Convergence Tests

12 Key excerpts on "Convergence Tests"

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  A Radical Approach to Real Analysis

  Second Edition

  • David Bressoud(Author)
  • 2006(Publication Date)
  • American Mathematical Society

    (Publisher)

  Its justification rests on the Cauchy criterion. 130 4 The Convergence of Infinite Series Theorem 4.6 (The Comparison Test). Let a 1 + a 2 + a 3 + · · · and b 1 + b 2 + b 3 + · · · be two series with summands that are greater than or equal to zero. We assume that each b i is greater than or equal to the corresponding a i : b 1 ≥ a 1 ≥ 0 , b 2 ≥ a 2 ≥ 0 , b 3 ≥ a 3 ≥ 0 , . . . If b 1 + b 2 + b 3 + · · · converges, then so does a 1 + a 2 + a 3 + · · · . If a 1 + a 2 + a 3 + · · · diverges, then so does b 1 + b 2 + b 3 + · · · . Proof: Let S n = a 1 + a 2 + · · · + a n and T n = b 1 + b 2 + · · · + b n . If m < n , then 0 ≤ S n − S m = a m + 1 + a m + 2 + · · · + a n ≤ b m + 1 + b m + 2 + · · · + b n = T n − T m , and so | S n − S m | ≤ | T n − T m | . (4.8) We assume the series b 1 + b 2 + b 3 + · · · converges. Given a positive bound , we have a response N . Equation (4.8) shows us that the same response will work for the series a 1 + a 2 + a 3 + · · · . The contrapositive of what we have just proven says that if a 1 + a 2 + a 3 + · · · diverges, then b 1 + b 2 + b 3 + · · · diverges. Q.E.D. The Ratio Test The ratio and root test rely on comparing our series to a geometric series. They are very simple and powerful techniques that quickly yield one of three conclusions: 1. the series in question converges absolutely, 2. the series in question diverges, or 3. the results of this test are inconclusive. It is the third possibility that is the principal drawback of these tests. The most interesting series mathematicians and scientists were encountering in the early 1800s all fell into category 3. Nevertheless, these tests are important because they are simple. Start with one of these tests, and move on to a more complicated test only if the results are inconclusive. Theorem 4.7 (The Ratio Test). Given a series with nonzero summands, a 1 + a 2 + a 3 + · · · , we consider the ratio r ( n ) = | a n + 1 /a n | .

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  Calculus

  Single Variable

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

    (Publisher)

  Later, we will use some of these tests to study the convergence of Taylor series. The Comparison Test We will begin with a test that is useful in its own right and is also the building block for other important Convergence Tests. The underlying idea of this test is to use the known convergence or divergence of a series to deduce the convergence or divergence of another series. 9.5 The Comparison, Ratio, and Root Tests 535 Theorem 9.5.1: Comparison Test Let ∑ ∞ k=1 a k and ∑ ∞ k=1 b k be series with nonnegative terms and suppose that a 1 ≤ b 1 , a 2 ≤ b 2 , a 3 ≤ b 3 , . . . , a k ≤ b k , . . . (a) If the “bigger series” Σb k converges, then the “smaller series” Σa k also converges. (b) If the “smaller series” Σa k diverges, then the “bigger series” Σb k also diverges. In Theorem 9.5.1, it is not essential that the condition a k ≤ b k hold for all k, as stated; the conclusions of the theorem remain true if this condition is eventually true. We have left the proof of this theorem for the exercises; however, it is easy to visualize why the theorem is true by interpreting the terms in the series as areas of rectangles (Figure 9.5.1). The comparison test states that if the total area ∑ b k is finite, then the total area ∑ a k must also be finite; and if the total area ∑ a k is infinite, then the total area ∑ b k must also be infinite. 1 . . . . . . . . . . . . a 1 b 1 2 a 2 b 2 3 a 3 b 3 4 a 4 b 4 5 a 5 b 5 k a k b k For each rectangle, a k denotes the area of the blue portion and b k denotes the combined area of the white and blue portions. FIGURE 9.5.1 Using the Comparison Test There are two steps required for using the comparison test to determine whether a series ∑ u k with positive terms converges: Step 1. Guess whether the series ∑ u k converges or diverges. Step 2. Find a series that proves the guess to be correct.

