Absolute and Conditional Convergence

7 Key excerpts on "Absolute and Conditional Convergence"

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  Calculus

  Late Transcendental

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

    (Publisher)

  As a matter of terminol- ogy, a series that converges but diverges absolutely is said to converge conditionally (or to be conditionally convergent). Thus, (8) is a conditionally convergent series. Example 5 In Example 1(b) we used the alternating series test to show that the se- ries ∞  k=1 (−1) k+1 k + 3 k(k + 1) converges. Determine whether this series converges absolutely or converges conditionally. Solution. We test the series for absolute convergence by examining the series of absolute values: ∞  k=1     (−1) k+1 k + 3 k(k + 1)     = ∞  k=1 k + 3 k(k + 1) Principle 9.5.3 suggests that the series of absolute values should behave like the divergent p-series with p = 1. To prove that the series of absolute values diverges, we will apply the limit comparison test with a k = k + 3 k(k + 1) and b k = 1 k We obtain ρ = lim k→+∞ a k b k = lim k→+∞ k(k + 3) k(k + 1) = lim k→+∞ k + 3 k + 1 = 1 Since ρ is finite and positive, it follows from the limit comparison test that the series of absolute values diverges. Thus, the original series converges and also diverges absolutely, and so converges conditionally. THE RATIO TEST FOR ABSOLUTE CONVERGENCE Although one cannot generally infer convergence or divergence of a series from absolute divergence, the following variation of the ratio test provides a way of deducing divergence from absolute divergence in certain situations. We omit the proof. 9.6 Alternating Series; Absolute and Conditional Convergence 559 9.6.5 THEOREM (Ratio Test for Absolute Convergence) Let ∑ u k be a series with nonzero terms and suppose that ρ = lim k →+∞ |u k+1 | |u k | (a) If ρ < 1, then the series ∑ u k converges absolutely and therefore converges. (b) If ρ > 1 or if ρ = +∞, then the series ∑ u k diverges. (c) If ρ = 1, no conclusion about convergence or absolute convergence can be drawn from this test. Example 6 Use the ratio test for absolute convergence to determine whether the se- ries converges.

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  Calculus Early Transcendentals

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

    (Publisher)

  For example, consider the two series 1 − 1 2 + 1 3 − 1 4 + ⋯ + (−1) k+1 1 k + ⋯ (8) − 1 − 1 2 − 1 3 − 1 4 − ⋯ − 1 k − ⋯ (9) Both of these series diverge absolutely, since in each case the series of absolute values is the divergent harmonic series 1 + 1 2 + 1 3 + ⋯ + 1 k + ⋯ However, series (8) converges, since it is the alternating harmonic series, and series (9) diverges, since it is a constant times the divergent harmonic series. As a matter of terminol- ogy, a series that converges but diverges absolutely is said to converge conditionally (or to be conditionally convergent). Thus, (8) is a conditionally convergent series. ▶ Example 5 The series ∞ ∑ k=1 (−1) k k + 4 k (k + 1) converges. Determine whether this series converges absolutely or converges conditionally. Solution We test the series for absolute convergence by examining the series of absolute values: ∞ ∑ k=1 | | | | (−1) k k + 4 k (k + 1) | | | | = ∞ ∑ k=1 k + 4 k (k + 1) 10.6 Alternating Series; Absolute and Conditional Convergence 693 Principle 10.5.3 suggests that the series of absolute values should behave like the divergent p-series with p = 1. To prove that the series of absolute values diverges, we will apply the limit comparison test with a k = k + 4 k (k + 1) and b k = 1 k We obtain  = lim k→+∞ a k b k = lim k→+∞ k (k + 4) k (k + 1) = lim k→+∞ k + 4 k + 1 = 1 Since  is finite and positive, it follows from the limit comparison test that the series of absolute values diverges. Thus, the original series converges and also diverges absolutely, and so converges conditionally. The Ratio Test for Absolute Convergence Although one cannot generally infer convergence or divergence of a series from absolute divergence, the following variation of the ratio test provides a way of deducing divergence from absolute divergence in certain situations.

