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30.5 Convergence Tests 971 Limit Comparison Test Suppose ∞ k=1 a k and ∞ k=1 b k are series whose terms are positive for all k ≥ N for some constant N. i. If lim k→∞ a k b k = L for 0 <L<∞, then both series converge or both series diverge. ii. If lim k→∞ a k b k = 0 and ∞ k=1 b k converges, then ∞ k=1 a n converges as well. iii. If lim k→∞ a k b k =∞and ∞ k=1 b k diverges, then ∞ k=1 a k diverges as well. Proof i. Choose positive constants α and β with 0 <α<L<β.Then for k sufficiently large α< a k b k <β and αb k <a k <βb k . Suppose ∞ k=1 b k diverges. Then ∞ k=1 αb k diverges as well. ∞ k=1 a k diverges by the Comparison Test. Suppose ∞ k=1 b k converges. Then ∞ k=1 βb k converges as well. ∞ k=1 a k con- verges by the Comparison Test. The proofs of parts (ii) and (iii) are left as an exercise for the reader. ◆ EXAMPLE 30.29 Determine whether ∞ k=2 2k 2 +3k k 4 −k+1 converges or diverges. SOLUTION For k large the terms of the series “look like” 2k 2 k 4 = 2 k 2 . ∞ k=2 2 k 2 = 2 ∞ k=2 1 k 2 converges. (It’sap-series with p>1.) We use the Limit Comparison Test to show ∞ k=2 2k 2 +3k k 4 −k+1 converges as well. lim k→∞ a k b k = lim k→∞ 2k 2 +3k k 4 −k+1 2 k 2 = lim k→∞ 2k 2 + 3k k 4 − k + 1 · k 2 2 = lim k→∞ 2k 4 + 3k 3 2k 4 − 2k + 2 = 1 Therefore ∞ k=2 2k 2 +3k k 4 −k+1 converges. ◆ EXERCISE 30.9 Use the Limit Comparison Test to determine whether the following series are convergent or divergent. (a) ∞ k=1 4 10 k −2 (b) ∞ k=2 √ k 3 +1 k(k 2 −1) Answers (a) Convergent: Compare with ∞ k=1 4 10 4 or ∞ k=1 1 10 k , convergent geometric series. (b) Convergent: “Looks like” ∞ k=2 k 3/2 k 3 = ∞ k=2 1 k 3/2 , a convergent p-series. In the next example we tackle a more challenging problem. 972 CHAPTER 30 Series ◆ EXAMPLE 30.30 Determine whether ∞ k=2 ln k k 2 converges or diverges. SOLUTION There are several options available for determining convergence. One option is to use the Integral Test. To do this, verify that f(x)= ln x x 2 is positive, decreasing, and continuous on [2, ∞) and then compute ∞ 2 ln x x 2 dx. This improper integral requires integration by parts. As an exercise, verify that ∞ 2 ln x x 2 dx = ln 2+1 2 and conclude that ∞ k=2 ln k k 2 converges. An alternative is to use the Limit Comparison Test. The challenge is to make a good choice for comparison. 1 < ln k<k for x ≥ 3so 1 k 2 < ln k k 2 < k k 2 = 1 k for x ≥ 3 The terms are greater than those of a known convergent series and less than those of a known divergent series! This doesn’t clarify the convergence issues. If we try to use limit comparison by comparing with ∞ k=3 1 k 2 we get lim k→∞ ln k k 2 1 k 2 = lim k→∞ ln k =∞, which is inconclusive. On the other hand, using limit comparison by comparing with ∞ k=3 1 k we get lim k→∞ ln k k 2 1 k = lim k→∞ ln k k 2 · k = lim k→∞ ln k k = 0, which is inconclusive. Let’s compare with ∞ k=3 1 k 1.5 . This is a convergent p-series. lim k→∞ ln k k 2 1 k 1.5 = lim k→∞ ln k k 2 · k 1.5 = lim k→∞ ln k k 0.5 We can compute this limit by computing lim x→∞ ln x x 1 2 using L’H ˆ opital’s Rule. 17 lim x→∞ ln x x 1 2 = lim x→∞ 1 x 1 2 1 √ x = lim x→∞ 1 x · 2 √ x 1 = lim x→∞ 2 √ x = 0 Therefore, lim k→∞ ln k k 0.5 = 0 and ∞ k=2 ln k k 2 converges. ◆ The Ratio and the Root Test In order for a series to converge its terms must be going toward zero sufficiently rapidly. Therefore, it is reasonable to think that valuable information can be obtained by looking at the ratio of successive terms, a k+1 /a k , for k