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  CounterExamples

  From Elementary Calculus to the Beginnings of Analysis

  • Andrei Bourchtein, Ludmila Bourchtein(Authors)
  • 2014(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  Numerical series: Convergence Tests Convergence of geometric series . A geometric series ∑ q n ( q = const ) is convergent if | q | < 1 and divergent if | q | ≥ 1. Convergence of p -series . A p -series ∑ 1 n p ( p = const ) is convergent if p > 1 and divergent if p ≤ 1. The Integral test . Suppose f ( x ) is positive and decreasing function on [1 , + ∞ ) and a n = f ( n ) , n ∈ N . Then the improper integral ∫ + ∞ 1 f ( x ) dx is convergent if, and only if, the series ∑ a n is convergent. The Comparison test . Suppose ∑ a n and ∑ b n are positive series (that is, series with positive terms). Then 1) if ∑ b n is convergent and a n ≤ b n , n ∈ N , then ∑ a n is also convergent; 2) if ∑ b n is divergent and a n ≥ b n , n ∈ N , then ∑ a n is also divergent. The Limit Comparison test . Suppose ∑ a n and ∑ b n are positive series. If lim n →∞ a n b n = c , where c is a positive constant, then either both series converge or both diverge. The Leibniz (alternating series) test . If an alternating series ∑ ( − 1) n − 1 b n , b n > 0 , n ∈ N satisfies the two conditions 1) b n +1 ≤ b n , n ∈ N 2) lim n →∞ b n = 0 , then the series is convergent. Absolute convergence . A series ∑ a n is called absolutely convergent if the series of the absolute values ∑ | a n | is convergent. Theorem . If a series is absolutely convergent, then it is convergent. Conditional convergence . A series is conditionally convergent if it is convergent, but not absolutely convergent. 178 Counterexamples: From Calculus to the Beginnings of Analysis The Cauchy (root) test . Let ∑ a n be a positive series and lim n →∞ a 1 n n = C . Then: 1) if C < 1, then ∑ a n converges; 2) if C > 1, then ∑ a n diverges; 3) if C = 1 or the limit does not exist, then the test is inconclusive. The D’Alembert (ratio) test . Let ∑ a n be a positive series and lim n →∞ a n a n +1 = D .

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  Calculus Early Transcendentals

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

    (Publisher)

  Later, we will use some of these tests to study the convergence of Taylor series. The Comparison Test We will begin with a test that is useful in its own right and is also the building block for other important Convergence Tests. The underlying idea of this test is to use the known convergence or divergence of a series to deduce the convergence or divergence of another series. In Theorem 10.5.1, it is not essential that the condition a k ≤ b k hold for all k, as stated; the conclusions of the theorem remain true if this condition is eventually true. Theorem 10.5.1: Comparison Test Let ∑ ∞ k=1 a k and ∑ ∞ k=1 b k be series with nonneg- ative terms and suppose that a 1 ≤ b 1 , a 2 ≤ b 2 , a 3 ≤ b 3 , …, a k ≤ b k , … (a) If the “bigger series” ∑ b k converges, then the “smaller series” ∑ a k also con- verges. (b) If the “smaller series” ∑ a k diverges, then the “bigger series” ∑ b k also diverges. We have left the proof of this theorem for the exercises; however, it is easy to visualize why the theorem is true by interpreting the terms in the series as areas of rectangles (Figure 10.5.1). The comparison test states that if the total area ∑ b k is finite, then the total area ∑ a k must also be finite; and if the total area ∑ a k is infinite, then the total area ∑ b k must also be infinite. 1 . . . . . . . . . . . . a 1 b 1 2 a 2 b 2 3 a 3 b 3 4 a 4 b 4 5 a 5 b 5 k a k b k For each rectangle, a k denotes the area of the blue portion and b k denotes the combined area of the white and blue portions. ▴ Figure 10.5.1 Using the Comparison Test There are two steps required for using the comparison test to determine whether a series ∑ u k with positive terms converges: Step 1. Guess whether the series ∑ u k converges or diverges. Step 2. Find a series that proves the guess to be correct. That is, if we guess that ∑ u k diverges, we must find a divergent series whose terms are “smaller” than

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  Mathematical Methods for Physicists

  • George B. Arfken(Author)
  • 2013(Publication Date)
  • Academic Press

    (Publisher)

