Ratio Test

12 Key excerpts on "Ratio Test"

• 

  eBook - PDF

  Calculus

  Resequenced for Students in STEM

  • David Dwyer, Mark Gruenwald(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  The details are provided in the proof of the Ratio Test, which follows. Theorem 12.6.1 The Ratio Test Let ∑ a n be a series with nonzero terms, and let L = lim n→∞    an+1 an   . 1. If L < 1, then ∑ a n is absolutely convergent. 2. If L > 1 (or L = ∞), then ∑ a n diverges. 3. If L = 1, then no conclusion can be drawn from the Ratio Test. Proof 1. Our strategy is to show that all but the first few terms of the series are less than the corresponding terms of a convergent geometric series, which enables us to apply the Direct Comparison Test. Let r be a number between L and 1, so that L < r < 1. Since lim n→∞ |a n+1 /a n | = L, the ratio |a n+1 /a n | is less than r for n sufficiently large. That is, there is an N such that     a n+1 a n     < r or, equivalently, |a n+1 | < r|a n | 12.6. RATIO AND ROOT TESTS 713 for n ≥ N . By repeatedly applying this inequality for n = N, N +1, N +2, . . ., we obtain |a N+1 | < r|a N | |a N+2 | < r|a N+1 | < r 2 |a N | |a N+3 | < r|a N+2 | < r 3 |a N | . . . Thus, |a N+k | < r k |a N | for k = 1, 2, 3, . . .. Now ∑ ∞ k=1 r k |a N | is a convergent geometric series since r < 1. It follows from the Direct Comparison Test that ∞ X k=1 |a N+k | = ∞ X n=N+1 |a n | converges also. Thus, ∑ ∞ n=1 |a n | converges, which means that ∑ a n is abso- lutely convergent. 2. If lim n→∞    an+1 an    = L > 1, then there is an N such that, for n ≥ N ,     a n+1 a n     > 1 or, equivalently, |a n+1 | > |a n | Thus, |a n | > |a N | for n ≥ N . But this means that lim n→∞ a n 6= 0, and so it follows from the nth-Term Test that ∑ a n diverges. 3. If we compute the limit of the ratio of consecutive terms for a p-series ∑ ∞ n=1 1 n p , we obtain L = lim n→∞     a n+1 a n     = lim n→∞ 1 (n+1) p 1 n p = lim n→∞ n p (n + 1) p = lim n→∞  n n + 1  p = 1 p = 1 Thus, all p-series yield a limit of 1.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

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

    (Publisher)

  If there are negative terms, check out Section 23.7 first to see how to handle them. Otherwise, if everything’s positive, proceed through the tests below. 23.3 How to Use the Ratio Test Use the Ratio Test whenever factorials are involved. Remember, facto-rials involve exclamation points, such as in n ! or (2 n +5)!. The Ratio Test is also often useful when there are exponentials around, such as 2 n or ( -5) 3 n . Here’s the statement of the test, summarized from what we found in Section 22.5.1 of the previous chapter: if L = lim n →∞ a n +1 a 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 Ratio Test, always start with the following framework: lim n →∞ a n +1 a n = lim n →∞ n th term with n replaced by ( n + 1) n th term . Make sure you use a bigass fraction bar, since you may have to write a fraction over a fraction. The n th term of the series is just a n , whereas if you replace n by ( n + 1) wherever you see it, you get a n +1 instead. Anyway, now you have to find the above limit; let’s say you’ve done that and got an answer L . There are three possibilities: 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 Ratio Test is useless. Try something else. Now let’s look at some examples. First, consider ∞ X n =1 n 1000 2 n . Section 23.3: How to Use the Ratio Test • 505 It’s not a geometric series because the numerator is a polynomial. Since exponentials grow faster than polynomials (see Section 21.3.3 in Chapter 21), the limit of the n th term is zero: lim n →∞ n 1000 2 n = 0 . So we can’t use the n th term test. Since the series involves exponentials, let’s try the Ratio Test. Following the standard framework, we start with lim n →∞ a n +1 a n = lim n →∞ ( n + 1) 1000 2 n +1 n 1000 2 n .

