Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed Sequences Def.: A sequence is a 1 whose domain is the integers or a subset of the integers. Note 1: For most of the examples that we will encounter, the domain of the sequence will be the integers: { 1 , 2 , 3 , 4 , …} Note 2: Although a sequence is a function, it is common to represent sequences with subscript notation, , rather than with the standard function notation, f(n). Exercise 1: Let {an} = {3n – 1} . (i.) Write the sequence in (the generally unused) function notation. (ii.) Make a table with the first four terms. input n output f(1) = a1 = 3(1) – 1 = 2 1 Note 3: {an} = {3n – 1} denotes the , whereas an = 3n – 1 denotes only the of the sequence. Limit of a Sequence Def.: Let L be a real number. L is the limit of a sequence {an}, written if for any > 0, there exists M > 0 such that for every . If the limit exists, we say that the sequence {an} If no limit exists, we say that the sequence {an} f(n) = an to L. . Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed 2 1 n Exercise 2: Consider the sequence {an} = . 2 (i.) Write out the first four terms of the sequence. What do you suppose the limit, L, is? (ii.) Suppose, for example, that = 0.01. How large must n be so that | an – L | < ? That is, how large must n be so that | (0.5)n – 0 | < 0.01 ? By trial and error, we see that for n = , | an – L | = | (0.5)n – 0 | = < 0.01 = (iii.) Suppose, for example, that = 0.0001. How large must n be so that | an – L | < ? That is, how large must n be so that | (0.5)n – 0 | < 0.0001 ? Theorem: Let f be a function of x. Suppose that If f(n) = an for every positive integer n, then . . Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed Exercise 3: Let f ( x) 3 6 x 2 5x 2 . 2x 2 (i.) Evaluate lim f ( x) . x (ii.) Write out the first four terms of {an}, where an = 6(1) 2 5(1) 2 a1 = = 2(1) 2 , 6n 2 5n 2 . 2n 2 6( ) 2 5( ) 2 a2 = = 2( ) 2 , a3 = , a4 = (iii.) Evaluate lim a n . n Other Properties of Limits of Sequences Theorem: Suppose that lim a n L and lim bn K . Then each of the following is true. n n (i.) (ii.) (iii.) (iv.) Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed Exercise 4: Suppose that lim a n 5 and that bn n 4 n an 2 . Evaluate lim bn . n 3n Exercise 5: Determine whether {an} = {1 + (-1)n } converges or diverges. a1 = 1 + (-1)1 = 1 – 1 by theT.I. 89 0 , a2 = 1 + (-1)2 = , a3 = , a4 = So, {an} = The terms alternate as 0 and 2. Therefore, the limit Squeeze Theorem Theorem: Suppose that lim a n L and lim bn L . Suppose also that n n Then Absolute Value Theorem Theorem: Suppose that lim an 0 . Then n Note: If lim an c 0 , then lim a n may not even exist. n Example: {an} = n . That is, {an} Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed 1 n Exercise 6: Consider the sequence {an} = . Write out the first four terms of 2 the sequence, and determine the limit of the sequence. Exercise 7: Write an expression for the nth term of the sequence, and then determine the limit of the sequence. (i.) 2 , 1 , 4 5 2 , , ,… 5 7 3 (ii.) 1 + 1 3 7 15 ,1+ ,1+ ,1+ ,… 2 4 8 16 5 Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed 6 Monotonic Sequences Def.: A sequence {an} is nondecreasing if Def.: A sequence {an} is strictly increasing if Def.: A sequence {an} is nonincreasing if Def.: A sequence {an} is strictly decreasing if , that is, if , that is, if , that is, if , that is, if Def.: A sequence {an} is monotonic if it is nondecreasing, strictly increasing, nonincreasing, or strictly decreasing. Exercise 8: Determine if the following sequences are monotonic, given the nth term. (i.) an = n3 3n 1 (ii.) bn = 3n 1 2n Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed Bounded Sequences Def.: A sequence {an} is bounded above if Def.: A sequence {an} is bounded below if Def.: A sequence {an} is bounded if it is Exercise 9: Determine if the following sequence is bounded, given the nth term. bn = 3n 1 2n 7 Bob Brown CCBC Dundalk Math 252 Calculus 2 Chapter 9, Section 1 Completed Theorem: If a sequence is bounded and monotonic, then it 8 . Exercise 10: Use this theorem to prove that the sequence whose nth term is bn = 3n 1 2n converges, without determining the limit. By Exercise 8, we showed that {bn} is By Exercise 9, we showed that {bn} is Therefore, by the theorem Exercise 11: Give an example of a sequence that is monotonic but not bounded. Exercise 12: Give an example of a sequence that is bounded, not monotonic, and which does not converge. Exercise 13: Give an example of a sequence that is bounded, not monotonic, yet which does converge.