math 131 infinite series, part v: the alternating series test 32

14.9 Absolute and Conditional Convergence

Sometimes series 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 Â 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 absolute value of the terms. This turns the sequence into a non-negative series and now we can apply many of our previous convergence tests. For example if we take the absolute value of the terms in the series above, we get

• sin n .  n2 n=1 Since sin n 1, then | | sin n 1 0 < < . n2 n2 • 1 • sin n But converges by the p-series test (p= 2 > 1), so converges  n2  n2 n=1 n=1 • 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 14.9.1 (The Absolute Convergence 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. | | 8a a = 0, if a < 0 < n n n : • 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 convergent series.

• • 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.

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• • DEFINITION 14.9.2.  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 14.9.3. Determine whether converges absolutely, conditionally, or not at  3 2 n=1 pn all. • ( 1)n • 1 Solution. First we check absolute convergence. = is a p-  p3 2  2/3 n=1 n n=1 n 2 series with p = 1. So the series of absolute values diverges. The original series 3  is not absolutely convergent. Since the series is alternating and not absolutely convergent, we check for condi- tional convergence using the alternating series test with a = 1 . Check the two n n2/3 conditions. 1 1. lim an = lim = 0. n • n • 2/3 ! ! n 2. Further a a because 1 < 1 . n+1  n (n+1)2/3 n2/3 • ( 1)n Since the two conditions of the alternating series test are satisfied, is  3 2 n=1 pn conditionally convergent by the alternating series test.

• cos n EXAMPLE 14.9.4. Determine whether  2 converges absolutely, conditionally, or not at n=1 n 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 Â 2 2 2 2 + 2 + . n=1 n ⇡ 1 2 3 4 ···

• cos n • 1 First we check absolute convergence. looks a lot like the p-series  n2  n2 n=1 n=1 with p = 2 > 1. We can use the direct comparison test. Since 0 cos n 1, | | cos n 1 0  n2  n2 • 1 • cos n for all n. Since the p-series converges, so does by the direct  n2  n2 n=1 n=1 comparison test (Theorem 14.7.1). So the series of absolute values converges. The original series is absolutely convergent. We need not check further.

• ( 1)n EXAMPLE 14.9.5. Determine whether converges absolutely, conditionally, or not  2 n=2 pn 1 at all. • ( 1)n • 1 Solution. First we check absolute convergence. = .  p 2  p 2 n=2 n 1 n=2 n 1 1 1 Notice that 2 . So let’s use the limit comparison test. The terms of the pn 1 ⇡ n series are positive and

a 1 n HPwrs n n lim n = lim = lim = lim = 1 > 0. n • bn n • pn2 1 · 1 n • pn2 n • n ! ! ! !

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• 1 • ( 1)n Since the harmonic series diverges (p-series with p = 1), then  n  p 2 n=2 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 con- 1 ditional convergence using the alternating series test with an = . Check the pn2 1 two conditions. 1 1. lim an = lim = 0. n • n • pn2 1 ! ! 1 1 2. Further an+1 an is decreasing because < . (You could  (n + 1)2 1 pn2 1 also show the derivative is negative.) Since the two conditions of the alternating p • ( 1)n series test are satisfied, is conditionally convergent by the alternating  2 n=2 pn 1 series test. • ( 1)n(2n4 + 7) EXAMPLE 14.9.6. Determine whether converges absolutely, conditionally,  6n9 2n n=1 or not at all. • ( 1)n(2n4 + 7) • 2n4 + 7 Solution. First we check absolute convergence. = .  6n9 2n  6n9 2n n=1 n=1 4 Notice that 2n +7 1 . So let’s use the limit comparison test. The terms of the se- 6n9 2n ⇡ n5 ries are positive and

4 5 9 5 9 a 2n + 7 n 2n + 7n HPwrs 2n 1 lim n = lim = lim = lim = > 0. n • b n • 6n9 2n · 1 n • 6n9 2n n • 6n9 3 ! n ! ! ! • 1 • ( 1)n(2n4 + 7) Since converges (p-series with p = 5 > 1), then  n5  6n9 2n n=1 n=1 converges by the limit comparison test. So the series converges absolutely. • ( 1)n EXAMPLE 14.9.7. Determine whether  converges absolutely, conditionally, or not at n=2 ln n all. • ( 1)n • 1 Solution. First we check absolute convergence. = . We use  ln n  ln n n=2 n=2 1 1 1 the direct comparison test with . Notice that 0 < because n > 1. n ln n n ln n  ln n • 1 • 1 1 • 1 Next  diverges (as we saw in earlier examples). Consequently  1 To check that di- n ln n ln n  n ln n n=2 n=2 n=2 diverges by the direct comparison test. So the series does not converge absolutely. verges, use the integral test and u-substitution with u = ln x. • Since the series is alternating and not absolutely convergent, we check for con- 1 b 1 dx = lim ln ln x = 2 x ln x b • | | 2 ditional convergence using the alternating series test with an = ln n . Check the two Z ! lim ln ln b ln(ln 2)=+•. b • | | conditions. ! 1 1. lim an = lim = 0. n • n • ln ! ! n ( ) 2 2. Further a is decreasing since f (x)= 1 =(ln x) 1 then f (x)= ln x < 0 n ln x 0 x 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 14.9.8. Determine whether  converges absolutely, conditionally, or not n=1 2n at all.

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• 2n • 2n Solution. First we check absolute convergence. = . Hey wait. . . does  2n  2n n=1 n=2 this series diverge? Use the nth term test: n x x 2 2 l’Ho 2 ln 2 lim = lim = lim = •. n • 2 x • 2 x • 2 ! n ! x ! Since limn • an = 0 the series automatically diverges and cannot converge abso- ! 6 lutely or conditionally.

Bonus Fact: The Ratio Test 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 14.9.9 (The Ratio Test Extension). Assume that  an is a series with non-zero terms n=1 a and let r = lim n+1 . n • a ! n • 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 14.9.10. Determine whether  n converges absolutely, conditionally, or n=1 3 not at all.

Solution. Here’s a perfect place to use the ratio test extension because there is a factorial. ( 1)n+1(n + 1)! 3n n + 1 r = lim = lim = •. n • 3n+1 · ( 1)nn! 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 14.9.11. Determine whether  n converges absolutely, conditionally, n=1 2 n or not at all.

Solution. First we check absolute convergence using the ratio test because of the factorial. n+1 n 2 ( 1) (n + 2)! 2 n (n + 2)n n + 2n HPwrs n r = lim = lim = lim = lim = •. n • 2n+1(n + 1) · ( 1)n(n + 1)! n • 2(n + 1) n • 2n + 2 n • 2 ! ! ! ! The (original) series diverges by the ratio test extension. • n( 2)n EXAMPLE 14.9.12. Determine whether  n+1 converges absolutely, conditionally, or not n=1 3 at all.

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Solution. Check absolute convergence using the ratio test extension.

n+1 n+1 (n + 1)( 2) 3 2(n + 1) HPwrs 2 r = lim = lim = < 1. n • 3n+2 · n( 2)n n • 3n 3 ! ! The (original) series converges absolutely by the ratio test extension. Here are lots of WeBWorK practice problems to try. webwork: Click to try Problems 179 through 190. Use guest login, if not in my course.

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