4/4/22, 8:12 PM DeltaMath https://www.deltamath.com/app/student/solve/15529307/_lbodner__ConditionalAndAbsoluteConvergence 1/2 Back Next Problem Record: 1/5 Score: 0.75 Penalty: 0.25 off Complete: 40% Sara Cui Conditional and Absolute Convergence Apr 04, 8:10:24 PM Determine if the series converges absolutely, converges conditionally, or diverges, and justify your answer. (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 Consider the absolute value of the series: (−1) = n =1 ∑ ∞ ∣ ∣ ∣ ∣ n +1 n + 1 2 3 6 n + 1 ∣ ∣ ∣ ∣ n =1 ∑ ∞ n + 1 2 3 6 n + 1 can be compared to the p-series with p = n =1 ∑ ∞ n + 1 2 3 6 n + 1 n =1 ∑ ∞ n 2 1 6 2 1 diverges since p ≤ n =1 ∑ ∞ n 2 1 6 1 Since diverges, diverges by the limit comparison test. n =1 ∑ ∞ n 2 1 6 n =1 ∑ ∞ n + 1 2 3 6 n + 1 4/4/22, 8:12 PM DeltaMath https://www.deltamath.com/app/student/solve/15529307/_lbodner__ConditionalAndAbsoluteConvergence 2/2 is an alternating series. Since the absolute value of the terms decrease to 0, convergest by the alternating series test. (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 Since diverges and converges, converges conditionally. n =1 ∑ ∞ n + 1 2 3 6 n + 1 (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 The series converges by the alternate series test since the absolute value of the terms decrease to 0. Consider the absolute value of the series: diverges by the limit comparison to the p-series with p = , which diverges since a p-series will diverge if and only if p ≤ 1. Since diverges and converges, converges conditionally. (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 n =1 ∑ ∞ n + 1 2 3 6 n + 1 2 1 n =1 ∑ ∞ n + 1 2 3 6 n + 1 (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 (−1) n =1 ∑ ∞ n +1 n + 1 2 3 6 n + 1 Your Solution: student pressed "see solution" instead of submitting an answer Privacy Policy Terms of Service Copyright © 2022 DeltaMath.com. All Rights Reserved.