In Exercises 5 – 12, use the Integral Test to determine the con- 22. n2 ln n vergence of the given series. n=2 ∑∞ 1 5. In Exercises 23 – 32, use the Limit Comparison Test to deter- 2n n=1 mine the convergence of the given series; state what series is ∑∞ used for comparison. 1 6. ∑∞ n4 1 n=1 23. n2 − 3n + 5 ∑∞ n=1 n 7. ∑∞ n2 + 1 1 n=1 24. 4n − n2 ∑∞ n=1 1 8. ∞ n ln n ∑ n=2 ln n 25. − ∑∞ n 3 1 n=4 9. 2 ∑∞ n + 1 1 n=1 26. √ ∑∞ n2 + n 1 n=1 10. n(ln n)2 ∑∞ n=2 1 27. √ ∑∞ n + n n n=1 11. 2n ∑∞ n=1 n − 10 28. ∑∞ n2 + 10n + 10 ln n n=1 12. 3 n ∞ n=1 ∑ ( ) 29. sin 1/n In Exercises 13 – 22, use the Direct Comparison Test to deter- n=1 mine the convergence of the given series; state what series is ∑∞ used for comparison. n + 5 30. ∑∞ n3 − 5 1 n=1 13. 2 − n + 3n 5 ∑∞ √ n=1 n + 3 ∑∞ 31. 2 1 n + 17 14. n=1 4n + n2 − n n=1 ∑∞ 1 ∑∞ 32. √ ln n n + 100 15. n=1 n n=1

417 ∑∞ In Exercises 33 – 40, determine the convergence of the given cos(1/n) 40. √ series. State the test used; more than one test may be appro- n n=1 priate. ∑∞ ∑∞ n2 33. 41. Given that a converges, state which of the following 2n n n=1 n=1 ∑∞ series converges, may converge, or does not converge. 1 34. (2n + 5)3 ∑∞ n=1 a (a) n ∑∞ n n! n=1 35. n 10 ∞ n=1 ∑ ∑∞ (b) anan+1 ln n 36. n=1 n! n=1 ∑∞ ∑∞ 2 1 (c) (an) 37. 3n + n n=1 n=1 ∞ ∑∞ ∑ n − 2 38. (d) nan 10n + 5 n=1 n=1 ∑∞ ∑∞ 3n 1 39. 3 (e) n an n=1 n=1

418 Solutions to Odd Exercises

41. Le to reader (b) The nth paral sum of the odd series is 1 + 1 + 1 + ··· + 1 . The nth paral sum of 1 plus the Secon 8.2 3 5 2n−1 + 1 + 1 + ··· + 1 even series is 1 2 4 2(n−1) . Each term of the 1. Answers will vary. even series is now greater than or equal to the corresponding term of the odd series, with equality only on { 3. One sequence is the sequence of terms∑a}. The other is the the first term. This gives us the result. th { } { n } sequence of n paral sums, Sn = i=1 ai . (c) If the odd series converges, the work done in (a) shows the th 5. F even series converges also. (The sequence of the n paral sum of the even series is bounded and , 5 , 49 , 205 , 5269 7. (a) 1 4 36 144 3600 monotonically increasing.) Likewise, (b) shows that if the (b) Plot omied even series converges, the odd series will, too. Thus if either series converges, the other does. 9. (a) 1, 3, 6, 10, 15 Similarly, (a) and (b) can be used to show that if either (b) Plot omied series diverges, the other does, too. 1 , 4 , 13 , 40 , 121 11. (a) 3 9 27 81 243 (d) If both the even and odd series converge, then their sum (b) Plot omied would be a convergent series. This would imply that the Harmonic Series, their sum, is convergent. It is not. Hence 13. (a) 0.1, 0.11, 0.111, 0.1111, 0.11111 each series diverges. (b) Plot omied Secon 8.3 15. lim an = ∞; by Theorem 63 the series diverges. n→∞ 1. connuous, posive and decreasing 17. lim an = 1; by Theorem 63 the series diverges. n→∞ 3. The Integral Test (we do not have a connuous definion of n! = yet) and the Limit Comparison Test (same as above, hence we 19. →∞lim an e; by Theorem 63 the series diverges. n cannot take its derivave). 21. Converges 5. Converges 23. Converges 7. Diverges 25. Converges 9. Converges 27. Converges 11. Converges 29. Diverges ∑∞ ( ) 1 2 13. Converges; compare to , as 1/(n2 + 3n − 5) ≤ 1/n2 for 31. (a) S = n(n+1) n2 n 2 n=1 all n > 1. (b) Diverges ∑∞ − n 1 33. (a) S = 5 1 1/2 15. Diverges; compare to , as 1/n ≤ ln n/n for all n ≥ 2. n 1/2 n n=1 (b) Converges to 10. ∑∞ 1 √ √ 1−(−1/3)n 2 2 − 35. (a) S = 17. Diverges; compare to . Since n = n > n 1, n 4/3 n √ n=1 (b) Converges to 3/4. 1/n ≤ 1/ n2 − 1 for all n ≥ 2. ( ) ∑∞ 37. (a) With paral fracons, a = 3 1 − 1 . Thus 1 ( n ) 2 n n+2 19. Diverges; compare to : n = 3 3 − 1 − 1 n=1 Sn 2 2 n+1 n+2 . 1 n2 n2 + n + 1 n2 + n + 1 (b) Converges to 9/4 = < < , ( ) n n3 n3 n3 − 5 39. (a) Sn = ln 1/(n + 1) for all n ≥ 1. (b) Diverges (to −∞). ∑∞ 1 1 41. (a) an = ; using paral fracons, the resulng 21. Diverges; compare to . Note that n(n+3) n telescoping( sum reduces to ) n=1 1 1 1 − 1 − 1 − 1 2 Sn = 1 + + + + + n n 1 1 3 2 3 n 1 n 2 n 3 = · > , 2 − 2 − (b) Converges to 11/18. n 1 n 1 n n ( ) n2 1 1 1 > ≥ 43. (a) With paral fracons, a = − . Thus as n2−1 1, for all n 2. ( n ) 2 n−1 n+1 ∑∞ S = 1 3/2 − 1 − 1 . 1 n 2 n n+1 23. Converges; compare to . n2 (b) Converges to 3/4. n=1 ∑∞ 45. (a) The nth paral sum of the odd series is ln n 1 1 ··· 1 th 25. Diverges; compare to . 1 + + + + − . The n paral sum of the even n 3 5 2n 1 n=1 series is 1 + 1 + 1 + ··· + 1 . Each term of the even 2 4 6 2n ∑∞ 1 series is less than the corresponding term of the odd 27. Diverges; compare to . series, giving us our result. n n=1