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  Calculus

  Single Variable

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

    (Publisher)

  As a matter of terminology, a series that converges but diverges absolutely is said to converge conditionally (or to be conditionally convergent). Thus, (8) is a conditionally convergent series. Example 5 In Example 1(b) we used the alternating series test to show that the series ∞  k=1 (−1) k+1 k + 3 k(k + 1) converges. Determine whether this series converges absolutely or converges conditionally. Solution We test the series for absolute convergence by examining the series of absolute values: ∞  k=1     (−1) k+1 k + 3 k(k + 1)     = ∞  k=1 k + 3 k(k + 1) Principle 9.5.3 suggests that the series of absolute values should behave like the divergent p-series with p = 1. To prove that the series of absolute values diverges, we will apply the limit comparison test with 9.6 Alternating Series; Absolute and Conditional Convergence 545 a k = k + 3 k(k + 1) and b k = 1 k We obtain ρ = lim k→+∞ a k b k = lim k→+∞ k(k + 3) k(k + 1) = lim k→+∞ k + 3 k + 1 = 1 Since ρ is finite and positive, it follows from the limit comparison test that the series of abso- lute values diverges. Thus, the original series converges and also diverges absolutely, and so converges conditionally. The Ratio Test for Absolute Convergence Although one cannot generally infer convergence or divergence of a series from absolute diver- gence, the following variation of the ratio test provides a way of deducing divergence from absolute divergence in certain situations. We omit the proof. Theorem 9.6.5: Ratio Test for Absolute Convergence Let ∑ u k be a series with nonzero terms and suppose that ρ = lim k →+∞ |u k+1 | |u k | (a) If ρ < 1, then the series ∑ u k converges absolutely and therefore converges. (b) If ρ > 1 or if ρ = +∞, then the series ∑ u k diverges. (c) If ρ = 1, no conclusion about convergence or absolute convergence can be drawn from this test. Example 6 Use the ratio test for absolute convergence to determine whether the series converges.

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  A Course in Real Analysis

  • Hugo D. Junghenn(Author)
  • 2015(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  6.4.1 Definition. A series ∑ a n is said to converge absolutely if ∑ | a n | converges. A convergent series that does not converge absolutely is said to converge conditionally . ♦ 6.4.2 Theorem. (a) If ∑ a n converges absolutely, then the series a n , a + n , and a -n converge and a n = a + n - a -n , | a n | = a + n + a -n . (b) If ∑ a n converges conditionally, then ∑ a + n and ∑ a -n diverge. 182 A Course in Real Analysis Proof. (a) If ∑ | a n | converges, then the inequalities 0 ≤ a ± n = 1 2 ( | a n | ± a n ) ≤ | a n | and the comparison test show that ∑ a + n and ∑ a -n converge. The remaining assertions in (a) follow from the identities a n = a + n -a -n and | a n | = a + n + a -n . (b) If ∑ a n and ∑ a -n converge, then ∑ | a n | = ∑ a n + 2 ∑ a -n converges. The same conclusion holds if ∑ a n and ∑ a + n converge. Hence if ∑ a n converges conditionally, then neither ∑ a + n nor ∑ a -n can converge. All series of the form ∑ ∞ n =1 ( -1) n +1 /n p , 0 < p ≤ 1 converge conditionally. This follows from the alternating series test given below. The following example is somewhat more interesting. 6.4.3 Example. We show that the series s := ∞ n =2 ( -1) n n p -1 -1 converges conditionally iff 1 / 2 < p ≤ 1 and absolutely iff p > 1. To see this, note first that if p < 0, then the n th term of the series does not tend to zero, and if p = 0 the series is undefined. So assume p > 0. If s n denotes the n th partial sum of the series, then s 2 n +1 = n k =1 1 (2 k ) p -1 -1 (2 k + 1) p + 1 = n k =1 ( α k + β k ) , (6.5) where α k := (2 k + 1) p -(2 k ) p (2 k ) p -1 (2 k + 1) p + 1 and β k := 2 (2 k ) p -1 (2 k + 1) p + 1 . By the mean value theorem applied to x p on the interval [2 k, 2 k + 1], α k = px p -1 k (2 k ) p -1 (2 k + 1) p + 1 , for some x k ∈ (2 k, 2 k + 1) . If 0 < p ≤ 1, then α k ≤ p (2 k ) 1 -p (2 k ) p -1 (2 k ) p + 1 = 1 (2 k ) 1 -p (2 k ) 2 p -1 ≤ 1 k p +1 the last inequality for sufficiently large k .

