# SP21 MATH 2024 - The Alternating Series Test.pdf - MATH...

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MATH 2024 The Alternating Series Test Adams Alternating Series So far, most series we have dealt with have had positive terms. Today, we begin to examine series that contain both positive and negative terms. The simplest such series is an alternating series , whose terms alternate in sign. For example, the geometric series X n =0 - 1 2 n = X n =0 ( - 1) n 1 2 n = 1 - 1 2 + 1 4 - 1 8 + . . . is an alternating geometric series with r = - 1 2 . Alternating series occur in two ways: either the odd terms are negative or the even terms are negative. Theorem: The Alternating Series Test Let a n > 0. The alternating series X n =1 ( - 1) n a n and X n =1 ( - 1) n +1 a n converge if the following two conditions are met: 1. lim n →∞ a n = 0 2. a n +1 a n for all n . Proof: Consider the alternating series X ( - 1) n +1 a n . For this series, the partial sum, where 2 n is even, S 2 n = ( a 1 - a 2 ) + ( a 3 - a 4 ) + ( a 5 - a 6 ) + . . . + ( a 2 n - 1 - a 2 n ) has all nonnegative terms, and therefore { S 2 n } is a nondecreasing sequence. But you can also write S 2 n = a 1 - ( a 2 - a 3 ) - ( a 4 - a 5 ) - ( a 6 - a 7 ) - . . . - ( a 2 n - 2 - a 2 n - 1 ) - a 2 n which implies that S 2 n a n for every integer n . So, { S 2 n } is a bounded, nondecreasing sequence that converges to some value L . Because S 2 n - 1 - a 2 n = S 2 n and a 2 n 0, you have lim n →∞ S 2 n - 1 = lim n →∞ S 2 n + lim n →∞ a 2 n = L + lim n →∞ a 2 n = L Because both S 2 n and S 2 n - 1 converge to the same limit L , it follows that { S 2 n } also converges to L . Consequently, the given alternating series converges. Page 1 of 5
MATH 2024 The Alternating Series Test Adams Example 1. Determine the convergence or divergence of X n =1 ( - 1) n +1 1 n .