Solutions

Math 1B: Calculus Spring 2020 Discussion 12 Solutions: Comparison Tests Instructor: Alexander Paulin Date: Feb 28, 2020

1 The Comparison Test

1. Do the following series converge or diverge?

∞ X 1 n2 + 6 n=0 ∞ X 1 We expect this series to converge because it looks like . So we need to compare it to a larger series. n2 n=1 ∞ 1 1 X 1 Note that ≤ for all n. Because is a p-series with p = 2 > 1 it converges, so the series n2 + 6 n2 n2 n=0 ∞ X 1 converges too by the Comparison Test. n2 + 6 n=0 ∞ X 1 n2 − 6 n=1 ∞ X 1 We expect this series to converge because it looks like . So we need to compare it to a larger series. n2 n=1 1 1 However, ≥ , so we will need a different approach. We can use the limit comparison test. n2 − 6 n2 2 1 1 an n + 6 6 Let an = 2 and bn = 2 . Now lim = lim 2 = lim (1 + 2 ) = 1 n n + 6 n→∞ bn n→∞ n n→∞ n ∞ X 1 and the series is a p-series with p = 2 > 1 so it converges, so the series n2 n=0 ∞ X 1 converges too by the Limit Comparison Test. n2 + 6 n=0 ∞ X 3 n − 2 n=1 3 ∞ X 1 3 1 We expect this series to diverge because it looks like the harmonic series . Note that ≥ . n n − 2 n n=1 3 ∞ ∞ X 1 X 1 Because the harmonic series diverges, the series diverges too by the Comparison Test. n n2 + 6 n=1 n=0

1 2 Discussion 12 Solutions: Comparison Tests

∞ X 1 6n + 27 n=1 ∞ X 1 We expect this series to diverge because it looks like . So we need to compare it to a smaller series. n n=1 1 1 However, ≤ , so we will need a different approach. We can use the limit comparison test. 6n + 27 n 1 1 an 6n + 27 27 Let an = and bn = . Now lim = lim = lim (6 + ) = 6 n 6n + 27 n→∞ bn n→∞ n n→∞ n ∞ ∞ X 1 X 1 and the harmonic series diverges, so the series diverges too by the Limit Comparison Test. n n2 + 6 n=0 n=0 ∞ X ln n n2 n=3 Intuitively, the function ln(n) grows very slowly, so it should not hurt us too much and the series should converge. √ √ Note that ln(n) ≤ n for n ≥ 1. We can see this by noting that ln(1) = 0 < 1 = 1, and that the derivatives 1 1 ≤ √ , so the function ln(n) is smaller at 1 and grows more slowly. n 2 n √ ln n n 1 Therefore, we have ≤ = , which is a p-series with p = 1.5 > 1 so it converges and so the series n2 n2 n1.5 ∞ X ln n converges too by the Comparison Test. n2 n=3

P∞ P∞ 2 2. (*) If n=1 an is a convergent series with positive terms, is it true that n=1(an) also converges? P∞ √ What about n=1 an?

P∞ 2 P∞ We claim that the series n=1(an) does indeed converge. Because the series n=1 an converges, we must have limn→∞ an = 0, so in particular there is some N after which an ≤ 1. Then for n ≥ N, we 2 P∞ 2 have (an) ≤ an, so the series n=1(an) converges by the Comparison Test.

P∞ √ On the other hand, the series n=1 an may converge or diverge depending on an. We can see this 1 most clearly by considering the p-series an = np . Recall that this series will converge whenever p > 1. √ 1 Then an = p . So for example, if p = 2 this series will diverge, but if p = 3 it will converge. n 2