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Home , Absolute convergence, Conditional convergence

math131 infiniteseries, part vii: absolute and 30

Absolute and Conditional Convergence

Sometimes have both positive and negative terms but they are not perfectly alternating like those in the previous section. For example

∞ sin n 0.841 0.909 0.141 0.757 0.279 0.657 ≈ + + − − + + ··· ∑ 2 n=1 n 1 4 9 16 25 36 is not alternating but does have both positive and negative terms. So how do we deal with such series? The answer is to take the of the terms. This turns the into a non-negative series and now we can apply many of our previous . For example if we take the absolute value of the terms in the series above, we get

∞ sin n ∑ 2 . n=1 n

Since | sin n| ≤ 1, then

sin n 1 0 < < . n2 n2 ∞ 1 ∞ sin n = > But ∑ 2 converges by the p-series test (p 2 1), so ∑ 2 converges n=1 n n=1 n ∞ sin n by comparison. But what about the original series ∑ 2 ? The next theorem n=1 n provides the answer: The series does converge.

∞ ∞ THEOREM 14.11 (The Test). If ∑ |an| converges so does ∑ an. n=1 n=1

∞ ∞ Proof. Given ∑ |an| converges. Define a new series ∑ bn, where n=1 n=1  an + an = 2an, if an ≥ 0 bn = an + |an| = ≥ 0. an − an = 0, if an < 0

∞ So 0 ≤ bn = an + |an| ≤ |an| + |an| = 2|an|. But ∑ 2|an| converges, hence by direct n=1 ∞ comparison ∑ bn converges. Therefore n=1

∞ ∞ ∞ ∞ ∞ ∑ an = ∑ [(an + |an|) − |an|] = ∑ bn − |an| = ∑ bn − ∑ |an| n=1 n=1 n=1 n=1 n=1 converges since it is the difference of two .

∞ ∞ Important Note. The converse is not true. If ∑ an converges, ∑ |an| may or may n=1 n=1 ∞ (−1)n not converge. For example, the alternating harmonic series ∑ converges, n=1 n ∞ 1 but if we take the absolute value of the terms, the harmonic series ∑ diverges. n=1 n This leads to the following definition. 31

∞ ∞ DEFINITION 14.5. ∑ an is absolutely convergent if ∑ |an| converges. n=1 n=1 ∞ ∞ ∞ ∑ an is conditionally convergent if ∑ an converges but ∑ |an| diverges. n=1 n=1 n=1

∞ (−1)n EXAMPLE 14.42. Determine whether √ converges absolutely, conditionally, or ∑ 3 2 n=1 n not at all. ∞ ∞ (−1)n 1 SOLUTION. First we check absolute convergence. √ = is a p-series ∑ 3 2 ∑ 2/3 n=1 n n=1 n 2 with p = 3 ≤ 1. So the series of absolute values diverges. The original series is not absolutely convergent. Since the series is alternating and not absolutely convergent, we check for condi- = 1 tional convergence using the test with an n2/3 . Check the two conditions. 1 1. lim an = lim = 0. n→∞ n→∞ n2/3 2. a ≤ a 1 < 1 Further n+1 n because (n+1)2/3 n2/3 . ∞ (−1)n Since the two conditions of the alternating series test are satisfied, √ is ∑ 3 2 n=1 n conditionally convergent by the alternating series test. ∞ cos n EXAMPLE 14.43. Determine whether ∑ 2 converges absolutely, conditionally, or n=1 n not at all.

SOLUTION. Notice that this series is not positive nor is it alternating since the first few terms are approximately ∞ cos n 0.540 0.416 0.990 0.284 ≈ − − + + ··· ∑ 2 2 2 2 2 . n=1 n 1 2 3 4 ∞ ∞ cos n 1 First we check absolute convergence. ∑ 2 looks a lot like the p-series ∑ 2 n=1 n n=1 n with p = 2 > 1. We can use the . Since 0 ≤ | cos n| ≤ 1, cos n 1 0 ≤ ≤ n2 n2 ∞ ∞ 1 cos n for all n. Since the p-series ∑ 2 converges, so does ∑ 2 by the direct compar- n=1 n n=1 n ison test (Theorem 14.6). So the series of absolute values converges. The original series is absolutely convergent. We need not check further. ∞ (−1)n EXAMPLE 14.44. Determine whether √ converges absolutely, conditionally, ∑ 2 n=2 n − 1 or not at all. ∞ ∞ (−1)n 1 SOLUTION. First we check absolute convergence. √ = √ . ∑ 2 ∑ 2 n=2 n − 1 n=2 n − 1 Notice that √ 1 ≈ 1 . So let’s use the limit comparison test. The terms of the series n2−1 n are positive and

a 1 n HPwrs n n lim n = lim √ · = lim √ = lim = 1 > 0. n→∞ bn n→∞ n2 − 1 1 n→∞ n2 n→∞ n ∞ ∞ 1 (−1)n Since the harmonic series diverges (p-series with p = 1), then √ ∑ ∑ 2 n=2 n n=2 n − 1 diverges by the limit comparison test. So the series does not converge absolutely. Since the series is alternating and not absolutely convergent, we check for condi- 1 tional convergence using the alternating series test with an = √ . Check the two n2−1 conditions. math131 infiniteseries, part vii: absolute and conditional convergence 32

1 1. lim an = lim √ = 0. n→∞ n→∞ n2 − 1 1 1 2. Further an+1 ≤ an is decreasing because p < √ . (You could (n + 1)2 − 1 n2 − 1 also show the derivative is negative.) Since the two conditions of the alternating ∞ (−1)n series test are satisfied, √ is conditionally convergent by the alternating ∑ 2 n=2 n − 1 series test. ∞ (−1)n(2n4 + 7) EXAMPLE 14.45. Determine whether converges absolutely, condi- ∑ 9 − n=1 6n 2n tionally, or not at all.