very large. The following convergence test, the 17 See Appendix F for a discussion of L’H ˆ opital’s Rule. 30.5 Convergence Tests 973 Ratio Test, does this. Its proof relies on our knowledge of geometric series, series for which the ratio of successive terms is fixed. The Ratio Test Let ∞ k=1 a k be a series such that a k > 0 for k sufficiently large. Suppose lim k→∞ a k+1 a k = L, where L is a real number or ∞. i. If L<1, then ∞ k=1 a k converges. ii. If L>1, then ∞ k=1 a k diverges. iii. If L = 1, the test is inconclusive. Proof i. Suppose L<1. We will compare our series with a convergent geometric series. Since 0 ≤ L<1, we can choose a constant r such that 0 ≤ L<r<1. ∞ k=1 r k converges. lim k→∞ a k+1 a k = L<r, sofor k sufficiently large, a k+1 a k <r. In other words, there exists an integer N such that a k+1 a k <r for all k ≥ N . Therefore, for k ≥ N , a k+1 <a k r, and a k+2 <a k+1 r<(a k r) · r = a k r 2 a k+3 <a k+2 r<(a k r 2 )·r=a k r 3 . . . The series ∞ k=1 a N+k = a N+1 + a N+2 + a N+3 +···converges by comparison to ∞ k=1 a N r k = a N r + a N r 2 + a N r 3 +···. This latter series is a geometric series with |r| < 1. Each of its terms is larger than the corresponding term in ∞ k=1 a N+k . a N+i <a N r i for i = 1, 2, 3, Because ∞ k=N+1 a k converges we conclude that ∞ k=1 a k converges. ii. Suppose lim k→∞ a k+1 a k = L, L>1, or lim k→∞ a k+1 a k =∞.Then there exists a number N such that a k+1 a k > 1 for all k ≥ N ; that is, 0 <a k <a k+1 for all k ≥ N . But then lim k→∞ a k = 0; the series ∞ k=1 a k diverges by the nth Term Test for Divergence. 974 CHAPTER 30 Series iii. To show that lim k→∞ a k+1 a k = 1 is inconclusive, we need only produce both convergent and divergent series for which the limit of the ratio of successive terms is 1. ∞ k=1 1 k diverges. lim k→∞ a k+1 a k = lim k→∞ 1 k+1 1 k = lim k→∞ k k + 1 = 1. ∞ k=1 1 k 2 converges. lim k→∞ a k+1 a k = lim k→∞ 1 (k+1) 2 1 k 2 = lim k→∞ k 2 (k + 1) 2 = 1. As demonstrated, the Ratio Test is not useful for p-series or series whose terms look like polynomials. It is a great test to use for series whose terms involve factorials and/or exponentials. ◆ EXAMPLE 30.31 Test the following series for convergence. (a) ∞ k=1 10 k k! (b) ∞ k=1 k k k! SOLUTION Both series have positive terms. We use the Ratio Test. (a) a k+1 a k = 10 k+1 (k+1)! 10 k k! = 10 k+1 (k + 1)! · k! 10 k = 10 k · 10 (k + 1)k! · k! 10 k = 10 k + 1 Compute: lim k→∞ a k+1 a k = lim k→∞ 10 k+1 = 0 < 1. Therefore, ∞ k=1 10 k k! converges. (b) a k+1 a k = (k+1) k+1 (k+1)! k k k! = (k + 1) k+1 (k + 1)! · k! k k = (k + 1) k (k + 1) · k! (k + 1)k!k k = k + 1 k k = 1 + 1 k k Compute: lim k→∞ 1 + 1 k k = e>1. Therefore, ∞ k=1 k k k! diverges. ◆ REMARK In Example 30.31 part (b) we showed that for the series ∞ k=1 k k k! , lim k→∞ a k+1 a k = e. It follows that for the series ∞ k=1 k! k k , lim k→∞ a k+1 a k = 1 e < 1. Thus ∞ k=1 k! k k converges and its terms must tend toward zero. We can conclude that k k grows much more rapidly than does k! From part (a) of Example 30.31 we can deduce that k!in turn grows much more rapidly than does b k for any constant b. The Ratio Test gives us an alternative way of proving the fact that lim k→∞ b k k! = 0 for every real number b. The following test gives us a convenient way of testing for convergence when the terms of a series are raised to the kth power. 