A.7 ∑∞ 1 sin n 29. Diverges; compare to . Just as lim = 1, (c) absolute n n→0 n n=1 7. (a) diverges (limit of terms is not 0) sin(1/n) lim = 1. (b) diverges n→∞ 1/n (c) n/a; diverges ∑∞ 1 31. Converges; compare to . 9. (a) converges n3/2 n=1 (b) diverges (Limit Comparison Test with 1/n) 33. Converges; Integral Test (c) condional th 35. Diverges; the n and Direct Comparison Test can be 11. (a) diverges (limit of terms is not 0) used. (b) diverges 37. Converges; the Direct Comparison Test can be used with sequence (c) n/a; diverges 1/3n. 13. (a) diverges (terms oscillate between 1) 39. Diverges; the nth Term Test can be used, along with the Integral Test. (b) diverges an < (c) n/a; diverges 41. (a) Converges; use Direct Comparison Test as n n. (b) Converges; since original series converges, we know 15. (a) converges limn→∞ an = 0. Thus for large n, anan+1 < an. (b) converges (Geometric Series with r = 2/3) 2 (c) Converges; similar logic to part (b) so (an) < an. (c) absolute

(d) May converge; certainly nan > an but that does not mean 17. (a) converges it does not converge. (b) converges (Rao Test) th (e) Does not converge, using logic from (b) and n Term Test. (c) absolute Secon 8.4 19. (a) converges (b) diverges (p-Series Test with p = 1/2) 1. algebraic, or polynomial. (c) condional 3. Integral Test, Limit Comparison Test, and 21. S5 = −1.1906; S6 = −0.6767; ∑∞ 5. Converges (−1)n −1.1906 ≤ ≤ −0.6767 7. Converges ln(n + 1) n=1

9. The Rao Test is inconclusive; the p-Series Test states it diverges. 23. S6 = 0.3681; S7 = 0.3679; ∑∞ 11. Converges (−1)n 0.3681 ≤ ≤ 0.3679 ∑∞ n! 2nn! n=0 13. Converges; note the summaon can be rewrien as , 3nn! 25. n = 5 n=1 from which the Rao Test can be applied. 27. Using the theorem, we find n = 499 guarantees the sum is within 15. Converges 0.001 of π/4. (Convergence is actually faster, as the sum is within ε of π/24 when n ≥ 249.) 17. Converges Secon 8.6 19. Diverges 21. Diverges. The Root Test is inconclusive, but the nth-Term Test 1. 1 2 shows . (The terms of the sequence approach e , not 3. 5 0, as n → ∞.) 5. 1 + 2x + 4x2 + 8x3 + 16x4 23. Converges 2 3 4 7. 1 + x + x + x + x 25. Diverges; Limit Comparison Test 2 6 24 9. (a) R = ∞ 27. Converges; Rao Test or Limit Comparison Test with 1/3n. (b) (−∞, ∞) 29. Diverges; nth-Term Test or Limit Comparison Test with 1. 11. (a) R = 1 31. Diverges; Direct Comparison Test with 1/n (b) (2, 4] 33. Converges; Root Test 13. (a) R = 2 Secon 8.5 (b) (−2, 2) 15. (a) R = 1/5 1. The signs of the terms do not alternate; in the given series, some terms are negave and the others posive, but they do not (b) (4/5, 6/5) necessarily alternate. 17. (a) R = 1 n 3. Many examples exist; one common example is an = (−1) /n. (b) (−1, 1) 5. (a) converges 19. (a) R = ∞ (b) converges (p-Series) (b) (−∞, ∞)

A.8