Root Test

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  So we have to try something else. It turns out that the integral test is the best one to use in this example—we’ll check it out a little later, in Section 23.5. 23.4 How to Use the Root Test Use the Root Test when there are a lot of tricky exponentials around involving functions of n . It’s especially useful when the terms of your series look like A B , where both A and B are functions of n . Here’s the statement of the test, fresh from Section 22.5.2 of the previous chapter: if L = lim n →∞ | a n | 1 /n , then ∞ X n =1 a n converges absolutely if L < 1, and diverges if L > 1; but if L = 1 or the limit doesn’t exist, then the ratio test tells you nothing. To use the Root Test, always start off with the following expression: lim n →∞ | a n | 1 /n , and then replace a n by the general term of the series. Find the limit (if it exists) and call it L . Then you have three possibilities, which are identical to the possibilities which arise in the ratio test. The conclusions are luckily the same as well: 1. If L < 1, then the original series ∑ ∞ n =1 a n converges; in fact, it converges absolutely. 2. If L > 1, then the original series diverges. 3. If L = 1, or the limit doesn’t exist, then the Root Test is useless. Try something else. For example, consider the series ∞ X n =1 1 -2 n n 2 . Since the terms involved have exponents involving powers of n , this series is just begging you to use the Root Test on it. Let’s try it: lim n →∞ | a n | 1 /n = lim n →∞ 1 -2 n n 2 1 /n = lim n →∞ 1 -2 n n 2 × 1 n = lim n →∞ 1 -2 n n = e -2 < 1 . Note that we removed the absolute values since everything’s positive, and used the important limit at the end of Section 22.1.2 of the previous chapter (with k replaced by -2). So the limiting ratio is e -2 , which is certainly less than 1; by the Root Test, the original series above converges.

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  Real Analysis

  With Proof Strategies

  • Daniel W. Cunningham(Author)
  • 2021(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  series a 1, a 2, …, a k, … by examining the long-term behavior of the value | a k | k. Again, if this rate is less than 1, then the series will converge. The proof of this test also depends on the direct comparison test and a geometric series. Theorem 7.2.12 (Root Test I). Let ∑ k = 1 ∞ a k be an infinite series. If there is a positive real number ρ < 1 and a natural number N such that | a k | k ≤ ρ for all k ≥ N, then ∑ k = 1 ∞ a k converges absolutely. If | a k | k ≥ 1 for infinitely many natural numbers k, then ∑ k = 1 ∞ a k diverges. Proof. To prove (1), let ρ ∈ ℝ + and N ∈ ℕ be such. that | a k | k ≤ ρ < 1 for all k ≥ N. It follows, by induction, that | a N + k | ≤ ρ N ρ k for all k ∈ ℕ. (7.7) Since 0 < ρ < 1, the series ∑ k = 1 ∞ ρ N ρ k converges, by Theorem 7.1.11. Thus, (7.7) and Theorem 7.2.1 imply that ∑ k = 1 ∞ | a N + k | converges. Since ∑ k = 1 ∞ | a N + k | is a tail of the series ∑ k = 1 ∞ | a k |, Theorem 7.1.13 and Theorem 7.2.8 imply that ∑ k = 1 ∞ a k converges absolutely. To prove (2), assume that | a k | k ≥ 1 for infinitely many natural numbers k. Thus, | a k | ≥ 1 for infinitely many k. So the sequence 〈 a k 〉 does not converge to 0. Hence, the series ∑ k = 1 ∞ a k diverges by Theorem 7.1.9. Theorem 3.8.16 and Theorem 7.2.12 imply the following modified version of the above Root Test (also see Exercise 2.) Corollary 7.2.13 (Root Test II). Let ∑ k = 1 ∞ a k be an infinite series. If lim ⁡ sup ⁡ k → ∞ | a k | k < 1, then ∑ k = 1 ∞ a k converges absolutely. If lim ⁡ sup ⁡ k → ∞ | a k | k > 1 or 〈 | a k | 〉 k is not bounded above, then ∑ k = 1 ∞ a k diverges. Theorem 3.8.17 implies that if ratio test II decides either the convergence or the divergence of a series, then Root Test II will arrive at the same conclusion. So Root Test II is considered to be a stronger test. However, ratio test II is often easier to apply. Recall that if a sequence 〈 s n 〉 of partial sums diverges, then the corresponding series ∑ k = 1 ∞ a k is said to diverge