30.5 Convergence Tests 975 The Root Test Let ∞ k=1 a k be a series such that a k > 0 for k sufficiently large. Suppose that lim k→∞ k √ a k = L, where L is a real number or ∞. i. If L<1, then ∞ k=1 a k converges. ii. If L>1, then ∞ k=1 a k diverges. iii. If L = 1, the test is inconclusive. The proof of the Root Test runs along the same lines as that of the Ratio Test, so it is omitted. EXERCISE 30.10 Use the Root Test to show that ∞ k=1 2n 3 +3n 3n 3 +1 n converges. Extending Our Results In this section we’ve presented several tests that can be used to determine the convergence or divergence of a series: The nth Term Test for Divergence, the Comparison and Limit Comparison Tests, the Integral Test, the Ratio Test, and the Root Test. Of these, all but the nth Term Test require that eventually all the terms of the series being tested must be positive. If, instead, all the terms of the series are negative, or eventually the terms are all negative, then we can factor out a (−1) and the tests can be applied. Otherwise, we can look at ∞ k=1 |a k |.If ∞ k=1 |a k |converges, not only does ∞ k=1 a k converge, but it converges absolutely.If ∞ k=1 |a k |diverges, then ∞ k=1 a k will either diverge or converge conditionally. 18 If the series is alternating we can try to apply the Alternating Series Test. The Ratio Test and Root Test can be easily generalized to apply to arbitrary series. Generalized Ratio Test Let ∞ k=1 a k be a series with a k = 0 for k sufficiently large. Suppose lim k→∞ |a k+1 | |a k | = L, where L is a real number or ∞. i. If L<1, then ∞ k=1 a n converges absolutely (and therefore converges). ii. If L>1, then ∞ k=1 a n diverges. iii. If L = 1, then the test is inconclusive. The proof is left as a problem at the end of the section. The Root Test can be similarly generalized. The generalized Ratio Test can be applied to a power series in order to find its radius of convergence and to show that in the interior of its interval of convergence (not including endpoints) the series converges absolutely. Let ∞ k=0 a k (x − b) k be a power series centered at x = b. Let w k = a k (x − b) k . Compute 18 Conditional convergence, defined in Section 30.4, means a k converges but |a k | diverges. 976 CHAPTER 30 Series lim k→∞ |w k+1 | |w k | = lim k→∞ |a k+1 (x − b) k+1 | |a k (x − b) k | = lim k→∞ |a k+1 | |a k | |x − b|= lim k→∞ |a k+1 | |a k | |x − b|. If lim k→∞ |a k+1 | |a k | = 0 then by the generalized Ratio Test the series converges absolutely for all x. If lim k→∞ |a k+1 | |a k | = Q, where Q is finite and nonzero, then, the generalized Ratio Test says the series converges absolutely for all x such that Q|x − b| < 1, i.e., |x − b| < 1 Q . The series diverges for Q|x − b| > 1, i.e., |x − b| > 1 Q . Therefore, the radius of convergence is 1 Q . If lim k→∞ |a k+1 | |a k | =∞,then, by the generalized Ratio Test, the series converges only for x = b. ◆ EXAMPLE 30.32 The Bessel function J 0 (x) = ∞ k=0 (−1) k x 2k (k!) 2 2 2k is a function defined as a power series. Find the set of all x for which the series converges absolutely. SOLUTION Apply the generalized Ratio Test. |w k+1 | |w k | = (−1) k+1 x 2(k+1) (k + 1)!(k + 1)!2 2(k+1) · (k!)