SalasSV_11_03 - 652 CHAPTER 11 INFINITE SERIES result in...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 652 CHAPTER 11 INFINITE SERIES result in Exercise 40, to give upper and lower bounds on R4 . (c) Use parts (a) and (b) to estimate the sum of the series. 41. 1 . 3 k =1 k ∞ ∞ 42. k =1 1 . k4 1 , what k2 k =1 ∞ In Exercises 43–46, use the error bounds given in Exercise 40. c 43. (a) If you were to use s100 to approximate would be the bounds on your error? (b) How large would you have to choose n to ensure that Rn is less than 0.0001? ∞1 c 44. (a) If you were to use s100 to approximate , what 3 k =1 k would be the bounds on your error? (b) How large would you have to choose n to ensure that Rn is less than 0.0001? ∞1 c 45. (a) How many terms of the series should you use to 4 k =1 k ensure that Rn is less than 0.0001? ∞1 to three decimal places. (b) Estimate 4 k =1 k ∞1 c 46. Repeat Exercise 45, for the series . 5 k =1 k Exercises 47 and 48 complete the limit comparison test. bk be series with positive terms. Suppose 47. Let ak and that ak /bk → 0. (a) Show that if bk converges, then ak converges. (b) Show that if ak diverges, then bk diverges. bk (c) Show by example that if ak converges, then may converge or diverge. (d) Show by example that if bk diverges, then ak may converge or diverge. [Parts (c) and (d) explain why we stipulated L > 0 in Theorem 11.2.6.] bk be series with positive terms. Suppose 48. Let ak and that ak /bk → ∞. (a) Show that if bk diverges, then ak diverges. (b) Show that if ak converges, then bk converges. (c) Show by example that if ak diverges, then bk may converge or diverge. ak (d) Show by example that if bk converges, then may converge or diverge. 49. Let ak be a series with nonnegative terms. (a) Prove that if ak converges, then a2 converges. k (b) Suppose that a2 converges. Does ak converge or k diverge? Prove or give a counterexample. 50. Let ak be a series with nonnegative terms. Prove that if a2 converges, then (ak /k ) converges. k 51. Let f be a continuous, positive, decreasing function on ∞ [1, ∞) such that 1 f (x) dx converges. Then we know that the series ∞ 1 f (k ) also converges. Prove that k= 0 < L − sn < n ∞ f (x) dx, where L is the sum of the series and sn is the nth partial sum. In Exercises 52 and 53, use the result of Exercise 51 to determine the smallest integer N such that the difference between the sum of the given series and the N th partial sum is less than 0.001. 52. 1 . k2 + 1 k =1 ∞ ∞ 53. k =1 k e −k . 2 54. (a) Use the method of the proof of the integral test to show that ln (n + 1) < 1 + 1 2 + 1 3 + ··· + 1 n < 1 + ln n. (b) How many terms of the harmonic series are needed to ensure that sn > 100? 55. This exercise demonstrates that we cannot always use the same testing series for both the basic comparison test and the limit comparison test. (a) Show that ln k √ kk converges by comparison with 1 . k 5/4 (b) Show that the limit comparison test does not apply. 56. Let p and q be polynomials with nonnegative coefficients. State conditions that will imply the convergence or divergence of the series p( k ) . q( k ) 11.3 THE ROOT TEST; THE RATIO TEST We continue our study of series with nonnegative terms. Comparison with the geometric series. xk leads to two important tests for convergence: the root test and the ratio test. 11.3 THE ROOT TEST; THE RATIO TEST 653 THEOREM 11.3.1 THE ROOT TEST Let ak be a series with nonnegative terms, and suppose that. (ak )1/k → ρ . (a) If ρ < 1, then ak converges. (b) If ρ > 1, then ak diverges. (c) If ρ = 1, then the test is inconclusive. The series may converge; it may diverge. We suppose first that ρ < 1 and choose µ so that ρ < µ < 1. Since (ak ) 1 /k PROOF → ρ , we have (ak )1/k < µ for all k sufficiently large. for all k sufficiently large. (explain) Thus ak < µk Since µk converges (a geometric series with 0 < µ < 1), we know by the basic comparison test that ak converges. We suppose now that ρ > 1 and choose µ so that ρ > µ > 1. Since (ak ) 1 /k → ρ , we have (ak )1/k > µ for all k sufficiently large. for all k sufficiently large. (explain) Thus ak > µk Since µk diverges (a geometric series with µ > 1), we know by the basic comparison test that ak diverges. To see the inconclusiveness of the root test when ρ = 1, consider the series (1/k 2 ) and (1/k ). The first series converges and the second series diverges. However, in each case we have (ak )1/k = (ak )1/k = 1 k2 1 k 1 /k = 1 k 1 /k 2 → 12 = 1 as k → ∞, as k → ∞. 1 /k = 1 →1 k 1/k [Recall that k 1/k → 1 as k → ∞, see (10. 4. 6). ] Applying the Root Test Example 1 For the series 1 , we have ( ln k )k ( a k ) 1 /k = 1 → 0. ln k The series converges. 654 CHAPTER 11 INFINITE SERIES Example 2 For the series 3 /k 2k , we have k3 1 k 1 /k 3 ( ak ) 1 /k 1 =2 k =2 =2 1 k 1 /k 3 → 2 · 13 = 2 as k → ∞. The series diverges. Example 3 In the case of 1− 1 k k , we have 1 → 1. k (ak )1/k = 1 − Here the root test is inconclusive. It is also unnecessary: since ak = (1−1/k )k converges to 1/e and not to 0 (10.4.7), the series diverges (11.1.6). THEOREM 11.3.2 THE RATIO TEST Let ak be a series with positive terms and suppose that ak +1 → λ. ak (a) If λ < 1, then ak converges. (b) If λ > 1, then ak diverges. (c) If λ = 1, then the test is inconclusive. The series may converge; it may diverge. PROOF We suppose first that λ < 1 and choose µ so that λ < µ < 1. Since a k +1 → λ, ak we know that there exists k0 > 0 such that if k ≥ k0 , This gives ak0 +1 < µak0 , and more generally, ak0 +j < µj ak0 , For k > k0 we have (1) ak < µk −k0 ak0 = ↑ ak0 k µ. µk0 j = 1, 2, . . . . ak0 +2 < µak0 +1 < µ2 ak0 , then ak +1 < µ. ak (explain) set j = k − k0 11.3 THE ROOT TEST; THE RATIO TEST 655 Since µ < 1, ak0 k ak µ = k0 k0 µ µ0 µk converges. ak converges. The Therefore, it follows from (1) and the basic comparison test that proof of the rest of the theorem is left to the Exercises. Remark Contrary to some people’s intuition, the root and ratio tests are not equivalent. See Exercise 50. Applying the Ratio Test Example 4 The ratio test shows that the series 1 converges: k! k! 1 1 a k +1 · = → 0. = ak (k + 1)! 1 k +1 k , we have 10k Example 5 For the series k + 1 10k 1 k +1 1 a k +1 = k +1 · = → . ak 10 k 10 k 10 The series converges.† Example 6 For the series kk , we have k! k +1 k k a k +1 (k + 1)k +1 k ! · = = ak (k + 1)! k k Since e > 1, the series diverges. Example 7 = 1+ 1 k k →e as k → ∞. For the series 1 , the ratio test is inconclusive: 2k + 1 a k +1 2k + 1 2k + 1 2 + 1/ k 1 · = = → 1 as k → ∞. = ak 2(k + 1) + 1 1 2k + 3 2 + 3/k Therefore, we have to look further. Limit comparison with the harmonic series shows that the series diverges: 1 k 1 1 ÷= → 2k + 1 k 2k + 1 2 and 1 diverges. k † This series can be summed explicitly. See Exercise 43. 656 CHAPTER 11 INFINITE SERIES Summary on Convergence Tests In general, the root test is used only if powers are involved. The ratio test is particularly effective with factorials and with combinations of powers and factorials. If the terms are rational functions of k , the ratio test is inconclusive and the root test is difficult to apply. Rational terms are most easily handled by comparison or limit comparison with a p-series, 1/k p . If the terms have the configuration of a derivative, you may be able to apply the integral test. Finally, keep in mind that, if ak →0, then there is no reason / to apply any special convergence test; the series diverges (Theorem 11.1.6). EXERCISES 11.3 Determine whether the series converges or diverges. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 28. 29. 31. 10k . k! 1 . kk k! . 100k k2 + 2 . k 3 + 6k k 2 3 k 33. 35. . 37. 2. 4. 6. 8. 10. 12. 14. 16. 18. 1/2 1 . k 2k k 2k + 1 ( ln k ) . k 1 . ( ln k )k 1 . ( ln k )10 √ 2k + k √. k3 + k k2 . ek 2k k ! . kk 1 k √ 1 ln k 1 k3 − 1 2 2 k k /2 . k! kk . 3( k 2 ) 1 2 4 8 + + 3 + 4 + ···. 2 32 4 5 34. 36. kk . (3k )2 √ √ k − k −1 k k . 38. 1 + 39. 40. 1·2 1·2·3 1·2·3·4 + + + ···. 13 1·3·5 1·3·5·7 1 1·3 1·3·5 1·3·5·7 + + + + ···. 4 4 · 7 4 · 7 · 10 4 · 7 · 10 · 13 2 2·4 2·4·6 2·4·6·8 + + + + ···. 3 3 · 7 3 · 7 · 11 3 · 7 · 11 · 15 . 1 √. 1+ k k! . 104k √ k . 2+1 k k! . (k + 2)! 1 k 1 ln k k k + 100 k −(1+1/k ) . ln k . ek ln k . k2 k! . 1 · 3 · · · (2k − 1) 2 · 4 · · · 2k . (2k )! k !(2k )! . (3k )! . k c 41. Use a CAS to implement the ratio test to determine whether the series converges or diverges. ∞ 2k ∞ 2k . (b) . (a) k =1 k ! k =1 k 3/2 c 42. Use a CAS to implement the root test to determine whether the series converges or diverges. . (a) 2k k +1 k =1 k ∞ ∞ (b) k =1 k +2 2k 1 10 k 20. 22. 24. 26. . 43. Find the sum of the series . (k !) . (2k )! 11 . 1 + 100−k k! . kk + 2 100 + 3 1000 + 4 10000 + ···. HINT: Exercise 70 of Section 11.1. 44. Complete the proof of the ratio test. (a) Prove that, if λ > 1, then ak diverges. (b) Prove that, if λ = 1, the ratio test is inconclusive. HINT: Consider 1/k and 1/k 2 . 45. Prove that the sequence the series k! . kk n! nn has limit 0. HINT: Consider 46. Let r be a positive number. Prove that the sequence 30. 32. (2k + 1)2k . (5k 2 + 1)k ln k . k 5/ 4 has limit 0. 47. Let p ≥ 2 be an integer. Find the values of p (if any) such (k !)2 converges. that (pk )! rn n! 11.4 ABSOLUTE AND CONDITIONAL CONVERGENCE; ALTERNATING SERIES 657 48. Let r be a positive number. For what values of r (if any) rk does converge? kr 49. Let {ak } be a sequence of positive numbers and take r > 0. Use the root test to show that, if (ak )1/k → ρ and ρ < 1/r , then ak r k converges. 1 1 50. Consider the series 1 + 1 + 1 + 1 + 32 + 16 + . . . formed 2 8 4 by rearranging a convergent geometric series. (a) Use the root test to show that the series converges. (b) Show that the ratio test does not apply. 11.4 ABSOLUTE AND CONDITIONAL CONVERGENCE; ALTERNATING SERIES In this section we consider series that have both positive and negative terms. Absolute and Conditional Convergence Let ak be a series with positive and negative terms. One way to show that converges is to show that the series of absolute values |ak | converges. ak THEOREM 11.4.1 If |ak | converges, then ak converges. PROOF For each k , −|ak | ≤ ak ≤ |ak | and therefore 0 ≤ ak + |ak | ≤ 2|ak |. If |ak | converges, then 2|ak | = 2 |ak | converges, and therefore, by the basic comparison theorem, (ak + |ak |) converges. Since ak = (ak + |ak |) − |ak |, we can conclude that ak converges. Series ak for which |ak | converges are called absolutely convergent. The theorem we have just proved says that (11.4.2) absolutely convergent series are convergent. As we will show presently, the converse is false. There are convergent series that are not absolutely convergent. Such series are called conditionally convergent. Example 1 Consider the series ∞ k =1 1 1 1 1 (−1)k +1 = 1 − 2 + 2 − 2 + 2 − ··· . k2 2 3 4 5 If we replace each term by its absolute value, we obtain the series ∞ k =1 1 1 1 1 1 = 1 + 2 + 2 + 2 + 2 + ··· . 2 k 2 3 4 5 ...
View Full Document

This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

Ask a homework question - tutors are online