11.6 Absolute Convergence and the Ratio and Root Tests

∞ ln(n) Therefore, by the Alternating Series Test, we have that ( 1)n con- − n n=1 verges.

Example 1. Can we apply any of the tests we’ve learned so far to the series ∞ cos(n) ? n2 n=1 Solution. If we write out the ﬁrst ﬁve terms we get cos(1) cos(2) cos(3) cos(4) cos(5) + + + + + ... 12 22 32 42 52 + + ... −−− where we have indicated the sign of each term with a + or below the term. Since some of the terms are negative, we cannot apply the− Comparison Tests or the Integral Test. Since the terms do not alternate perfectly, we cannot apply the alternating series test.

∞ cos(n) Example 2. [Continued] Show that is absolutely convergent. n2 n=1 Solution. We take absolute values of each term

∞ cos(n) ∞ cos(n) = | |. n2 n2 n=1 n=1 This series is better than the original one because now we can apply the comparison test. cos(n) | | 0 n2 ≥ 1 0 n2 ≥ cos(n) 1 | | n2 ≤ n2 1 converges, p-series test, p =2 n2 cos(n) Therefore, by the Comparison Test, converges. n2 cos(n) Therefore, by Theorem ??, converges too. In other words, n2 cos(n) is absolutely convergent. n2 CHAPTER 11. SEQUENCES AND SERIES 92

Example 3. Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞ 1 ( 1)n . − √n n=1 Solution. The divergence test does not apply here. Thus, we think the series might have some sort of convergence.

Original series 1 ( 1)n − √n 1 0 √n ≥ 1 ? 1 ? ,i.e.√n +1 √n, n +1 n. √n ≥ √n +1 ≥ ≥ 1 1 lim = =0. n √n →∞ ∞ ∞ 1 ( 1)n converges by AST. − √n n=1 Absolute convergence

∞ ∞ 1 ∞ 1 a = ( 1)n = | n| − √n √n n=1 n=1 n=1 Diverges, p-series test, p =1/2.

Summary

∞ 1 The series ( 1)n is conditionally convergent. − √n n=1 ∞ 10n Example 4. Apply the Ratio Test to ( 1)n . − n! n=1 Solution. The main trick here is to simplify the factorials:

n+1 ( 1)n+1 10 − (n+1)! L =lim 10n n ( 1)n →∞ − n! ( 1)n+1 10n +1 n! =lim − n ( 1)n · (n +1)! · 10n →∞ − 10n+1 n! =lim( 1) n − 10n · (n +1)! →∞ 1 =lim1 101 n →∞ · · n +1 CHAPTER 11. SEQUENCES AND SERIES 93

=1 101 0 · · =0

∞ 10n Since L =0,and0< 1, the Ratio Test implies that ( 1)n is Abs Conv. − n! n=1 ∞ n10 Example 5. Apply the Ratio Test to ( 1)n . − 1.5n n=1 Solution.

n+1 (n+1)10 ( 1) 1.5n+1 L =lim − 10 n n n ( 1) n →∞ − 1.5 ( 1)n+1 (n +1)10 1.5n =lim − n ( 1)n 1.5n+1 · n10 →∞ − 1.5n (n +1)10 =lim( 1) n − · 1.5n+1 · n10 →∞ 1 (n +1)10 =lim1 n 10 →∞ · 1.5 · n 1 =1 1 · 1.5 · 1 = 1.5

1 1 ∞ n10 Since L = ,and < 1, the Ratio Test implies that ( 1)n is 1.5 1.5 − 1.5n n=1 Abs. Conv. Example 6. What happens when you apply the Ratio test to the following series:

∞ ln(n) n2 n=1 Solution. ln(n+1) 2 L =lim (n+1) n ln(n) →∞ n2 ln(n +1) n2 =lim n 2 →∞ (n +1) · ln(n) ln(n +1) n2 =lim n 2 →∞ ln(n) · (n +1) We analyse each the limit of each factor here by itself, starting with the simpler n2 one, , (n +1)2 n2 n2 lim =lim =1 n 2 n 2 →∞ (n +1) →∞ n CHAPTER 11. SEQUENCES AND SERIES 94

ln(n +1) Now we look at the limit of ln(n) ln(n +1) ln( ) lim = ∞ n ln(n) ln( ) →∞ ∞ = ∞ use L’Hospital ∞ 1 ln(n +1) lim LH=lim n+1 n n 1 →∞ ln(n) →∞ n 1 n =lim n n +11 →∞ n =lim n n +1 →∞ n =lim =1 n →∞ n Combining the limits we’ve found with our formula for L above, we have

L =1 1 · So, this test tells us nothing about this series.

Example 7. Determine if the following series converges:

n ∞ 2n3 + n − 3n3 + n2 +1 n=1 Solution.

2n3 + n n L =lim n − n 3n3 + n2 +1 →∞ 2 2n + n =lim − n 3n3 + n2 +1 →∞ 2n2 =lim− n 3n3 →∞ 2 = − 3 2 = 3

n 2 2 ∞ 2n3 + n Since L = ,and < 1, the Root Test implies that − Make sure you see that the 3 3 3n3 + n2 +1 absolute values make the ﬁnal n=1 2 2 converges. limit here, not .Inthis 3 − 3 example, the diﬀerence doesn’t matter, but it would matter a lot if we had an example with L = 3/2 . |− |