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Anton's Calculus

  Early Transcendentals

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

    (Publisher)

  57. Writing What does the Ratio Test tell you about the conver- gence of a geometric series? Discuss similarities between geometric series and series to which the Ratio Test applies. 58. Writing Given an infinite series, discuss a strategy for deciding what convergence test to use. QUICK CHECK ANSWERS 9.5 1. diverges; 1∕k 2∕3 2. converges; ratio 3. diverges; ratio 4. converges; root 9.6 ALTERNATING SERIES; ABSOLUTE AND CONDITIONAL CONVERGENCE Up to now we have focused exclusively on series with nonnegative terms. In this section we will discuss series that contain both positive and negative terms. ALTERNATING SERIES Series whose terms alternate between positive and negative, called alternating series, are of special importance. Some examples are ∞ ∑ k = 1 (−1) k+1 1 k = 1 − 1 2 + 1 3 − 1 4 + 1 5 − ⋯ ∞ ∑ k = 1 (−1) k 1 k = −1 + 1 2 − 1 3 + 1 4 − 1 5 + ⋯ In general, an alternating series has one of the following two forms: ∞ ∑ k = 1 (−1) k+1 a k = a 1 − a 2 + a 3 − a 4 + ⋯ (1) ∞ ∑ k = 1 (−1) k a k = −a 1 + a 2 − a 3 + a 4 − ⋯ (2) where the a k ’s are assumed to be positive in both cases. The following theorem is the key result on convergence of alternating series. 9.6.1 theorem (Alternating Series Test) An alternating series of either form (1) or form (2) converges if the following two conditions are satisfied: (a) a 1 ≥ a 2 ≥ a 3 ≥ ⋯ ≥ a k ≥ ⋯ (b) lim k →+∞ a k = 0 proof We will consider only alternating series of form (1). The idea of the proof is to show that if conditions (a) and (b) hold, then the sequences of even-numbered and odd- numbered partial sums converge to a common limit S. It will then follow from Theorem 9.1.4 that the entire sequence of partial sums converges to S. It is not essential for condition (a) in Theorem 9.6.1 to hold for all terms; an alternating series will converge if condition (b) is true and condition (a) holds eventually.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  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.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  A Mathematical Bridge

  An Intuitive Journey in Higher Mathematics

  • Stephen Hewson(Author)
  • 2009(Publication Date)
  • WSPC

    (Publisher)

  Thus the sum of the a n 0 + r is absolutely convergent. To conclude, consider the following decomposition N summationdisplay n =0 a n = n 0 summationdisplay n =1 a n + N summationdisplay n = n 0 +1 a n This converges because the first piece is just a summation of a finite number of terms, whereas we have just shown that the second piece is absolutely convergent. Thus we have demonstrated that if the ratio of the terms is less than 1, then the series converges. Similar manipulations are used to prove divergence if the modulus of the ratio is greater than 1. square Two comments concerning the Ratio Test are in order. First, the limit of the ratio of the terms must be strictly less than or greater than 1 for the Ratio Test to work. For example, we cannot use the Ratio Test to determine the convergence or divergence of the series ∑ n α for fixed α . To see why, look at the ratio vextendsingle vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle vextendsingle = ( n + 1) α n α = parenleftbigg 1 + 1 n parenrightbigg α Although the right-hand side is not equal to 1 for any finite value of n , in the infinite limit the ratio actually becomes equal to 1, as may be proved 146 A Mathematical Bridge by expanding out the bracket. Therefore, since the limit of the ratio of the terms is 1, the Ratio Test tells us nothing about the convergence properties: as far as the Ratio Test is concerned the series may either converge or diverge for any value of α . To see the positive power of the Ratio Test we look at the expansion for the very important exponential series 10 S ( z ) = ∞ summationdisplay n =0 z n n ! z ∈ C This is an interesting series to investigate because for large n and | z | we have very huge values for both | z n | and n !. It is not at all clear as to whether this sum is finite or not.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Introduction to Analysis

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

    (Publisher)

