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Calculus 2: Infinite Series

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Calculus 2: Infinite Series Infinite Series – Ratio and Root Tests for Convergence In this section we’ll learn two additional convergence tests; the ratio test and the root test. After learning these tests, we will summarize all of the tests we have learned and provide strategies for determining which test to apply to a given series. Ratio Test The ratio test can be useful for series containing factorials and constants to the power 푛. The test is formally stated below. Ratio Test Given the series ∑ 푎푛 we define the following 푎푛+1 퐿 = 푙푖푚 | | 푛→∞ 푎푛 Then, • If 퐿 < 1 then ∑ 푎푛 converges absolutely. • If 퐿 > 1 then ∑ 푎푛 diverges. • If 퐿 = 1 the test is inconclusive. Let’s do a few examples. Example 1: Apply the ratio test to the following series. ∞ 2푛 ∑ 푛! 푛=1 We start by defining the following. 2푛 2푛+1 푎 = 푎 = 푛 푛! 푛+1 (푛 + 1)! Next, we apply the ratio test 2푛+1 푛! 퐿 = 푙푖푚 | ∙ | 푛→∞ (푛 + 1)! 2푛 2푛21 (푛) ∙ (푛 − 1) ∙ (푛 − 2). (2) ∙ (1) = 푙푖푚 | ∙ | 푛→∞ (푛 + 1) ∙ (푛) ∙ (푛 − 1) ∙ (푛 − 2). (2) ∙ (1) 2푛 2 = 푙푖푚 | | = 0 푛→∞ (푛 + 1) Since 퐿 < 0, the series converges. Example 2: Apply the ratio test to the following alternating series. ∞ 푛! ∑(−1)푛 1000푛 푛=1 Directly applying the ratio test we have (푛 + 1)! 1000푛 퐿 = 푙푖푚 | ∙ | 푛→∞ 1000푛+1 푛! (푛 + 1) ∙ (푛) ∙ (푛 − 1) ∙ (푛 − 2). (2) ∙ (1) 1000푛 = 푙푖푚 | ∙ | 푛→∞ 1000푛10001 (푛) ∙ (푛 − 1) ∙ (푛 − 2). (2) ∙ (1) (푛 + 1) = 푙푖푚 | | = ∞ 푛→∞ 1000 In this case 퐿 > 0, and therefore the series diverges. Example 3: Evaluate whether the following two series converge by first using the ratio test. ∞ ∞ 1 ∑ 푛2 ∑ 푛2 푛=1 푛=1 We apply the ration test to both series below. (푛 + 1)2 푛2 퐿 = 푙푖푚 | | 퐿 = 푙푖푚 | | 푛→∞ 푛2 푛→∞ (푛 + 1)2 푛2 + 2푛 + 1 푛2 = 푙푖푚 | | = 푙푖푚 | | 푛→∞ 푛2 푛→∞ 푛2 + 2푛 + 1 푛2 푛2 = 푙푖푚 | | = 1 = 푙푖푚 | | = 1 푛→∞ 푛2 푛→∞ 푛2 Unfortunately, the ratio test in inconclusive in both cases since 퐿 = 1. However, we can easily determine from the nth term divergence test that the first series diverges and that the second series in a convergent p-series. As you can see from this example a series can either diverge or converge when the ratio test results in 퐿 = 1. Root Test The root test can be useful for series containing terms for the form 푓(푛)(푛). The test is formally stated below. Root Test Given the series ∑ 푎푛 we define the following 푛 퐿 = 푙푖푚 ( √|푎푛|) 푛→∞ Then, • If 퐿 < 1 then ∑ 푎푛 converges absolutely. • If 퐿 > 1 then ∑ 푎푛 diverges. • If 퐿 = 1 the test is inconclusive. Let’s do a few examples. Example 4: Apply the root test to the following series. ∞ 푛 푛 ∑ ( ) 2푛 + 3 푛=1 푛 We identify the sequence, 푎 , has the form 푓(푛)(푛), where 푓(푛) = and 푔(푛) = 푛. 푛 2푛+3 Therefore, we can directly apply the root test. 