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  Calculus, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  7. Root Test If a n is of the form sb n d n , then the Root Test may be useful. 8. Integral Test If a n - f snd, where y ` 1 f s xd dx is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisfied). In the following examples we don’t work out all the details but simply indicate which tests should be used. EXAMPLE 1 o ` n-1 n 2 1 2n 1 1 Since a n l 1 2 ± 0 as n l `, we should use the Test for Divergence. ■ EXAMPLE 2 o ` n-1 sn 3 1 1 3n 3 1 4n 2 1 2 Since a n is an algebraic function of n, we compare the given series with a p-series. The comparison series for the Limit Comparison Test is  b n , where b n - sn 3 3n 3 - n 3y2 3n 3 - 1 3n 3y2  ■ EXAMPLE 3 o ` n-1 ne 2n 2 Since the integral y ` 1 xe 2x 2 dx is easily evaluated, we use the Integral Test. The Ratio Test also works. ■ EXAMPLE 4 o ` n-1 s21d n n 2 n 4 1 1 Since the series is alternating, we use the Alternating Series Test. We can also observe that  | a n | converges (compare to  1yn 2 ) so the given series converges absolutely and hence converges. ■ EXAMPLE 5 o ` k-1 2 k k! Since the series involves k!, we use the Ratio Test. ■ EXAMPLE 6 o ` n-1 1 2 1 3 n Since the series is closely related to the geometric series  1 y3 n , we use the Direct Comparison Test or the Limit Comparison Test. ■ Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 11.8 Power Series 819 11. 7 Exercises 1–8 Two similar-looking series are given. Test each one for convergence or divergence. 1. (a) o ` n-1 1 5 n (b) o ` n-1 1 5 n 1 n 2.

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  Single Variable Calculus

  Early Transcendentals, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  7. Root Test If a n is of the form sb n d n , then the Root Test may be useful. 8. Integral Test If a n - f snd, where y ` 1 f s xd dx is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisfied). In the following examples we don’t work out all the details but simply indicate which tests should be used. EXAMPLE 1 o ` n-1 n 2 1 2n 1 1 Since a n l 1 2 ± 0 as n l `, we should use the Test for Divergence. ■ EXAMPLE 2 o ` n-1 sn 3 1 1 3n 3 1 4n 2 1 2 Since a n is an algebraic function of n, we compare the given series with a p-series. The comparison series for the Limit Comparison Test is  b n , where b n - sn 3 3n 3 - n 3y2 3n 3 - 1 3n 3y2  ■ EXAMPLE 3 o ` n-1 ne 2n 2 Since the integral y ` 1 xe 2x 2 dx is easily evaluated, we use the Integral Test. The Ratio Test also works. ■ EXAMPLE 4 o ` n-1 s21d n n 2 n 4 1 1 Since the series is alternating, we use the Alternating Series Test. We can also observe that  | a n | converges (compare to  1yn 2 ) so the given series converges absolutely and hence converges. ■ EXAMPLE 5 o ` k-1 2 k k! Since the series involves k!, we use the Ratio Test. ■ EXAMPLE 6 o ` n-1 1 2 1 3 n Since the series is closely related to the geometric series  1 y3 n , we use the Direct Comparison Test or the Limit Comparison Test. ■ Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 11.8 Power Series 781 11. 7 Exercises 1–8 Two similar-looking series are given. Test each one for convergence or divergence. 1. (a) o ` n-1 1 5 n (b) o ` n-1 1 5 n 1 n 2.

