Lecture44-rev

Lecture44-rev - k = 2 ` + 1 is odd, we have ( k !) 1 /k =...

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MAT A26 Lecture 44 1 Root Test for Convergence proposition 1.1. (The Root Test) Suppose that lim k → ∞ | a k | 1 /k = r 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 the proof of the case | r | < 1 only. Since | r | < 1 we may choose s ( r, 1). Note that 0 < r < 1 and then 0 < | a k | 1 /k < s for k N (for some N ). Thus | a k | < s k for k N and so X k = N | a k | < X k = N s k = s k 1 - s So by the comparison test, k = N a k converges absolutely, and hence so does k =1 a k . ± example 1.2. Does the root test apply to the series X k =1 x k k ! solution. We have | a k | 1 /k = ± ± ± ± x k k ! ± ± ± ± 1 /k = | x | ( k !) 1 /k The limit r = lim k → ∞ | x | ( k !) 1 /k can indeed be found as follows (Note: Our method here is by brute force; using the the ratio test is easier.) Assuming 1
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k = 2 ` is even we have ( k !) 1 /k = (1 · 2 · 3 ··· ` · ( ` + 1) ··· k ) | {z } k = 2 ` factors 1 /k (( ` + 1) ··· k ) | {z } ` factors 1 /k ` `/k = ( 1 2 k ) 1 2 On the other hand if
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Unformatted text preview: k = 2 ` + 1 is odd, we have ( k !) 1 /k = (1 2 3 ` ( ` + 1) k ) | {z } k = 2 ` + 1 factors 1 /k (( ` + 1) k ) | {z } ` + 1 factors 1 /k ( ` + 1) ` +1 k ( 1 2 k ) 1 2 (since ` + 1 = k +1 2 &gt; 1 2 k ` + 1 k &gt; 1 2 ) In either case, ( k !) 1 /k ( 1 2 k ) 1 2 as k . Thus r = 0 and we have convergence. example 1.3. See whether or not the the root test applies to the series X k =1 1 2 k +(-1) k . solution. We have | a k | 1 /k = 1 2 k +(-1) k 1 /k = 1 2 ( k +(-1) k ) /k = 1 2 1+ 1 k (-1) k 1 2 as k Since lim k | a k | 1 /k = 1 2 &lt; 1 the series converges. 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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Lecture44-rev - k = 2 ` + 1 is odd, we have ( k !) 1 /k =...

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