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(a) For all n N and a # 0, show that ak an k=1 (b) For a sequence (xn) in R, suppose that there exists an b € (0, 1) such that |xn+1-xn| ≤ b for all n € N. M Use part (a) to prove that xnx for some x € R. Note: If 0 < b < 1 then b = for some a > 1. Then use part (a), since a 1 + 0.

(a) For all n N and a # 0, show that ak an k=1 (b) For a sequence (xn) in R, suppose that there exists an b € (0, 1) such that |xn+1-xn| ≤ b for all n € N. M Use part (a) to prove that xnx for some x € R. Note: If 0 < b < 1 then b = for some a > 1. Then use part (a), since a 1 + 0.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 23E
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a-1 ak 1 an = 1 (a) For all ne N and a # 0, show that k=1 (b) For a sequence (xn) in R, suppose that there exists an b € (0, 1) such that |xn+1-Xn ≤ b for all n € N. Use part (a) to prove that xnx for some x € R. Note: If 0 < b < 1 then b = for some a > 1. Then use part (a), since a > 1 + 0.
Transcribed Image Text:a-1 ak 1 an = 1 (a) For all ne N and a # 0, show that k=1 (b) For a sequence (xn) in R, suppose that there exists an b € (0, 1) such that |xn+1-Xn ≤ b for all n € N. Use part (a) to prove that xnx for some x € R. Note: If 0 < b < 1 then b = for some a > 1. Then use part (a), since a > 1 + 0.
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