  It therefore becomes a matter of extreme importance to be able to tell whether a given series is convergent. We shall develop a number of possible tests, starting with the simple and relatively insen-sitive tests and working up to the more complicated but quite sensitive tests. For the present let us consider a series of positive terms, a n > 0, postponing negative terms until the next section. Comparison Test If term by term a series of terms u n < a n , in which the a n form a convergent series, the series £„ u n is also convergent. Symbolically, we have Convergence Tests 281 Σ α η = α 1 + α 2 +03 + n £u„ = Ul +u 2 + u 3 + convergent, If u n < a n for all n, then £„ u n < £„ α η and £„ u n therefore is convergent. If term by term a series of terms v n >b n , in which the b n form a divergent séries, the series £„ v n is also divergent. Note that comparisons of u n with b n or v n with a„ yield no information. Here we have ZK = b 1 +b 2 + b 3 + n Σ ν η = νΐ +V 2 + V 3 + divergent, If v n > b n for all n, then £„ v n > ^„ b n and ^„ v n therefore is divergent. For the convergent series a n we already have the geometric series, whereas the harmonic series will serve as the divergent series b n . As other series are identified as either convergent or divergent, they may be used for the known series in this comparison test. All tests developed in this section are essentially comparison tests. Figure 5.1 exhibits these tests and the interrelationships. ( Cauchy ^V f root J V Kummer. (Comparison with geometric series) a n — 1 ( D'Alembert, A Cauchy ratio J (Also by comparison with geometric series) C ) Γ Euler-Maclaurin I integral ) a n (Comparison with integral) Raabe a n = n In n 3 Gauss J> FIG. 5.1 Comparison tests EXAMPLE 5.21 The p Series Test Ση η ~ Ρ > P = 0.999, for convergence. Since n ~ 0 9 > n~ and b n = n _1 forms the divergent harmonic series, the comparison test shows that X„«~ 0 ' 999 is divergent.

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  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.

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  Calculus

  Late Transcendental

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

    (Publisher)

  The underlying idea of this test is to use the known convergence or divergence of a series to deduce the convergence or divergence of another series. It is not essential in Theorem 9.5.1 that the condition a k ≤ b k hold for all k, as stated; the conclusions of the theorem remain true if this condition is eventu- ally true. 9.5.1 THEOREM (The Comparison Test) Let ∑ ∞ k=1 a k and ∑ ∞ k=1 b k be series with non- negative terms and suppose that a 1 ≤ b 1 , a 2 ≤ b 2 , a 3 ≤ b 3 , . . . , a k ≤ b k , . . . (a) If the “bigger series” Σb k converges, then the “smaller series” Σa k also con- verges. (b) If the “smaller series” Σa k diverges, then the “bigger series” Σb k also diverges. We have left the proof of this theorem for the exercises; however, it is easy to visual- ize why the theorem is true by interpreting the terms in the series as areas of rectangles (Figure 9.5.1). The comparison test states that if the total area ∑ b k is finite, then the total Figure 9.5.1 area ∑ a k must also be finite; and if the total area ∑ a k is infinite, then the total area ∑ b k must also be infinite. USING THE COMPARISON TEST There are two steps required for using the comparison test to determine whether a series ∑ u k with positive terms converges: Step 1. Guess at whether the series ∑ u k converges or diverges. Step 2. Find a series that proves the guess to be correct. That is, if we guess that ∑ u k diverges, we must find a divergent series whose terms are “smaller” than the corresponding terms of ∑ u k , and if we guess that ∑ u k converges, we must find a convergent series whose terms are “bigger” than the corresponding terms of ∑ u k . In most cases, the series ∑ u k being considered will have its general term u k expressed as a fraction. To help with the guessing process in the first step, we have formulated two principles that are based on the form of the denominator for u k . These principles sometimes suggest whether a series is likely to converge or diverge.

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  Calculus

  Concepts and Contexts, Enhanced Edition

  • James Stewart(Author)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  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. 580 CHAPTER 8 INFINITE SEQUENCES AND SERIES Note: The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series The inequality is useless as far as the Comparison Test is concerned because is convergent and . Nonetheless, we have the feeling that ought to be convergent because it is very similar to the convergent geometric series . In such cases the fol-lowing test can be used. The Limit Comparison Test Suppose that and are series with positive terms. If where c is a finite number and , then either both series converge or both diverge. Although we won’t prove the Limit Comparison Test, it seems reasonable because for large . Using the Limit Comparison Test Test the series for convergence or divergence. SOLUTION We use the Limit Comparison Test with and obtain Since this limit exists and is a convergent geometric series, the given series con-verges by the Limit Comparison Test. Estimating the Sum of a Series Suppose we have been able to use the Integral Test to show that a series is convergent and we now want to find an approximation to the sum of the series. Of course, any par-tial sum is an approximation to because .

