The Comparison, Limit Comparison, Ratio, & Root Tests SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference your lecture notes and the relevant chapters in a textbook/online resource. EXPECTED SKILLS: • Use the following tests to make a conclusion about the convergence of series with no negative terms: - Comparison Test - Limit Comparison Test - Ratio Test - Root Test PRACTICE PROBLEMS: For problems 1 & 2, apply the Comparison Test to determine if the series con- verges. Clearly state to which other series you are comparing. 1 X 1 1. 3k + 5 k=1 1 1 < for k ≥ 1. 3k + 5 3k 1 1 X 1 X 1 Since converges (geometric series, r = 1 ), must converge. 3k 3 3k + 5 k=1 k=1 1 X 1 2. 3(k − 2) k=3 1 1 1 = > for k ≥ 3. 3(k − 2) 3k − 6 3k 1 1 X 1 X 1 Since diverges (p-series with p = 1), must diverge as well. 3k 3(k − 2) k=3 k=3 For problems 3 & 4, apply the Limit Comparison Test to determine if the series converges. Clearly state to which other series you are comparing. 1 1 X 1 3. 3(k + 2) k=1 1 k 1 lim 3(k+2) = lim = , which is finite and nonzero. k!1 1 k!1 k 3k + 6 3 1 1 X 1 X 1 So since diverges (Harmonic Series), must diverge as well. k 3(k + 2) k=1 k=1 1 X 1 4. 3k − 5 k=2 1 k k 3 lim 3 −5 = lim = 1, which is finite and nonzero. k!1 1 k!1 k 3k 3 − 5 1 1 X 1 X 1 So since converges (geometric series, r = 1 ), must converge. 3k 3 3k − 5 k=2 k=2 For problems 5 { 7, apply the Ratio Test to determine if the series converges. If the Ratio Test is inconclusive, apply a different test. 1 X 1 5. k! k=0 The series converges by the Ratio Test. 1 X 1 6. 3k k=0 The series converges by the Ratio Test. [Of course this just confirms what we already 1 knew as this is a geometric series with r = 3 .] 1 X 1 7. 3(k + 2) k=1 The Ratio Test is inconclusive; however, as shown in #3 above, the series diverges by the Limit Comparison Test. For problems 8 { 10, apply the Root Test to determine if the series converges. If the Root Test is inconclusive, apply a different test. 1 X 1 8. 3k k=0 The series converges by the Root Test. [Of course this just confirms what we already 1 knew as this is a geometric series with r = 3 .] 2 1 k X 1 9. 1 + k k=1 The Root Test is inconclusive; however, as shown in the previous assigned problems, Convergence Tests #7, the series diverges by the Divergence Test. 1 X k 10. 7k k=1 The series converges by the Root Test.; Detailed Solution: Here For problems 11 { 22, apply the Comparison Test, Limit Comparison Test, Ratio Test, or Root Test to determine if the series converges. State which test you are using, and if you use a comparison test, state to which other series you are comparing to. 1 X 1 p 11. 3 k=1001 k − 10 1 X 1 p The series diverges by the Comparison Test. Compared to 3 . k=1001 k 1 X 4k2 + 5k 12. p 5 k=1 10 + k 1 X 1 The series diverges by the Limit Comparison Test. Compared to p .; k=1 k Detailed Solution: Here 1 X 2k + 1 13. (2k)! k=0 The series converges by the Ratio Test. 1 X k2 cos2 k 14. 2 + k5 k=1 1 X 1 The series converges by the Comparison Test. Compared to . k3 k=1 1 X 1 15. k(5k) k=1 The series converges by the Root Test. 3 1 X 2 + 2k 16. 3 + 3k k=0 1 k X 2 The series converges by the Limit Comparison Test. Compared to . 3 k=0 1 X 6(k+1) 17. k! k=0 The series converges by the Ratio Test. 1 X 6k + k 18. [Hint:Use the result from the previous problem.] k! + 6 k=0 1 X 6(k+1) The series converges by the Comparison Test. Compared to .; Detailed Solution: Here k! k=0 1 X k3 − 2 19. 2 2 k=2 (k + 1) 1 X 1 The series diverges by the Limit Comparison Test. Compared to . k k=2 1 X arctan k 20. k1:5 k=1 1 X π=2 The series converges by the Comparison Test. Compared to . k1:5 k=1 1 X 7k 21. k2 + j sin kj k=1 1 X 1 The series diverges by the Limit Comparison Test. Compared to . k k=1 1 X (2k)! 22. (k!)2 k=0 The series diverges by the Ratio Test. 4.