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  Introduction to Real Analysis

  • Robert G. Bartle, Donald R. Sherbert(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  To prove this assertion, we first note that a conditionally convergent series must contain infinitely many positive terms and infinitely many negative terms (see Exercise 1), and that both the series of positive terms and the series of negative terms diverge (see Exercise 2). To construct a series converging to c, we take positive terms until the partial sum is greater than c, then we take negative terms until the partial sum is less than c, then we take positive terms until the partial sum is greater than c, then we take negative terms, etc. In our manipulations with series, we generally want to be sure that rearrangements will not affect the convergence or the value of the series. That is why the following result is important. 9.1.5 Rearrangement Theorem Let P x n be an absolutely convergent series in R . Then any rearrangement P y k of P x n converges to the same value. Proof. Suppose that P x n converges to x 2 R . Thus, if e > 0, let N be such that if n, q > N and s n :¼ x 1 þ    þ x n , then jx  s n j < e and X q k¼Nþ1 jx k j < e : Let M 2 N be such that all of the terms x 1 , . . . , x N are contained as summands in t M :¼ y 1 þ    þ y M . It follows that if m  M, then t m  s n is the sum of a finite number of terms x k with index k > N. Hence, for some q > N, we have jt m  s n j  X q k¼Nþ1 jx k j < e: Therefore, if m  M, then we have jt m  xj  jt m  s n j þ js n  xj < e þ e ¼ 2e: Since e > 0 is arbitrary, we conclude that P y k converges to x. Q.E.D. Exercises for Section 9.1 1. Show that if a convergent series contains only a finite number of negative terms, then it is absolutely convergent. 2. Show that if a series is conditionally convergent, then the series obtained from its positive terms is divergent, and the series obtained from its negative terms is divergent. 9.1 ABSOLUTE CONVERGENCE 269 3. If P a n is conditionally convergent, give an argument to show that there exists a rearrangement whose partial sums diverge to 1.

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  Mathematical Methods for Mathematicians, Physical Scientists and Engineers

  • Jeremy Dunning-Davies(Author)
  • 2003(Publication Date)
  • Woodhead Publishing

    (Publisher)

  If a convergent series Σ*.^,. is such that Σ * Μ |ΐί Γ | is also con-vergent, the original series is said to be absolutely convergent. Again, a series Σ Γ -1 ι * γ is convergent if X ' j u ^ is convergent since, if S n = Σ Γ _,μ γ and T n = Σ Ι ,_ 1 |«^| then for a large enough value of η and for all positive integral values of ρ S n + P - S n =£ T n+p - Τ π < ε . Therefore, Σ r l u r does converge. The sum of an infinite series is defined as a limit, and so is different from the sum of a finite series which is obtained by adding together a finite number of terms. Hence, operations, which are justifiable when applied to finite series, may not be applied to infinite series without further investigation. The impor-tant property of absolutely convergent series is that, in some ways, they may be treated as if they were finite series: an absolutely convergent series may be rearranged without altering the fact that it is an absolutely convergent series

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  Mathematical Methods for Physicists

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

    (Publisher)

  This condition, however, is not sufficient to guarantee convergence. Equation 5.2 is usually written in formal mathematical notation: The condition for the existence of a limit S is that for each ε > 0, there is a fixed N such that S -Si < , for i > N. This condition is often derived from the Cauchy criterion applied to the partial sums s f . The Cauchy criterion is: A necessary and sufficient condition that a sequence (s f ) converge 277 278 INFINITE SERIES is that for each ε > 0 there is a fixed number N such that SJ — s t < ε for all ij > N. This means that the individual partial sums must cluster together as we move far out in the sequence. The Cauchy criterion may easily be extended to sequences of functions. We see it in this form in Section 5.5 in the definition of uniform convergence and in Section 9.4 in the development of Hubert space. Our partial sums s f may not converge to a single limit but may oscillate, as in the case 00 £u„ = 1-1 + 1-1 + 1 + ( -1 ) + . . . . (5.3) n = l Clearly, s f = 1 for i odd but 0 for i even. There is no convergence to a limit, and series such as this one are labeled oscillatory. For the series 1 + 2 + 3 + · · · + « + we have _ n(n + 1) As n -oo, lim s n = oo. (5.6) n-*oo Whenever the sequence of partial sums diverges (approaches + oo), the infinite series is said to diverge. Often the term divergent is extended to include oscil-latory series as well. Because we evaluate the partial sums by ordinary arithmetic, the convergent series, defined in terms of a limit of the partial sums, assume a position of supreme importance. Two examples may clarify the nature of convergence or divergence of a series and will also serve as a basis for a further detailed investiga-tion in the next section. EXAMPLE 5.1.1 The Geometric Series The geometrical sequence, starting with a and with a ratio r (r > 0), is given by a, ar, ar 2 , ar 3 , . .., ar 1 , . . . . The nth partial sum is given by 1 1 — r n s n = a-.

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