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  Introduction to Analysis

  • Corey M. Dunn(Author)
  • 2017(Publication Date)
  • CRC Press

    (Publisher)

  We consider the series � 1 , � ( − 1) n , and � 1 Our results concerning p − series (Theorem 4.15 ) n n n 2 . and the Alternating Series Test (Theorem 4.24 ) conclude that � n 1 diverges, � ( − n 1) n conditionally converges, and � n 1 absolutely converges. 2 We attempt to apply the Ratio Test (Theorem 4.31 ) to these series. If 1 or ( − 1) n , then a n = n n a n +1 n | | a n | | = n + 1 → 1 . If a n = n 1 2 , then 2 | a n +1 | = n 1 as well . | a n | ( n + 1) 2 → Now we attempt to apply the Root Test (Theorem 4.32 ) to these series. Recall from Theorem 3.55 that √ n 1. If a n = n n n 1 or ( − 1) n , then → � 1 n | a n | = √ n → 1 . n If a n = n 1 , then 2 � 1 1 n | a n | = √ n 2 = √ n 2 → 1 as well . n n Thus if L = 1 in either of these tests, there is nothing certain that can be said about � a n . squaresolid Since each of the Ratio and Root Tests utilizes a similar comparison in its proofs, it would seem that the two are equivalent in the sense that both will be conclusive or both will be inconclusive. We close this section with one final example from [8] that is designed to illustrate that this is not the case: the Root Test is a stronger result. The interested reader will find the Ratio and Root Tests stated more generally and a result indicating the dominance of the Root Test in Theorem 3.37 on page 68 in [24] . Example 4.36. This example will illustrate an instance where the Ratio Test fails, but the Root Test does not. Let 0 < r < 1, and consider the series � a n = 2 r + r 2 + 2 r 3 + r 4 + . In an attempt to apply the Ratio Test · · · (Theorem 4.31 ), | a n +1 | = 2 1 r if n is even , | a n | 2 r if n is odd . � 243 Series of Real Numbers Here, L = lim n →∞ | a n +1 | does not exist. However, our attempt to use the | a n | Root Test will be more useful: � � r if n is even , n 2 1 /n | a n | = r if n is odd . Here, we find that lim n →∞ n | a n | = r ∈ (0 , 1).

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  Real Infinite Series

  • Daniel D. Bonar, Michael J. Khoury Jr.(Authors)
  • 2018(Publication Date)
  • American Mathematical Society

    (Publisher)

  2 More Sophisticated Techniques It is an unfortunate truth that most students of calculus who encounter infinite series are given only a small collection of theorems about the convergence and divergence of infinite series. Students typically learn the Divergence Test, the Alternating Series Test, the Ratio Test, the Root Test, and some version of the Comparison Tests. Generally the message is that beyond these preliminary results the study of infinite series becomes very hard very fast. However, there are some easily constructed series that do not yield to any of the tests just listed. It is the learned response of calculus students to throw up their hands when the Ratio Test and Root Test give the inconclusive limit of 1. The authors certainly acknowledge that there does not exist a test or set of tests that can systematically determine whether every infinite series converges or diverges. However, we believe strongly that there are several more sophisticated and more powerful tests that are just as easy to use and that can decide the convergence of nearly all series that arise in practice. For whatever reason, these tests are not well known except among specialists of infinite series. In this short chapter we include a small collection of more sophisticated tests. While some of these tests are slightly more difficult to apply than the tests of Chapter 1, these tests are generally both easy to use and powerful. Series that defy all of these tests do exist, but are generally quite pathological. It is our hope that the reader will no longer be in a position of despair when confronting a series for which the Root Test fails. 2.1 The Work of Cauchy Theorem 2.1 (Cauchy Criterion) Let an ^e an infinite series. Then an con~ verges if and only if for every e > 0, there exists a positive integer N with the property that, for every n > no > N, < €. n=n0 A series that satisfies this criterion is said to be a Cauchy series.