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  Introduction to Analysis

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

    (Publisher)

  4 Series of Real Numbers Chapter Summary: The theory of infinite series is a natural next topic, since convergence of these series is defined by convergence of their associated sequence of partial sums. We introduce infinite series in Section 4.1, present various Convergence Tests in Sections 4.2 (for nonnegative terms) and 4.3. We close the chapter by presenting information concerning rearrangements of series, including Rie-mann’s Rearrangement Theorem in Section 4.4. Section 4.1: Infinite Series. Infinite series are defined here. Telescop-ing and geometric series are discussed, along with the Test for Divergence (Theorem 4.6 ), a result concerning sums, differences, and constant multiples of series (Theorem 4.9 ), and the Cauchy Criterion (Theorem 4.11 ). Section 4.2: Convergence Tests for Series with Nonnegative Terms. We present Convergence Tests for series with nonnegative terms. We use the Cauchy Condensation Test (Theorem 4.14 ) to determine the conver-gence or divergence of p − series in Theorem 4.15, and then present the Direct and Limit Comparison Tests (Theorems 4.17 and 4.20 ). Section 4.3: Other Convergence Tests for Series. We discuss con-vergence of series whose terms need not be nonnegative. The Alternating Se-ries Test (Theorem 4.24 ) follows immediately from Dirichlet’s Test (Theorem 4.23 ). Absolute and conditional convergence is discussed, and the Ratio and Root Tests (Theorems 4.31 and 4.32 ) are presented. Section 4.4: Rearrangements. Riemann’s Rearrangement Theorem (Theorem 4.40 ) is presented. In addition, absolutely convergent series remain absolutely convergent under rearrangement and converge to the same value (Theorem 4.38 ). Section 4.5: Troubleshooting Guide for Series of Real Numbers. A list of frequently asked questions and their answers, arranged by section. Come here if you are struggling with any material from this chapter. 219

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  Real and Complex Analysis

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

    (Publisher)

  The following test for convergence is of great use in proving theorems. Of course, it can also be used as a practical tool to show that a series converges. Proposition 4.21 (The Comparison Test) Suppose 0 ≤ x j ≤ y j for j = 1, 2, . . . . Then the following are true: a ) If ∑ ∞ j =1 y j converges, then ∑ ∞ j =1 x j converges, and ∑ ∞ j =1 x j ≤ ∑ ∞ j =1 y j . b ) If ∑ ∞ j =1 x j diverges, then ∑ ∞ j =1 y j diverges. P ROOF To prove a ), let s n = ∑ n j =1 x j , and t n = ∑ n j =1 y j be the n th par-tial sums of the two series, and assume that ∑ ∞ j =1 y j converges. By Propo-sition 2.2, the sequence of partial sums { t n } is bounded above. Clearly, we have s n ≤ t n for any n , and so the sequence { s n } is also bounded above. But since x j ≥ 0 for all j , it follows that the sequence { s n } is increasing, and so it must be convergent according to the monotone sequence theorem. It easily follows that ∑ ∞ j =1 x j ≤ ∑ ∞ j =1 y j . The proof of b ) is a direct consequence of part a ), since, according to a ) if ∑ ∞ j =1 y j converges, then ∑ ∞ j =1 x j converges. This means that the divergence of ∑ ∞ j =1 x j implies the divergence of ∑ ∞ j =1 y j . 3.47 Show that the condition “ 0 ≤ x j ≤ y j for j = 1, 2, . . . in the comparison theorem can be weakened to “ 0 ≤ x j ≤ y j for j ≥ N for some N ∈ N and the same conclusion still holds. Example 4.22 Consider the series given by ∑ ∞ j =1 ( j 2 j +1 ) j . Clearly for all j ≥ 1 we have 0 ≤ ( j 2 j +1 ) j ≤ ( 1 2 ) j , where ( 1 2 ) j are the terms of a convergent ge-ometric series. Therefore, by the comparison test, ∑ ∞ j =1 ( j 2 j +1 ) j converges. We now state a particularly potent test for convergence. Its proof relies on the comparison test. Theorem 4.23 (The Cauchy Condensation Test) Suppose { x j } is a nonincreasing sequence, and x j ≥ 0 for all j . Then the series ∑ ∞ j =1 x j and ∑ ∞ j =0 2 j x 2 j both converge, or both diverge.

