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consider the power series "(x-2)"" √n+2 n=o 1. Use Limit Comparison Test to show that the series (-3)^(x-2)^ √n+2 (-3)^(-2)^ √n+2 (-3)^(-1)^ n+2 x=¾: 2 11 11 n³0 10 no √n+2 +7

consider the power series "(x-2)"" √n+2 n=o 1. Use Limit Comparison Test to show that the series (-3)^(x-2)^ √n+2 (-3)^(-2)^ √n+2 (-3)^(-1)^ n+2 x=¾: 2 11 11 n³0 10 no √n+2 +7

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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please show step by step solution on how to get the circled part with complete explanation. do this typewritten and i will upvote. skip if you already did this

(-3)^(x-2)^ Consider the power series. √n+2 n=o 1. Use Limit Comparison Test to show that the series diverges when x=-=- X= x = 3: (-3)^(x-2)^ √n+2 (-3)^(-2)^ = √n+2 (-3)^(-1)^ √n +2 at x = n=0 ⇒diverges lim 1 > 0 - 13/12 Now, the series is divergent at x = - Since an diverges, therefore by Limit Comparison Test, bn also diverges. 348 = Now, bn = n Hence, an <bn Using Limit Comparison Test √√n+2 lim an lim 0+2 = n+f bo عالم = 10 no √n+2 √nt2 n+ 2 n=o lim n+ (1) √n+2 nt 2
Transcribed Image Text:(-3)^(x-2)^ Consider the power series. √n+2 n=o 1. Use Limit Comparison Test to show that the series diverges when x=-=- X= x = 3: (-3)^(x-2)^ √n+2 (-3)^(-2)^ = √n+2 (-3)^(-1)^ √n +2 at x = n=0 ⇒diverges lim 1 > 0 - 13/12 Now, the series is divergent at x = - Since an diverges, therefore by Limit Comparison Test, bn also diverges. 348 = Now, bn = n Hence, an <bn Using Limit Comparison Test √√n+2 lim an lim 0+2 = n+f bo عالم = 10 no √n+2 √nt2 n+ 2 n=o lim n+ (1) √n+2 nt 2
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