1041 11.1 Sequences and Series EXAMPLE 1 FindingTerms of Sequences Write the first five terms of each sequence. (a) an = n + 1 n + 2 (b) an = 1-12n # n (c) a n = 2n + 1 n2 + 1 SOLUTION (a) Replace n in an = n + 1 n + 2 with 1, 2, 3, 4, and 5. n = 1: a1 = 1 + 1 1 + 2 = 2 3 n = 2: a2 = 2 + 1 2 + 2 = 3 4 n = 3: a3 = 3 + 1 3 + 2 = 4 5 n = 4: a4 = 4 + 1 4 + 2 = 5 6 n = 5: a5 = 5 + 1 5 + 2 = 6 7 (b) Replace n in an = 1-12n # n with 1, 2, 3, 4, and 5. n = 1: a1 = 1-121 # 1 = -1 n = 2: a2 = 1-122 # 2 = 2 n = 3: a3 = 1-123 # 3 = -3 n = 4: a4 = 1-124 # 4 = 4 n = 5: a5 = 1-125 # 5 = -5 (c) For an = 2n + 1 n2 + 1 , the first five terms are as follows. a1 = 3 2 , a2 = 1, a3 = 7 10 , a4 = 9 17 , and a5 = 11 26 Replace n with 1, 2, 3, 4, and 5. S Now Try Exercises 13, 17, and 19. 012345678910 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 n an an = 1n Figure 3 If the terms of an infinite sequence get closer and closer to some real number, the sequence is said to be convergent and to converge to that real number. For example, the sequence defined by an = 1 n approaches 0 as n becomes large. Thus, an is a convergent sequence that converges to 0. A graph of this sequence for n = 1, 2, 3, c , 10 is shown in Figure 3. The terms of an approach the horizontal axis. A sequence that does not converge to any number is divergent. The first nine terms of the sequence an = n2 are 1, 4, 9, 16, 25, 36, 49, 64, 81, c . This sequence is divergent because as n becomes large, the values of an do not approach a fixed number—rather, they increase without bound. Some sequences are defined by a recursive definition, one in which each term after the first term or first few terms is defined as an expression involving the previous term or terms. The sequences in Example 1 were defined explicitly, with a formula for an that does not depend on a previous term.
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