(k!)2 2k (−1) k x 2k = |x 2k+2 |2 2k |x 2k |(k + 1)(k + 1)2 2k+2 = |x 2 | (k + 1) 2 2 2 = x 2 4 1 (k + 1) 2 lim k→∞ x 2 4 · 1 (k + 1) 2 = 0 < 1. The Bessel function J 0 (x) converges absolutely for all x. ◆ ◆ EXAMPLE 30.33 Find the interval of convergence of the power series ∞ k=2 (x+3) k ln k . SOLUTION Begin by applying the generalized Ratio Test. Compute |w k+1 | |w k | . (x+3) k+1 ln(k+1) (x+3) k ln k = |x + 3|| ln k| | ln(k + 1)| lim k→∞ |w k+1 | |w k | = lim k→∞ |x + 3| ln k ln(k + 1) =|x+3| The series converges for |x + 3| < 1 and diverges for |x + 3| > 1. We must check the endpoints of the interval of convergence, x =−4 and x =−2, independently. When x = 4wehave ∞ k=2 (−4+3) k ln k = ∞ k=2 (−1) k ln k . This converges by the Alternating Series Test, because the terms are decreasing in magnitude and tending toward zero. When x =−2wehave ∞ k=2 (−2+3) k ln k = ∞ k=2 1 ln k . This series diverges by comparison with the harmonic series. (See Example 30.28 for details.) The interval of convergence is centered at x =−3, the center of the series. The interval of convergence is [−4, −2). The series converges absolutely on (−4, −2) and converges conditionally at x =−4. ◆ 30.5 Convergence Tests 977 Summary of Convergence 19 Criteria for a k Geometric Series: ar k converges for |r|< 1 diverges for |r|≥1 p-series: 1 k p converges for p>1 diverges for p ≤ 1 nth Term Test for Divergence If lim k→∞ a k = 0, then the series diverges. Comparison Test terms all positive If a k ≤ b k for all k and b k converges, so does a k . If b k ≤ a k for all k and b k diverges, so does a k . Limit Comparison Test terms all positive Suppose lim k→∞ a k b k = L, L possibly infinite. If 0 <L<∞, a k and b k either both converge or both diverge. If L =0, and b k converges, so does a k . If L =∞,and b k diverges, so does a k . Integral Test terms positive If f(k)=a k and f(x) is positive, continuous, and decreasing on [1, ∞), then ∞ 1 f(x)dx and f(k)either both converge or both diverge. Ratio Test particularly useful with factorials and exponents Compute lim k→∞ |a k+1 | |a k | = L, L possibly infinite. If L<1, a k converges absolutely. If L>1, a k diverges. If L =1, the test is inconclusive. Root Test particularly useful when a k looks like (−) k Compute lim k→∞ k √ |a k |=L,Lpossibly infinite. If L<1, a k converges absolutely. If L>1, a k diverges. If L =1, the test is inconclusive. Alternating Series Test terms alternate sign If the series is alternating, the terms are decreasing in magnitude, and the terms tend toward zero, then the series converges. PROBLEMS FOR SECTION 30-5 For Problems 1 and 2, write out the first three terms of the series and then answer the question posed. 1. For what values of c will the series ∞ k=1 c k converge? Explain. 2. For what values of w will the series ∞ k=1 k w converge? Explain. 3. Suppose 0 ≤ a k ≤ b k ≤ c k for all k. Consider ∞ k=1 a k , ∞ k=1 b k and ∞ k=1 c k . What conclusions can be drawn if you know that ∞ k=1 b k (a) converges. (b) diverges. 19 Here we use a k to mean ∞ k=1 a k . The starting value of k is irrelevant since convergence is determined by the tail of the series. 978 CHAPTER 30 Series In Problems 4 through 19, determine whether the series converges or diverges. It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and Root Tests. 4. ∞ k=2 3 √ k 5. ∞ k=10 10 k √ k 6. ∞ k=1 ln k k 7. ∞ n=5 n −9/10 8. ∞ k=1 k e k 9. ∞ k=2 2k −10/9 10. ∞ k=1 e −2k 11. ∞ k=1 k+2 3k 2 12. ∞ k=1 2e −0.1k 13. ∞ k=2 1 k ln k 14. ∞ k=1 ke −k 2 15. ∞ k=1 k k 3 +k+1 16. ∞ k=1 2 3 k +1 17. ∞ k=2 5 k−0.5 18. ∞ n=1 3 n 2 n −1 19. ∞ n=1 1 e n +e 20. Suppose that a k = f(k) for k = 1, 2, 3, ,where f(x) is positive, decreasing, and continuous on [1, ∞). Put the following expressions in order, from smallest to largest. Explain your reasoning with a picture or two. n−1 k=2 a k , n k=3 a k , n 2 f(x)dx 21. Explain why the hypothesis that f(x)is decreasing is important in the Integral Test. 22. Use your knowledge of improper integrals to give an upper and lower bound for ∞ k=1 1 k 2 . 