  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). Thus, this series absolutely converges according to the Root Test, even though the Ratio Test is inconclu-sive. squaresolid 4.3.4 Review of important concepts from Section 4.3 1. Dirichlet’s Test (Theorem 4.23 ) is our first convergence test for se-ries whose terms need not be nonnegative. The Alternating Series Test (Theorem 4.24 ) is an easy corollary of Dirichlet’s Test. 2. The notion of Absolute and Conditional Convergence is defined in Definition 4.27. Absolute convergence implies convergence (Theo-rem 4.28 ). 3. The Ratio and Root Tests (Theorems 4.31 and 4.32 ) are common ways of testing for absolute convergence. Exercises for Section 4.3 1. Determine if the following converge conditionally, converge abso-lutely, or diverge. (a) � ( − 1) n ln n (d) � ( − n )2 n 3 − n n π n (b) � ( − 1) n sin( n n 2 2 ) ( n +1) n n (e) � ( − 1) n n ! (c) � r n n ! for any r ∈ R (f) � n n 2. Give an example showing that the assumption that ( b n ) is decreas-ing cannot be removed from the Alternating Series Test (Theorem 4.24 ). 3. Suppose that b n 0, and that ( b n ) is decreasing. Show that the → following series converges: b 1 + b 2 − b 3 − b 4 + b 5 + b 6 − b 7 − b 8 + . · · · 2 4. Is it always the case that if � a n converges, then � a n converges? (Compare this with Exercise 6 from the last section.) 5. The following is known as Abel’s Test : Suppose that � a n con-verges and ( b n ) is monotone and bounded. Prove � a n b n converges. [Hint: Use Dirichlet’s Test (Theorem 4.23 ).] 244 Introduction to Analysis 6. Suppose a n > 0, and that � a n converges.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Real and Complex Analysis

