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Lecture43-rev

# Lecture43-rev - vergence although it does aﬀect what the...

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MAT A26 Lecture 43 1 Ratio Test for Convergence proposition 1.1. (The Ratio Test ) Suppose that r = lim k → ∞ a k +1 a k exists. Then r < 1 k =1 a k converges absolutely r > 1 k =1 a k diverges r = 1 nothing at all (test is inconclusive in this case) proof. We prove the first case only. Since 0 r < 1 we may choose s > 0 such that r < s < 1. Note that 0 < r < s and lim k → ∞ a k +1 a k = r . Then there is a positive integer N such that k N 0 < a k +1 a k < s . Thus, for k > N , we have a k a N = a k a k - 1 a k - 1 a k - 2 · · · a N +1 a N k - N factors < s k - N Thus k > N ⇒ | a k | < | a N | s k - N . Now k = N +1 | a N | s k - N = | a N | ( s + s 2 + s 3 + · · · ) = | a N | s 1 - s (since | s | < 1) By the comparison test k = N +1 | a k | also converges. Therefore k =1 | a k | converges too. (Adding finitely many terms does not affect the fact of con- vergence , although it does affect what the series converges

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Unformatted text preview: vergence , although it does aﬀect what the series converges to .) ± example 1.2. Does the the ratio test applies to the series ∞ X k =1 x k k ! 1 solution. We have a k +1 a k = x k +1 ( k +1)! x k k ! = x k + 1 → 0 as k → ∞ Thus r = 0 and we have convergence for every x . example 1.3. Does the the ratio test applies to the series ∞ X k =1 1 2 k +(-1) k solution. We have a k +1 a k = 1 2 k +1+(-1) k +1 1 2 k +(-1) k = 2 k +(-1) k 2 k +1+(-1) k +1 = 1 2 · 2 2(-1) k = ± 2 if k is even 1 8 if k is odd Thus lim k → ∞ a k +1 a k doesn’t exist, and the test is inconclusive. 2...
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