23. Let ∞ k=1 a k be a series and S k = a 1 + a 2 + ···+a k its kth partial sum, where k = 1, 2, 3, Let L be a constant, 0 <L<1. 30.5 Convergence Tests 979 (a) If lim k→∞ a k = L, what can you conclude about ∞ k=1 a k ? (b) If lim k→∞ S k = L, what can you conclude about ∞ k=1 a k ? 24. Let ∞ k=1 a k be a series and S n = n k=1 a k its nth partial sum, where n = 1, 2, 3, Foreach of the following, decide whether or not enough information is given to assure that ∞ k=1 a k converges. M and m are constants. Explain your reasoning. (a) a k > 0 for all k and S n >mfor all n. (b) a k > 0 for all k and S n <M for all n. (c) a k < 0 for all k and S n >mfor all n. (d) m<S n <M for all n. In Problems 25 through 32, determine whether the series converges or diverges. In this set of problems knowledge of the Limit Comparison Test is assumed. 25. ∞ k=1 3 2 k −1 26. ∞ k=1 1 e k −1 27. ∞ n=2 n−1 2n 2 −n 28. ∞ k=1 2k 2 −k 3k 4 +1 29. ∞ k=3 k 2k 3 −2 30. ∞ n=2 1 √ n 2 −n 31. ∞ n=2 n+1 ln n 32. ∞ k=2 2 k 5 k −5 33. (a) Let ∞ k=1 a k be a convergent series with 0 <a k <1for k = 1, 2, 3, i. Show that ∞ k=1 a 2 k converges ii. Show that ∞ k=1 1 a k diverges. (b) Let ∞ k=1 b k be a convergent series with 0 <b k for k = 1, 2, 3, Argue that ∞ k=1 b 2 k converges. (Use the results of part (a)). In Problems 34 through 41, determine whether the series converges or diverges. In this set of problems knowledge of all the convergence tests from the chapter is assumed. 34. ∞ k=1 3k k! 35. ∞ n=1 n 3 2 n 36. ∞ n=1 3 n 2 n 37. ∞ k=1 k 3 k! 980 CHAPTER 30 Series 38. ∞ k=1 k! k 3 3 k 39. ∞ k=2 2 (ln k) k 40. ∞ k=1 1 + 1 k k 41. ∞ k=1 k 2 −3k 5k 2+1 k 42. For what values of n, n a positive integer, does ∞ k=1 k n k! converge? 43. For what values of r does ∞ k=1 r k k! converge? In Problems 44 through 51, determine whether the series converges absolutely, con- verges conditionally, or diverges. Explain your reasoning carefully. 44. ∞ k=1 (−1) k k! (k+1)! 45. ∞ k=1 (−1) k+1 k √ 2k 46. ∞ k=1 cos k k 3 47. ∞ k=1 sin(2k) 2 k 48. ∞ n=3 (−1) n 10 √ n 49. ∞ k=1 (−1) k 5 k k! 50. ∞ k=1 (−k) k k! 51. ∞ k=2 (−1) k 3k 3 +3 52. Does the series ∞ k=2 (−1) k ln k k converge absolutely, converge conditionally, or diverge? Explain your reasoning carefully and justify your assertions. 53. Prove the following version of the Integral Test. (It’s a slightly weaker version than the one stated in this section.) Let ∞ k=1 a k be a series such that a k = f(k)for k = 1, 2, 3, where the function f is positive, continuous, and decreasing on [1, ∞). (a) If ∞ 1 f(x)dx converges, then ∞ k=1 a k converges. (b) If ∞ 1 f(x)dx diverges, then ∞ k=1 a k diverges. In Problems 54 through 59, use the Ratio Test or Root Test to find the radius of convergence of the power series given. 54. ∞ k=1 (−1) k (2x) k k! . tend toward zero. We can conclude that k k grows much more rapidly than does k! From part (a) of Example 30.31 we can deduce that k!in turn grows much more rapidly than does b k for any constant. similarly generalized. The generalized Ratio Test can be applied to a power series in order to find its radius of convergence and to show that in the interior of its interval of convergence (not including endpoints). decreasing in magnitude, and the terms tend toward zero, then the series converges. PROBLEMS FOR SECTION 30-5 For Problems 1 and 2, write out the first three terms of the series and then answer the question
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