  • Christopher Apelian, Steve Surace(Authors)
  • 2009(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  To prove part b ), note that there exists a positive integer N such that for j > N , | x j +1 | | x j | -r < r -1 2 . This in turn implies that for j > N , | x j +1 | > 1 + r 2 | x j | , or equivalently, for t ≡ 1+ r 2 > 1, | x j | > t j -N -1 | x N +1 | for j ≥ N + 2. From this it follows that lim x j = 0. Therefore, by the test for divergence, the series ∑ ∞ j =1 x j diverges. Example 4.30 Fix z 0 ∈ C . We will show that the series ∑ ∞ n =0 z n 0 n ! converges absolutely by use of the Ratio Test. Simply consider the limit lim z n +1 0 / ( n + 1)! z n 0 /n ! = lim z 0 n + 1 = 0. Note that since z 0 ∈ C was arbitrary, this result holds for any z 0 ∈ C . The idea behind the truth of parts a ) and b ) of the Ratio Test is that, in each of these two cases, the tails of the given series can be seen to “look like” that of a convergent or a divergent geometric series, respectively. Since the convergence behavior of a series depends only on the behavior of its tail, we 122 LIMITS AND CONVERGENCE obtain a useful test for convergence in these two cases. Case c ) is a case that the Ratio Test is not fine enough to resolve. Other methods need to be tried in this case to properly determine the series” behavior. Similar reasoning gives rise to another useful test that is actually more general than the Ratio Test, although not always as easy to apply. That test is the root test. Our version of the root test defines ρ as a lim sup rather than as a limit, which gives the test more general applicability since the lim sup always exists (if we include the possible value of ∞ ), whereas the limit might not. Of course, when the limit does exist, it equals the lim sup . Theorem 4.31 (The Root Test) Let { x j } be a sequence of nonzero elements of X , and suppose lim sup j | x j | = ρ exists. Then, a ) ρ < 1 ⇒ ∑ ∞ j =1 | x j | converges, and hence, ∑ ∞ j =1 x j converges. b ) ρ > 1 ⇒ ∑ ∞ j =1 x j diverges. c ) ρ = 1 ⇒ The test is inconclusive.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Calculus Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Explain your answer. ■ 21. The limit comparison test decides convergence based on a limit of the reciprocal of the quotient of consecutive terms in a series. 22. If lim k→∞ u k+1 u k = 7, then ∑ u k diverges. 23. If → ∞k 2 u k = 7, then ∑ u k converges. 24. The Ratio Test decides convergence based on a limit of kth roots of terms in the sequence of partial sums for a series. 25–49 Use any method to determine whether the series con- verges. ■ ∞ ∑ k=0 12 k k! 25. ∞ ∑ k=1 √ (2k)! (k + 5)! 26. ∞ ∑ k=1 (k + 3) 4 2 k 27. ∞ ∑ k=1 2k 2 + 3 3k 3 − 2 28. ∞ ∑ k = 1 k 100 e −k 29. ∞ ∑ k=1 e k − k + 1  k + k 30. ∞ ∑ k = 1 √ k k 3 + 3 31. ∞ ∑ k=1 3 √ k 5 − 1 2k 3 + k 32. ∞ ∑ k=1 5k 4 − √ 3k 12k 5 + 7k 33. ∞ ∑ k=1 k 5∕2 + 5 (2k − 1) √ 9k 3 + k 34. ∞ ∑ k=1 3 + √ k (k + 1) 3 − 1 35. ∞ ∑ k=1 2 k sin 2 (k) 3 k−4 + 5 36. ∞ ∑ k=1 3 k (k!) 2 (2k)! 37. ∞ ∑ k = 1 (k + 1)! k k 38. ∞ ∑ k=1 2 ln k e k 39. ∞ ∑ k = 1 k! e k 3 40. ∞ ∑ k=0 (k + 3)! 3!k!3 k 41. ∞ ∑ k=1 ( 2k k + 1 ) k 2 42. ∞ ∑ k = 1 1 8 + 2 −k 43. ∞ ∑ k=1 tan −1 (k!) √ k + 2 44. ∞ ∑ k=1 (3k 2 + 1) k∕2 (2k − 1) k 45. ∞ ∑ k=2 1 ln (k!) 46. ∞ ∑ k=1 2 (k!) 2 (2k)! 47. ∞ ∑ k=1 [e(k + 1)] k k k+1 48. 49. ∞ ∑ k=1 ln k 5 k 50. For what values of  does the series ∑ ∞ k=1 ( k ∕k  ) converge? 51–52 Find the general term of the series and use the Ratio Test to show that the series converges. ■ 51. 1 + 1 ⋅ 2 1 ⋅ 3 + 1 ⋅ 2 ⋅ 3 1 ⋅ 3 ⋅ 5 + 1 ⋅ 2 ⋅ 3 ⋅ 4 1 ⋅ 3 ⋅ 5 ⋅ 7 + ⋯ 52. 1 + 1 ⋅ 3 3! + 1 ⋅ 3 ⋅ 5 5! + 1 ⋅ 3 ⋅ 5 ⋅ 7 7! + ⋯ 53. Show that ln x < √ x if x > 0 and use this result to investi- gate the convergence of ∞ ∑ k=1 ln k 3 a. ∞ ∑ k=2 1 (ln k) 2 b. 10.6 Alternating Series; Absolute and Conditional Convergence 687 FOCUS ON CONCEPTS 54. a. Make a conjecture about the convergence of the series ∑ ∞ k=1 sin(2∕k) by considering the local linear approximation of sin x at x = 0. b. Try to confirm your conjecture using the limit com- parison test.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Real Analysis

  A Constructive Approach Through Interval Arithmetic

  • Mark Bridger(Author)
  • 2019(Publication Date)
  • American Mathematical Society

    (Publisher)

  Later, we will be able to use the integral test (Theorem 5.3.27) to deal with series of this type. Proposition 3.3.25 (Ratio Comparison Test) Suppose for su ! ciently large n the numbers v n and e n are non-negative, and v n +1 v n e n +1 e n . If the series 4 X n =0 e n converges, so does 4 X n =0 v n . Proof. For n some Q , v n +1 e n +1 v n e n , so the sequence of ratios v n e n is decreasing. If = v Q e Q , then v n e n for n A Q ; so 4 X n =0 v n converges by the Comparison Test. So far we have been dealing with series whose terms are either all positive or at least eventually positive. Now we turn our attention to more general series whose terms don’t satisfy these conditions. Definition 3.3.26 P 4 n =0 v n is said to be absolutely convergent if P 4 n =0 | v n | con-verges. If a series converges but is not absolutely convergent, it is called conditionally convergent. Proposition 3.3.27 Every absolutely convergent series is convergent. SERIES OF NUMBERS 121 Proof. Let V q = P q n =0 v n and W q = P q n =0 | v n | . Since ( W q ) converges, it satis es the Cauchy criterion; we must show that ( V q ) does as well. | V q V p | = ¯ ¯ ¯ ¯ ¯ q X n = p +1 v n ¯ ¯ ¯ ¯ ¯ q X n = p +1 | v n | (by the triangle inequality) = | W p W q | Since ( W q ) is a Cauchy sequence, there is an Q ( ) that makes | W p W q | ? for all p> q A Q ( ) , so the same Q ( ) works for | V p V q | . Corollary 3.3.28 (Absolute Ratio Test) Suppose lim n $4 ¯ ¯ ¯ ¯ v n +1 v n ¯ ¯ ¯ ¯ = u ? 1 . Then 4 X n =0 v n is absolutely convergent, so it converges. Example 3.3.29 For the series 4 X n =0 ( 1) n 2 n { n n , we have lim n $4 ¯ ¯ ¯ ¯ v n +1 v n ¯ ¯ ¯ ¯ = lim n $ 0 n n + 1 | 2 { | = | 2 { | = When 1 @ 2 ? { ? 1 @ 2 , the series converges absolutely. When { = 1 @ 2 , we get the harmonic series, which diverges. When { = 1 @ 2 we get the alternating harmonic series which converges, as we shall see below.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  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.