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  (b) If  > 1 or  = +∞, the series diverges. (c) If  = 1, the series may converge or diverge, so that another test must be tried. Example 4 Use the Root Test to determine whether the following series converge or diverge. (a) ∞ ∑ k=2 ( 4k − 5 2k + 1 ) k (b) ∞ ∑ k = 1 1 (ln(k + 1)) k Solution (a). The series diverges, since  = lim k →+∞ (u k ) 1∕k = lim k →+∞ 4k − 5 2k + 1 = 2 > 1 Solution (b). The series converges, since  = lim k →+∞ (u k ) 1∕k = lim k →+∞ 1 ln(k + 1) = 0 < 1 QUICK CHECK EXERCISES 9.5 (See page 553 for answers.) 1–4 Select between converges or diverges to fill the first blank. 1. The series ∞ ∑ k=1 2k 2 + 1 2k 8∕3 − 1 by comparison with the p-series ∑ ∞ k=1 . 2. Since lim k →+∞ (k + 1) 3 ∕ 3 k+1 k 3 ∕ 3 k = lim k →+∞ ( 1 + 1 k ) 3 3 = 1 3 the series ∑ ∞ k=1 k 3 ∕ 3 k by the test. 3. Since lim k →+∞ (k + 1)! ∕ 3 k+1 k! ∕ 3 k = lim k →+∞ k + 1 3 = +∞ the series ∑ ∞ k=1 k! ∕ 3 k by the test. 4. Since lim k →+∞ ( 1 k k∕2 ) 1∕k = lim k →+∞ 1 k 1∕2 = 0 the series ∑ ∞ k=1 1 ∕ k k∕2 by the test. EXERCISE SET 9.5 1–2 Make a guess about the convergence or divergence of the series, and confirm your guess using the comparison test. 1. (a) ∞ ∑ k = 1 1 5k 2 − k (b) ∞ ∑ k = 1 3 k − 1 8 2. (a) ∞ ∑ k = 2 k + 1 k 2 − k (b) ∞ ∑ k = 1 4 k 4 + k 3. In each part, use the comparison test to show that the series converges. (a) ∞ ∑ k = 1 1 3 k + 11 (b) ∞ ∑ k = 1 5 sin 2 k k! 4. In each part, use the comparison test to show that the series diverges. (a) ∞ ∑ k = 1 ln k k (b) ∞ ∑ k = 1 k k 3∕2 − 1 4 552 Chapter 9 / Infinite Series 5–10 Use the limit comparison test to determine whether the series converges. 5. ∞ ∑ k = 1 4k 3 − 6k + 5 8k 7 + k − 8 6. ∞ ∑ k = 1 1 7k + 2 7. ∞ ∑ k = 1 7 3 k + 1 8. ∞ ∑ k = 1 k(k + 3) (k + 1)(k + 2)(k + 4) 9. ∞ ∑ k = 1 1 3 √ 8k 2 − 5k 10. ∞ ∑ k = 1 1 (2k + 3) 13 11–16 Use the ratio test to determine whether the series con- verges. If the test is inconclusive, then say so. 11. ∞ ∑ k = 1 99 k k! 12.