푛 퐿 = 푙푖푚 ( √|푎푛|) 푛→∞ 푛 푛 푛 = 푙푖푚 ( √|( ) |) 푛→∞ 2푛 + 3 푛 1 = 푙푖푚 ( ) = 푛→∞ 2푛 + 3 2 Since 퐿 < 1 the series converges. Example 5: Apply the root test to the following series. ∞ 2푛 ∑ 푛2푛 푛=1 In this case we start by rewriting the series as follows ∞ ∞ 2푛 2 푛 ∑ = ∑ ( ) 푛2푛 푛2 푛=1 푛=1 We can now directly apply the root test. 푛 푛 2 퐿 = 푙푖푚 ( √|( ) |) 푛→∞ 푛2 2 = 푙푖푚 ( ) = 0 푛→∞ 푛2 Again, since 퐿 < 1 the series converges. The summary at the end of this section includes general guidelines for determining which test to apply when first looking at an arbitrary series. You should read over that before attempting the next set of examples, Example 6: Evaluate the following series using any test we learned so far. a. b. c. ∞ ∞ ∞ 푛 푠푖푛(푛) 푛! ∑ ∑ ∑ 2푛 + 1 푛2 (2푛)! 푛=1 푛=1 푛=1 d. e. f. ∞ ∞ ∞ 푛3 1 ∑ ∑ ∑(0.8)−푛푛−0.8 5푛 √푛3 − 푛2 푛=1 푛=2 푛=1 g. ∞ h. i. 1 ∞ ∞ ∑ 푠푖푛 ( ) (2푛)! 푒4푛 푛2 ∑ ∑ 푛=1 5푛 + 1 (푛 − 2)! 푛=1 푛=2 ∞ 푛 ∑ a. 2푛 + 1 푛=1 푡ℎ We start by checking the 푛 term divergence test, which states that if 푙푖푚(푎푛) ≠ 0 then the 푛→∞ series diverges. 푛 푛 1 푙푖푚 ( ) = 푙푖푚 ( ) = ≠ 0 푛→∞ 2푛 + 1 푛→∞ 2푛 2 Therefore, the series diverges. ______________________________________________________________________________ ∞ 푠푖푛(푛) ∑ b. 푛2 푛=1 Since this series in non-positive we check for absolute convergence using the direct comparison 1 test with 푏 = . Since ∑∞ 푏 is a convergent p-series, we need to show that 푎 < 푏 for all 푛 푛2 푛=1 푛 푛 푛 푛 > 푀. 0 ≤ |푠푖푛(푛)| ≤ 1 |푠푖푛(푛)| 1 0 ≤ ≤ 푛2 푛2 |푠푛(푛)| 1 Therefore since ≤ the series converges absolutely. Furthermore, since the series 푛2 푛2 converges absolutely the original no-positive series also converges. ______________________________________________________________________________ ∞ 푛! ∑ c. (2푛 )! 푛=1 In this case we directly apply the ratio test. (푛 + 1)! (2푛)! 퐿 = 푙푖푚 | ∙ | 푛→∞ (2(푛 + 1))! 푛! (푛 + 1)(푛)(푛 − 1) … (1) (2푛)(2푛 − 1)(2푛 − 1). (1) = 푙푖푚 | ∙ | 푛→∞ (2푛 + 2)(2푛 + 1)(2푛)(2푛 − 1)(2푛 − 2) … (1) (푛)(푛 − 1) … (1) (푛 + 1) = 푙푖푚 | | 푛→∞ (2푛 + 2)(2푛 + 1) (푛 + 1) = 푙푖푚 | | 푛→∞ (4푛2 + 5푛 + 2) 푛 = 푙푖푚 | | = 0 < 1 푛→∞ 4푛2 Therefore, the series converges. ∞ 푛3 ∑ d. 5푛 푛=1 We again apply the ratio test. (푛 + 1)3 5푛 퐿 = 푙푖푚 | ∙ | 푛→∞ 5푛+1 푛3 푛3 + 3푛2 + 3푛 + 1 5푛 = 푙푖푚 | ∙ | 푛→∞ 5 ∙ 5푛 푛3 푛3 + 3푛2 + 3푛 + 1 = 푙푖푚 | | 푛→∞ 5푛3 푛3 1 = 푙푖푚 | | = < 1 푛→∞ 5푛3 2 Therefore, the series converges. ______________________________________________________________________________ ∞ 1 e. ∑ √푛3 − 푛2 푛=2 1 In this case we use the limit comparison test with 푏 = . Since ∑∞ 푏 . 푛 √푛3 푛=1 푛 1 √푛3 퐿 = 푙푖푚 ( ∙ ) 푛→∞ √푛3 − 푛2 1 푛3 = 푙푖푚 (√ ) 푛→∞ 푛3 − 푛2 푛3 = √푙푖푚 ( ) 푛→∞ 푛3 − 푛2 = √푙푖푚(1) = √1 = 1 푛→∞ 1 1 Therefore, the series exhibit the same behavior. Since ∑∞ = ∑∞ is a convergent p- 푛=2 √푛3 푛=2 푛1.5 series, the original series converges. ______________________________________________________________________________ ∞ −푛 −0.8 f. ∑(0.8) 푛 푛=1 We’ll use the ratio test, but first we’ll rearrange the sequence as follows: ∞ ∞ 10푛 ∑(0.8)−푛푛−0.8 = ∑ 8푛(푛4⁄5) 푛=1 푛=1 Now applying the ratio test we have: 10푛+1 8푛(푛4⁄5) 퐿 = 푙푖푚 | ∙ | 푛→∞ 8푛+1((푛 + 1)4⁄5) 10푛 10 (푛4⁄5) = 푙푖푚 | ∙ | 푛→∞ 8((푛 + 1)4⁄5) 1 5 5 푛 4 = ( ) 푙푖푚 |√( ) | 4 푛→∞ 푛 + 1 5 5 푛 4 = ( ) √푙푖푚 ( ) 4 푛→∞ 푛 + 1 5 5 = ( ) 5√1 = > 1 4 4 In this case since 퐿 > 1 the series diverges. ______________________________________________________________________________ ∞ 1 ∑ 푠푖푛 ( ) g. 