  Sign up to read

  Learn more about book
• 

  eBook - ePub

  Infinite Sequences and Series

  • Konrad Knopp(Author)
  • 2012(Publication Date)
  • Dover Publications

    (Publisher)

  Cauchy 1821):

  whose more detailed formulation asserts: If , from a certain stage on, does not exceed a fixed positive number a < 1, then Σ

  aν

  is convergent. We expressly emphasize that it is a fixed number a , which is less than 1, that is not to be exceeded, because beginners very often overlook this.7

  Likewise, if we take

  cν

  =

  aν

  , 3.1,(2) yields:

  Here again the divergence half is trivial, for it asserts that the sequence {

  aν

  } of terms of the series increases monotonically, and therefore is certainly no null sequence. The hereby acquired convergence criterion

  is commonly designated as the Ratio Test (Cauchy 1821).

  Before applying these tests to given series, we shall make them handier by means of a couple of remarks :

  1. It is often not at all easy to decide for a sequence such as or whether its terms exceed a fixed proper fraction8 a from a certain stage on. It is usually easier, however, to ascertain the limit of such a sequence, or, if it has no limit, its principal limits. In terms of these we have: If

  then Σ

  aν is convergent

  .

  Indeed, if , and if we set, say, , then 0 < a < 1, and, from a certain stage on, (cf 2.2 ,4 ), so that Σ

  aν

  is convergent. We call (3) the limit form of the radical test. Analogously we have: If

  then Σ

  aν is convergent

  . The matter is less simple for the corresponding divergence tests. We have:

  For this inequality means that infinitely often , and hence also

  aν

  > 1, and therefore {

  aν

  } is not a null sequence. No decision, however, is afforded by , (or even lim exists and = 1). For if we take

  aν

  = 1/

  να

  , then, no matter what value a may denote, (see 2.4 ,4 ), whereas Σ

  aν

  converges for α > 1, diverges for α < 1.

  3.2.1. Examples. In the following series, let x be a positive number. It does not matter whether the summation begins with ν = 0 or ν = 1, because we are only interested in investigating the convergence of the series. In each example we denote the terms of the series under investigation by

  aν

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  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)

  Notice that most of the series in Exercises 11.4 have this form. (The value of p should be chosen as in Section 11.4 by keeping only the highest powers of n in the numerator and denominator.) The comparison tests apply only to series with positive terms, but if  a n has some negative terms, then we can apply a comparison test to  | a n | and test for absolute convergence. 11.7 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. 780 CHAPTER 11 Sequences, Series, and Power Series 5. Alternating Series Test If the series is of the form  s21d n21 b n or  s21d n b n , then the Alternating Series Test is an obvious possibility. Note that if  b n converges, then the given series is absolutely convergent and therefore convergent. 6. Ratio Test Series that involve factorials or other products (including a constant raised to the nth power) are often conveniently tested using the Ratio Test. Bear in mind that | a n11 ya n | l 1 as n l ` for all p-series and therefore all rational or algebraic functions of n. Thus the Ratio Test should not be used for such series. 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.

  Sign up to read

  Learn more about book