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  Mathematical Analysis Explained

  • Neil A Watson(Author)
  • 1993(Publication Date)
  • WSPC

    (Publisher)

  Chapter 5 FURTHER RESULTS ON INFINITE SERIES Tests for convergence To use the comparison test (Theorem 2.9) on a particular series, you need to have a second series that you already know about and can relate to the given one. The first two theorems of this chapter give tests for convergence which do not explicitly require knowledge of a second series; that knowl-edge is built into the proofs, where the comparison test is used. The first theorem requires knowledge of n-th roots, so we discuss them first. Given any n G N , let f(x) = x n for all x > 0. For each 6 > 0, / is continuous on [0,6] and differentiable on ]0,6[, with f'(x) = nx n ~ l > 0 . Therefore, by Theorem 4.6, Corollary 2, / is strictly increasing on [0,6]. By Theorem 3.10, / therefore has an inverse function g, which is continuous and strictly increasing on [0,6 n ], such that g(y) n = y for all ye[0,b n ] , and g(x n ) = x for all a: €[0,6] . Since this holds for every 6 > 0, g is defined on [0, oo[, and these identities are valid for all re, y G [0, oo[. We denote g(y) by y 1 / 7 1 , so that the identities become (y 1 / n ) n = y a n ( j ( r c n j l / n = a ? > The function g is called the n-th root function, and y 1 ! 71 is called the n-th root of y. By Theorem 4.3, g is differentiable on ]0, oo[ with >( = 1 = i = y 1/n = y ( 1 / w M f'(9(y)) n f o 1 / * ) * -! n(y l l n ) n n Theorem 5.1 (Cauchy's n-th root test.) Suppose that u n >0 for all n. (i) If ul/ n — • / < 1, then Y) Un is convergent. (ii) If i4/ n — • / > 1, or Un ~~ * oo, then Yl u n is divergent. 85 86 INFINITE SERIES The idea of the proof of (i) is to compare Y^Un with a convergent geometric series. We have to use the information that I < 1. Now 1 — / > 0, so we could use that as an € and get | t i £ / n -l | < l -J for all large n, which implies that v^ n < 1, or Un < 1, which does not help.

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  Introduction to Mathematics with Maple

  • P Adams, K Smith;R V??born??;;(Authors)
  • 2004(Publication Date)
  • WSPC

    (Publisher)

  These tests are based on comparison with another series known to be convergent. The slower the convergence of this latter series, the finer the test derived. We mention here a few more tests. 11.8.1 M o r e Convergence Tests. For series with non-negative and decreasing terms the following theorem is rather strong but has a simple elementary proof. 304 Introduction to Mathematics with Maple Ti 11 Theorem 11.12 (Cauchy's condensation test) a3 2 - - - 2 0, then the series C ai converges if and only if does. If a1 2 a2 2 Pa2i Proof. I. Assume 2ia2i converges. We obviously have i=O Consequently the partial sums of Cai are bounded and this series con- verges. 11. Assume C ai converges and put C ai = S. Then 00 i=l 2 a1 + f C 2aa2i 2 - C 2aa2i. i=2 k+l i=O 2 Consequently the partial sums of 22a2i are bounded and this series con- verges. Example 11.19 We consider again En- for a E Q (Examples 11.8, 11.9 and 11.10). Applying Cauchy's condensation test yields the result that C n- converges if and only if 2n(1-) does. This latter series converges if and only if 2l- < 1, that is, if and only if a! > 1. Series 305 Theorem 11.13 (Raabe’s test) r > 1 and N E N such that Let an > 0. If there exists an n ( 3 -1 ) >_r (11.24) for all n > N , then the series Cui is convergent. If, however, there exists an N1 E N such that n(-&l) N1 then Cui is divergent. Proof. It follows from (11.24) that nun 2 (r + n)an+l > (n + l)an+l, and Hence the sequence n H nun is strictly decreasing after its Nth term and it is also bounded below. Consequently it converges, which by the result of Exercise 11.1.2 implies the convergence of the series Since T - 1 > 0 it follows from (11.26) by the Comparison Test that the series C ak+l also converges. Turning our attention to divergence, we see that Inequality (11.25) im- plies that nun 5 (n + l)an+l and consequently ( N + l ) a ~ + l 5 nun for n 2 N . It follows that there is a c such that 0 < c < nun for all n.

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