BC 3 Advanced Tests for Covergence Name: Now that we see that convergence of improper integrals and series can be thought of as the same problem under appropriate conditions, we immediately get the following theorem for series. Limit Comparison Test: Suppose that 0 ak and 0 bk for all k and lim n an L , with bn 0 L . Then a n 1 n converges if and only if n 1 If L 1 , then If L 1 , then a If L 1 , then and lim n converges . n diverges . a n 1 iii. converges . n 1 ii. n an 1 L . Then n a n Ratio Test: Suppose that 0 ak for all k i. b a n 1 n may converge or diverge, so we must look for a different method to determine convergence. Root Test: Suppose that 0 ak for all k iv. If L 1 , then If L 1 , then a n converges . n diverges . a n 1 vi. If L 1 , then k n 1 v. and lim k ak L . Then a n 1 n may converge or diverge, so we must look for a different method to determine convergence. IMSA Sequences and Series 4.1 Fall 14 The Grand Challenge: Using any of the tests that we have developed so far, determine whether each of the series converges or diverges. Your reasoning should be easy to follow. Be careful to refer to which test you are employing. n! n 1 (2 n )! (3) 3 1 n n 1 (2) 2n (4) (5) IMSA n sin 2 n 3 n 1 n n (1) 5k k k 2k k 0 ( k 1) 5n n 1 n ! (6) k 3 3k 21 5 3 k 2 5k 6k k 1 Sequences and Series 4.2 Fall 14 (7) n 1 (9) k 2 IMSA 1 n sin n 1 sin k (8) k ! 2 (2k )! k 1 (10) 1 n n 1 5n Sequences and Series 4.3 n Fall 14 1 converges, then find the value of this sum with an error of less than k! k 2 .0001. Explain your error analysis. (11) Show that 2 k (12) Find all values of x for which the series k 0 IMSA xk k! Sequences and Series 4.4 Fall 14