1065 11.3 Geometric Sequences and Series 0 0 2 6.1 This graph of the first six values of Sn in Example 7 suggests that its value is approaching 3 2. If a geometric series has first term a1 and common ratio r, then Sn = a111 - r n2 1 - r 1where r ≠12, for every positive integer n. If 0 r 0 61, then lim nS∞ rn = 0, and lim nS∞ Sn = a111 - 02 1 - r = a1 1 - r . The resulting quotient a1 1 - r is the sum of the terms of an infinite geometric sequence. The limit lim nuH Sn can be expressed as SH or aH i=1 ai. LOOKING AHEAD TO CALCULUS In calculus, functions are sometimes defined in terms of infinite series. Here are three functions we studied earlier in the text defined that way. ex = x0 0! + x1 1! + x2 2! + x3 3! + g ln11 + x2 = x - x2 2 + x3 3 - g for x in 1-1, 12 1 1 + x = 1 - x + x2 - x3 + g for x in 1-1, 12 EXAMPLE 7 Evaluating an Infinite Geometric Series Evaluate 1 + 1 3 + 1 9 + 1 27 + g. SOLUTION S1 = 1, S2 = 4 3 , S3 = 13 9 , S4 = 40 27 Using the formula for the sum of the first n terms of a geometric sequence, we obtain the following, in general. Sn = a111 - r n2 1 - r Formula for Sn Sn = 1C 1 - A1 3B nD 1 - 1 3 Let a1 = 1 andr = 1 3. The table shows the value of A1 3B n for larger and larger values of n. n 1 10 100 200 a1 3b n 1 3 1.69 * 10-5 1.94 * 10-48 3.76 * 10-96 As n increases without bound, A1 3B n approaches 0. That is, lim nS∞a 1 3b n = 0, making it reasonable that lim nS∞ Sn = lim nS∞ 1C 1 - A1 3B nD 1 - 1 3 = 111 - 02 1 - 1 3 = 1 2 3 = 3 2 . Simplify the complex fraction. Hence, 1 + 1 3 + 1 9 + 1 27 + g= 3 2 . S Now Try Exercise 51.
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