Series (2), a geometric series with ratio r = -1 -4 2, converges to -2 theorem 15the alternating series test The series q a n = 1 (-1)n + 1un = u1 u2 + u3 u4 + g converges if all three of the following conditions are satisfied: 1. The un>s are all positive. 2. The positive un>s are (eventually) nonincreasing: un un + 1 for all n N, for some integer N. 3. un S 0. Proof Assume N = 1. If n is an even integer, say n = 2m, then the sum of the first n terms is s2m = (u1 u2) + (u3 u4) + g + (u2m 1 u2m) = u1 (u2 u3) (u4 u5) g (u2m 2 u2m 1) u2m. The first equality shows that s2m is the sum of m nonnegative terms, since each term in parentheses is positive or zero. Hence s2m + 2 s2m, and the sequence s2m is nondecreasing. The second equality shows that s2m u1. Since is nondecreasing and 6 bounded from above, it has a limit, say s2m 5 lim mSq

5 6 If n is an odd integer, say n = 2m + 1, then the sum of the first n terms is s2m + 1 = s2m + u2m + 1. Since un S 0, lim mSq u2m + 1 = 0 and, as m S q, s2m + 1 = s2m + u2m + 1 S L + 0 = L. Combining the results of Equations (4) and (5) gives limnSq sn = L (Section 10.1, Exercise 133). example 1 The alternating harmonic series q a n = 1 (-1)n + 1 1 n = 1 1 2 + 1 3 1 4 + g clearly satisfies the three requirements of Theorem 15 with N = 1; it therefore converges. Rather than directly verifying the definition un un + 1, a second way to show that the sequence {un} is nonincreasing is to define a differentiable function (x) satisfying (n) = un. That is, the values of match the values of the sequence at every positive 599 (4) (5) 09/11/13 4:05 PM 600

Chapter 10: Infinite Sequences and Series +u1 u2 integer n. If (x) 0 for all x greater than or equal to some positive integer N, then (x) is nonincreasing for x N. It follows that (n) (n + 1), or un un + 1, for n N. +u3 0 s2 s4 L u4 s3 s1 x example 2 10x Consider the sequence where un = 10n (n2 + 16). Define (x) = (x2 + 16). Then from the Derivative Quotient Rule, > 10(16 x2) (x2 + 16)2 0 whenever x 4. > (x) = Figure 10.13 The partial sums of an alternating series that satisfies the hypotheses of Theorem 15 for N = 1 straddle the limit from the beginning. It follows that un un + 1 for n 4. That is, the sequence n 4. 5

13) shows how an alternating series converges to its limit L when the three conditions of Theorem 15 are satisfied with N = 1. Starting from the origin of the x-axis, we lay off the positive distance s1 = u1. To find the point corresponding to s2 = u1 u2, we back up a distance equal to u2. Since u2 u1, we do not back up any farther than the origin. We continue in this seesaw fashion, backing up or going forward as the signs in the series demand. But for n N, each forward or backward step is shorter than (or at most the same size as) the preceding step because un + 1 un. And since the nth term approaches zero as n increases, the size of step we take forward or backward gets smaller and smaller. We oscillate across the limit L, and the amplitude of oscillation approaches zero. The limit L lies between any two successive sums sn and sn + 1 and hence differs from sn by an amount less than un + 1.