∞ (−1)n(2n4 + 7) ∞ 2n4 + 7 SOLUTION. First we check absolute convergence. = . ∑ 9 − ∑ 9 − n=1 6n 2n n=1 6n 2n 2n4+7 ≈ 1 Notice that 6n9−2n n5 . So let’s use the limit comparison test. The terms of the series are positive and

a 2n4 + 7 n5 2n9 + 7n5 2n9 1 n = · = HPwrs= = > lim lim 9 lim 9 lim 9 0. n→∞ bn n→∞ 6n − 2n 1 n→∞ 6n − 2n n→∞ 6n 3

∞ 1 ∞ (−1)n(2n4 + 7) Since converges (p-series with p = 5 > 1), then converges ∑ 5 ∑ 9 − n=1 n n=1 6n 2n by the limit comparison test. So the series converges absolutely.

∞ (−1)n EXAMPLE 14.46. Determine whether ∑ converges absolutely, conditionally, or n=2 ln n not at all. ∞ ∞ (−1)n 1 SOLUTION. First we check absolute convergence. ∑ = ∑ . We use the n=2 ln n n=2 ln n 1 < 1 ≤ 1 > direct comparison test with n ln n . Notice that 0 n ln n ln n because n 1. Next ∞ ∞ 1 1 1 ∞ diverges (as we saw in earlier examples). Consequently diverges 1 1 ∑ n ln n ∑ ln n To check that ∑ di- n=2 n=2 n=2 n ln n by the direct comparison test. So the series does not converge absolutely. verges, use the test and Since the series is alternating and not absolutely convergent, we check for condi- u-substitution with u = ln x. Z ∞ 1 1 b tional convergence using the alternating series test with an = . Check the two dx = lim ln | ln x| = ln n 2 x ln x b→∞ 2 conditions. lim ln | ln b| − ln(ln 2) = +∞. b→∞ 1 1. lim an = lim = 0. n→∞ n→∞ ln n −2 ( ) = 1 = ( )−1 0( ) = − (ln x) < 2. Further an is decreasing since f x ln x ln x then f x x 0 for x ≥ 2. ∞ (−1)n Since the two conditions of the alternating series test are satisfied, ∑ is n=2 ln n conditionally convergent by the alternating series test.

∞ (−1)n2n EXAMPLE 14.47. Determine whether ∑ converges absolutely, conditionally, n=1 2n or not at all. ∞ ∞ 2n 2n SOLUTION. First we check absolute convergence. ∑ = ∑ . Hey wait. . . does n=1 2n n=2 2n this series diverge? Use the nth term test:

n x x 2 2 l’Ho 2 ln 2 lim = lim = lim = ∞. n→∞ 2n x→∞ 2x x→∞ 2

Since limn→∞ an 6= 0 the series automatically diverges and cannot converge absolutely or conditionally. 33

Bonus Fact: The Extension

When we test for absolute convergence using the ratio test, we can say more. If the ratio r is actually greater than 1, the series will diverge. We don’t even need to check conditional convergence.

∞ THEOREM 14.12 (The Ratio Test Extension). Assume that ∑ an is a series with non-zero terms n= 1 an+1 and let r = lim . n→∞ an ∞ 1. If r < 1, then the series ∑ an converges absolutely. n=1 ∞ 2. If r > 1 (including ∞), then the series ∑ an diverges. n=1 3. If r = 1, then the test is inconclusive. The series may converge or diverge.

This is most helpful when the series diverges. It says we can check for absolute convergence and if we find the absolute value series diverges, then the original series diverges. We don’t have to check for conditional convergence. Huzzah!

∞ (−1)nn! EXAMPLE 14.48. Determine whether ∑ n converges absolutely, conditionally, n=1 3 or not at all.

SOLUTION. Here’s a perfect place to use the ratio test extension because there is a .

(−1)n+1(n + 1)! 3n n + 1 = · = = r lim n+ n lim ∞. n→∞ 3 1 (−1) n! n→∞ 3

The (original) series diverges by the ratio test extension. That was easy!! The ratio test extension says we don’t have to check for conditional convergence.

∞ (−1)n(n + 1)! EXAMPLE 14.49. Determine whether ∑ n converges absolutely, condi- n=1 2 n tionally, or not at all.

SOLUTION. First we check absolute convergence using the ratio test because of the factorial.

(−1)n+1(n + 2)! 2nn (n + 2)n n2 + 2n n = · = = HPwrs= = r lim n+ n lim lim lim ∞. n→∞ 2 1(n + 1) (−1) (n + 1)! n→∞ 2(n + 1) n→∞ 2n + 2 n→∞ 2

The (original) series diverges by the ratio test extension.

∞ n(−2)n EXAMPLE 14.50. Determine whether ∑ n+1 converges absolutely, conditionally, n=1 3 or not at all.

SOLUTION. Check absolute convergence using the ratio test extension.

(n + 1)(−2)n+1 3n+1 2(n + 1) 2 = · = HPwrs= < r lim n+2 n lim 1. n→∞ 3 n(−2) n→∞ 3n 3

The (original) series converges absolutely by the ratio test extension.