121616949-math.271 - series were positive in general we...

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10.7 The Ratio and Root Tests 257 Because lim n →∞ a n +1 a n = 1 5 , then when n is big enough, say n N for some N , a n +1 a n < 1 2 and a n +1 < a n 2 . So a N +1 < a N / 2, a N +2 < a N +1 / 2 < a N / 4, a N +3 < a N +2 / 2 < a N +1 / 4 < a N / 8, and so on. The general form is a N + k < a N / 2 k . So if we look at the series a N + a N +1 + a N +2 + a N +3 + · · · + a N + k + · · · , its terms are less than or equal to the terms of the sequence a N + a N 2 + a N 4 + a N 8 + · · · + a N 2 k + · · · = X k =0 a N 2 k = 2 a N . So by the comparison test, X k =0 a N + k converges, and this means that X n =0 a n converges, since we’ve just added the fixed number a 0 + a 1 + · · · + a N - 1 . Under what circumstances could we do this? What was crucial was that the limit of a n +1 /a n , say L , was less than 1 so that we could pick a value r so that L < r < 1. The fact that L < r (1 / 5 < 1 / 2 in our example) means that we can compare the series a n to r n , and the fact that r < 1 guarantees that r n converges. That’s really all that is required to make the argument work. We also made use of the fact that the terms of the
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Unformatted text preview: series were positive; in general we simply consider the absolute values of the terms and we end up testing for absolute convergence. THEOREM 10.39 The Ratio Test Suppose that lim n →∞ | a n +1 /a n | = L . If L < 1 the series ∑ a n converges absolutely, if L > 1 the series diverges, and if L = 1 this test gives no information. Proof. The example above essentially proves the first part of this, if we simply replace 1 / 5 by L and 1 / 2 by r . Suppose that L > 1, and pick r so that 1 < r < L . Then for n ≥ N , for some N , | a n +1 | | a n | > r and | a n +1 | > r | a n | ....
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