푛 2 푛=1 1 In this case we test for absolute convergence using the limit comparison test with 푏 = . 푛 푛2 1 푠푖푛 ( ) 푛2 퐿 = 푙푖푚 | | 푛→∞ 1 푛2 To evaluate this limit, we apply Hopital’s rule as shown. 1 푐표푠 ( ) ∙ (−2푛−3) 푛2 퐿 = 푙푖푚 | | 푛→∞ (−2푛−3) 1 = 푙푖푚 |푐표푠 ( )| 푛→∞ 푛2 = 푙푖푚|푐표푠(0)| = 1 푛→∞ ∞ The series have similar behavior and since ∑푛=1 푏푛 converges, the original series converges. ______________________________________________________________________________ ∞ (2푛)! ∑ h. 5푛 + 1 푛=1 We’ll use the ratio test. (2푛 + 2)! 5푛 + 1 퐿 = 푙푖푚 | ∙ | 푛→∞ 5푛 + 6 (2푛)! (2푛 + 2)(2푛 + 1)(2푛)(2푛 − 1). (2)(1) 5푛 + 1 = 푙푖푚 | ∙ | 푛→∞ 5푛 + 6 (2푛)(2푛 − 1). (2)(1) (2푛 + 2)(2푛 + 1)(5푛 + 1) = 푙푖푚 | | = ∞ 푛→∞ 5푛 + 6 Therefore, the series diverges. ______________________________________________________________________________ ∞ 푒4푛 ∑ i. (푛 − 2)! 푛=1 We’ll use the ratio test. 푒4푛+4 (푛 − 2)! 퐿 = 푙푖푚 | ∙ | 푛→∞ (푛 − 1)! 푒4푛 푒4푒4푛 (푛 − 2)(푛 − 3)(푛 − 4). (2)(1) = 푙푖푚 | ∙ | 푛→∞ (푛 − 1)(푛 − 2)(푛 − 3)(푛 − 4). (2)(1) 푒4푛 푒4 = 푙푖푚 | | = 0 푛→∞ (푛 − 1) Therefore, the series converges. Final Summary for Infinite Series Summary of Tests 풏th Term Divergence Test If 푙푖푚(푎푛) ≠ 0 then the series 푛→∞ ∞ ∑푛=1 푎푛, diverges If 푙푖푚(푎푛) = 0 then the test is inconclusive. 푛→∞ Direct Comparison Test Assume that there exists 푀 > 0 such that 0 ≤ 푎푛 ≤ 푏푛 for all 푛 ≥ 푀. ∞ ∞ i. If ∑푛=1 푏푛 converges, then ∑푛=1 푎푛 also converges. ∞ ∞ ii. If ∑푛=1 푎푛 diverges, then ∑푛=1 푏푛 also diverges. Limit Comparison Test Let 푎푛 and 푏푛 be positive sequences and let the following limit exist. 푎푛 퐿 = 푙푖푚 ( ) 푛→∞ 푏푛 i. If 퐿 > 0 then 푎푛 ≅ 퐿푏푛 for large 푛, and both series either converge or diverge. ∞ ∞ ii. If 퐿 = 0 then 푏푛 ≫ 푎푛 for large 푛, and if ∑푛=1 푏푛 converge, so does ∑푛=1 푎푛 ∞ ∞ iii. If 퐿 = ∞ then 푎푛 ≫ 푏푛 for large 푛, and if ∑푛=1 푎푛 converge, so does ∑푛=1 푏푛. Integral Test Let 푎푛 = 푓(푛), where 푓 is a positive, decreasing, and continuous function of 푥 for 푥 ≥ 1. ∞ • 푓(푥)푑푥 ∑∞ 푎 If ∫1 converges, then 푛=1 푛 converges. ∞ 푓(푥)푑푥 ∑∞ 푎 If ∫1 diverges, then 푛=1 푛 diverges. Ratio Test Given the series ∑ 푎푛 we define the following 푎푛+1 퐿 = 푙푖푚 | | 푛→∞ 푎푛 Then, • If 퐿 < 1 then ∑ 푎푛 converges absolutely. • If 퐿 > 1 then ∑ 푎푛 diverges. • If 퐿 = 1 the test is inconclusive. Root Test Given the series ∑ 푎푛 we define the following 푛 퐿 = 푙푖푚 ( √|푎푛|) 푛→∞ Then, • If 퐿 < 1 then ∑ 푎푛 converges absolutely. • If 퐿 > 1 then ∑ 푎푛 diverges. • If 퐿 = 1 the test is inconclusive. Alternating Series Test An alternating series of the form ∞ 푛−1 ∑(−1) 푏푛 푛=1 Converges if • 푏푛 ≥ 0, Non-negative • 푏푛+1 < 푏푛, Decreasing 푙푖푚(푏푛) = 0 푛→∞ Absolute Convergence The series ∑ 푎푛 converges absolutely if ∑|푎푛| converges.
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