Lecture43-rev

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

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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 ± ± ± ± | {z } k - N factors < s k - N Thus k > N ⇒| a k | < | a N | s k - N . Now X 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-

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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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## This note was uploaded on 03/04/2012 for the course MATH 10250 taught by Professor Himonas during the Fall '08 term at Notre Dame.

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Lecture43-rev - vergence , although it does aﬀect what...

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