Joe Foster Direct and Limit Comparison Tests We have seen that a given improper integral converges if its integrand is less than the integrand of another integral known to converge. Similarly, a given improper integral diverges if its integrand is greater than the integrand of another integral known to diverge. We now apply the same idea to infinite series instead. Direct Comparison Test for Series: If0 a b for all n N, for some N, then, ≤ n ≤ n ≥ ∞ ∞ 1. If bn converges, then so does an. nX=1 nX=1 ∞ ∞ 2. If an diverges, then so does bn. nX=1 nX=1 a The Limit Comparison Test: Suppose a > 0 and b > 0 for all n. If lim n = L, where L is finite and L> 0, then n n →∞ n bn the two series an and bn either both converge or both diverge. X X ∞ 1 Example 1: Determine whether the series converges or diverges. 2n + n nX=1 1 1 We have for all n 1. So, 2n + n ≤ 2n ≥ ∞ ∞ 1 1 . 2n + n ≤ 2n nX=1 nX=1 1 Since the series on the right is a geometric series with r = , it converges. So we determine that our series of interest also 2 converges. ∞ 1 Example 2: Determine whether the series converges or diverges. n2 + n +1 nX=1 1 1 We have for all n 1. So, n2 + n +1 ≤ n2 ≥ ∞ ∞ 1 1 . n2 + n +1 ≤ n2 nX=1 nX=1 Since the series on the right is a p series with p = 2 > 1, it converges. So we determine that our series of interest also converges. ∞ √n4 1 Example 3: Use the Direct Comparison Test to determine if − converges or diverges. n5 +3 nX=1 √n4 1 √n4 n2 n2 1 We have − < = < = . So, n5 +3 n5 +3 n5 +3 n5 n3 ∞ ∞ √n4 1 1 − . n5 +3 ≤ n3 nX=1 nX=1 Since the series on the right is a p series with p = 3 > 1, it converges. So we determine that our series of interest also converges. Page 1 of 3 MATH 142 - Comparison Tests for Series Joe Foster ∞ 1 Example 4: Determine whether the series converge or diverges. n2 5 nX=3 − 1 1 We can see that the direct comparison test will not work here. So let’s try the limit comparison test. We have . n2 1 ≈ n2 Further, − 1 2 2− n 5 lim n 5 = lim = lim 1+ =1. →∞ 1 →∞ 2 →∞ 2 n 2 n n 5 n n 5 n − − ∞ 1 ∞ 1 Since is a p-series with p =2 > 1, it converges. Thus by the limit comparison test, converges also. n=3 n2 n=5 n2 ∞ 5 P n3 2n P − Example 5: Determine whether the series − converges or diverges. n4 +3 nX=2 n3 2n n3 1 We have − = . Further, n4 +3 ≈ n4 n 3 n −2n 3 4 2 2 n4+3 n(n 2n) n 2n 3+2n lim 1 = lim − = lim − = lim 1 =1. n→∞ n→∞ n4 n→∞ n4 n→∞ n4 n +3 +3 − +3 3 ∞ 1 ∞ n 2n Since is a p-series with p = 1, it diverges. Thus by the limit comparison test, − diverges also. n=2 n n=2 n4 +3 P P Practice Problems Using one of the comparison tests, determine whether the following series converge or diverge. ∞ ∞ ∞ n3 √n n2 5n 1. 6. 11. − n4 1 n 1 n3 + n +1 nX=2 − nX=2 − nX=1 ∞ ∞ ∞ n 1 1 n +5 2. − 7. 12. n2√n √n3 √3 n7 n2 nX=1 nX=1 +1 nX=1 + ∞ ∞ ∞ 4+3n 1 21/n 3. 8. 13. 2n 2n +3 n nX=1 nX=1 nX=1 ∞ ∞ ∞ n2 1 n +4n n! 4. − 9. 14. 3n4 +1 n +6n nn nX=1 nX=1 nX=1 ∞ ∞ ∞ 1 + sin(n) n +2 1 5. 10. 15. 10n (n + 1)3 n1+1/n nX=0 nX=3 nX=1 Page 2 of 3 MATH 142 - Comparison Tests for Series Joe Foster Hints to Practice Problems ∞ ∞ ∞ 1 1 1 1. Compare with 6. Compare with 11. Compare with n √n n nX=2 nX=2 nX=1 ∞ ∞ ∞ 1 1 1 2. Compare with 7. Compare with 12. Compare with n1.5 n1.5 n4/3 nX=2 nX=1 nX=1 ∞ ∞ ∞ 3 n 1 1 3. Compare with 8. Compare with 13. Compare with 2 n n nX=1 nX=1 nX=1 ∞ ∞ ∞ n 1 2 n √2 4. Compare with 9. Compare with 14. Compare with n2 3 2 nX=1 nX=1 nX=1 ∞ ∞ ∞ 2 1 1 5. Compare with 10. Compare with 15. Compare with 10n n2 n nX=0 nX=3 nX=1 Answers to Practice Problems 1. Diverges 6. Diverges 11. Diverges 2. Converges 7. Converges 12. Converges 3. Diverges 8. Diverges 13. Diverges 4. Converges 9. Converges 14. Converges 5. Converges 10. Converges 15. Diverges Page 3 of 3.