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  Calculus

  One and Several Variables

  • Saturnino L. Salas, Garret J. Etgen, Einar Hille(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  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. Series with rational terms are most easily handled by limit comparison with a p-series, a series of the form ∑ 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 a k / →0, then there is no reason to try any convergence test; the series diverges. [(12.2.6).] EXERCISES 12.4 Exercises 1–40. Determine whether the series converges or diverges. 1. ∑ 10 k k ! . 2. ∑ 1 k 2 k . 3. ∑ 1 k k . 4. ∑  k 2k + 1  k . 5. ∑ k ! 100 k . 6. ∑ (ln k ) 2 k . 7. ∑ k 2 + 2 k 3 + 6k . 8. ∑ 1 (ln k ) k . 9. ∑ k  2 3  k . 10. ∑ 1 (ln k ) 10 . 11. ∑ 1 1 + √ k . 12. ∑ 2k + √ k k 3 + √ k . 13. ∑ k ! 10 4k . 14. ∑ k 2 e k . 15. ∑ √ k k 2 + 1 . 16. ∑ 2 k k ! k k . 17. ∑ k ! (k + 2)! . 18. ∑ 1 k  1 ln k  3/2 . 19. ∑ 1 k  1 ln k  1/2 . 20. ∑ 1 √ k 3 − 1 . 21. ∑  k k + 100  k . 22. ∑ (k !) 2 (2k )! . 23. ∑ k −(1+1/ k) . 24. ∑ 11 1 + 100 −k . 25. ∑ ln k e k . 26. ∑ k ! k k . 27. ∑ ln k k 2 . 28. ∑ k ! 1 · 3 · · · (2k − 1) . 29. ∑ 2 · 4 · · · 2k (2k )! . 30. ∑ (2k + 1) 2k (5k 2 + 1) k . 31. ∑ k !(2k )! (3k )! . 32. ∑ ln k k 5/4 . 33. ∑ k k/2 k ! . 34. ∑ k k (3 k ) 2 . 35. ∑ k k 3 k 2 . 36. ∑  √ k − √ k − 1  k . 37. 1 2 + 2 3 2 + 4 4 3 + 8 5 4 + · · · . 12.5 ABSOLUTE CONVERGENCE AND CONDITIONAL CONVERGENCE; ALTERNATING SERIES ■ 597 38. 1 + 1 · 2 1 · 3 + 1 · 2 · 3 1 · 3 · 5 + 1 · 2 · 3 · 4 1 · 3 · 5 · 7 + · · · . 39. 1 4 + 1 · 3 4 · 7 + 1 · 3 · 5 4 · 7 · 10 + 1 · 3 · 5 · 7 4 · 7 · 10 · 13 + · · · . 40. 2 3 + 2 · 4 3 · 7 + 2 · 4 · 6 3 · 7 · 11 + 2 · 4 · 6 · 8 3 · 7 · 11 · 15 + · · · . 41. Find the sum of the series ∞  k=1 k 10 k . HINT: Exercise 36 of Section 12.2.

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  Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  The following example illustrates this principle. Example 2 Use the limit comparison test to determine whether the following series converge or diverge. (a) ∞  k=1 1 √ k + 1 (b) ∞  k=1 1 2k 2 + k (c) ∞  k=1 3k 3 − 2k 2 + 4 k 7 − k 3 + 2 Solution (a). As in Example 1, Principle 9.5.2 suggests that the series is likely to behave like the divergent p-series (1). To prove that the given series diverges, we will apply the limit comparison test with a k = 1 √ k + 1 and b k = 1 √ k We obtain ρ = lim k →+∞ a k b k = lim k →+∞ √ k √ k + 1 = lim k →+∞ 1 1 + 1 √ k = 1 Since ρ is finite and positive, it follows from Theorem 9.5.4 that the given series diverges. Solution (b). As in Example 1, Principle 9.5.3 suggests that the series is likely to behave like the convergent series (2). To prove that the given series converges, we will apply the limit comparison test with a k = 1 2k 2 + k and b k = 1 2k 2 We obtain ρ = lim k →+∞ a k b k = lim k →+∞ 2k 2 2k 2 + k = lim k →+∞ 2 2 + 1 k = 1 Since ρ is finite and positive, it follows from Theorem 9.5.4 that the given series converges, which agrees with the conclusion reached in Example 1 using the comparison test. Solution (c). From Principle 9.5.3, the series is likely to behave like ∞  k=1 3k 3 k 7 = ∞  k=1 3 k 4 (3) which converges since it is a constant times a convergent p-series. Thus, the given series is likely to converge. To prove this, we will apply the limit comparison test to series (3) and the given series. We obtain ρ = lim k →+∞ 3k 3 − 2k 2 + 4 k 7 − k 3 + 2 3 k 4 = lim k →+∞ 3k 7 − 2k 6 + 4k 4 3k 7 − 3k 3 + 6 = 1 Since ρ is finite and nonzero, it follows from Theorem 9.5.4 that the given series converges, since (3) converges. THE RATIO TEST The comparison test and the limit comparison test hinge on first making a guess about convergence and then finding an appropriate series for comparison, both of which can be difficult tasks in cases where Principles 9.5.2 and 9.5.3 cannot be applied. In such cases

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