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Step 1 |an + 1 an + 1 n+ 1 < 1, then Fa, converges absolutely. If lim an Recall the Ratio Test which states that, if a, is a series with nonzero terms, and lim > 1, or lim = 00, then n00 an Sa, diverges. For any fixed value of x such that x * 0, let a, : = n. (-3x)" - - (-3x)^ – 1 (n + 1) Then, (n + 1)(-3x)" an + 1 n+ 2 lim = lim n(-3x)^ - n00 n+1 n+ 1 lim (n + 1)(-3x)" . n + 2 n - 1 n -3x -3.r 3 x Step 2 By the Ratio Test, the series converges if lim n +1 < 1. Therefore, the series converges for x such that 3|x| < 1. n00 an Step 3 Solve to find the values of x such that the series converges. 3|x| < 1 |x| < 1/3V 1/3 Therefore, the radius of convergence is R = 1/3 1/3

Step 1 |an + 1 an + 1 n+ 1 < 1, then Fa, converges absolutely. If lim an Recall the Ratio Test which states that, if a, is a series with nonzero terms, and lim > 1, or lim = 00, then n00 an Sa, diverges. For any fixed value of x such that x * 0, let a, : = n. (-3x)" - - (-3x)^ – 1 (n + 1) Then, (n + 1)(-3x)" an + 1 n+ 2 lim = lim n(-3x)^ - n00 n+1 n+ 1 lim (n + 1)(-3x)" . n + 2 n - 1 n -3x -3.r 3 x Step 2 By the Ratio Test, the series converges if lim n +1 < 1. Therefore, the series converges for x such that 3|x| < 1. n00 an Step 3 Solve to find the values of x such that the series converges. 3|x| < 1 |x| < 1/3V 1/3 Therefore, the radius of convergence is R = 1/3 1/3

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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Step 1 Recall the Ratio Test which states that, if ) an + 1 < 1, then Sa, converges absolutely. If lim n→ 00 a n + 1 a n + 1 is a series with nonzero terms, and lim > 1, or lim = 0, then a an an an Σ Ea, diverges. n - 1 (-3x)^ – (n + 1) For any fixed value of x such that x # 0, let a, Then, (n + 1)(-3x)" 'n + 1 n + 2 lim n> 00 lim n> 00 1 n(-3x)" n - n + 1 n + 1 lim (n + 1)(-3x)" n - 1 %3D in -3x n- 00 n + 2 -3.x 3 x Step 2 an + 1 By the Ratio Test, the series converges if lim < 1. Therefore, the series converges for x such that 3|x|| 1. an Step 3 Solve to find the values of x such that the series converges. 3|x| < 1 |x| < 1/3 1/3 Therefore, the radius of convergence is R = 1/3 1/3
Transcribed Image Text:Step 1 Recall the Ratio Test which states that, if ) an + 1 < 1, then Sa, converges absolutely. If lim n→ 00 a n + 1 a n + 1 is a series with nonzero terms, and lim > 1, or lim = 0, then a an an an Σ Ea, diverges. n - 1 (-3x)^ – (n + 1) For any fixed value of x such that x # 0, let a, Then, (n + 1)(-3x)" 'n + 1 n + 2 lim n> 00 lim n> 00 1 n(-3x)" n - n + 1 n + 1 lim (n + 1)(-3x)" n - 1 %3D in -3x n- 00 n + 2 -3.x 3 x Step 2 an + 1 By the Ratio Test, the series converges if lim < 1. Therefore, the series converges for x such that 3|x|| 1. an Step 3 Solve to find the values of x such that the series converges. 3|x| < 1 |x| < 1/3 1/3 Therefore, the radius of convergence is R = 1/3 1/3
Step 4 1 E. Now we have to test the endpoints of the interval of convergence, x = 0 + 3 1 and x = 0 3 1 The radius of convergence is R 3 3 3 Substitute x = and simplify. 1 n - 1 n -31 Σ (n+1) (n + 1) n +1 n = 1 n = 1 Step 5 Recall the nth Term Test. If lim a, + 0, the series diverges. n> 00 in lim n- 0 n + 1 1 1 n(-3x)" (n + 1) 1 diverges at x = - n - Therefore, the series converges X n = 1 Step 6 1 the series is (-1)" – 1 n n n Let an Σ When x = Find the limit of the term of the series. = n + 1 n + 1 n = 1 lim 1 1 an n> 00 By the Alternating Series Test, the series convergesX diverges for x = Step 7 The radius of convergence is R = 1/3| 1/3 and the series diverges for x = ± Find the interval of convergence. (If the interval of convergence is an interval, enter your answer 3 using interval notation. If the interval of convergence is a finite set, enter your answer using set notation.)
Transcribed Image Text:Step 4 1 E. Now we have to test the endpoints of the interval of convergence, x = 0 + 3 1 and x = 0 3 1 The radius of convergence is R 3 3 3 Substitute x = and simplify. 1 n - 1 n -31 Σ (n+1) (n + 1) n +1 n = 1 n = 1 Step 5 Recall the nth Term Test. If lim a, + 0, the series diverges. n> 00 in lim n- 0 n + 1 1 1 n(-3x)" (n + 1) 1 diverges at x = - n - Therefore, the series converges X n = 1 Step 6 1 the series is (-1)" – 1 n n n Let an Σ When x = Find the limit of the term of the series. = n + 1 n + 1 n = 1 lim 1 1 an n> 00 By the Alternating Series Test, the series convergesX diverges for x = Step 7 The radius of convergence is R = 1/3| 1/3 and the series diverges for x = ± Find the interval of convergence. (If the interval of convergence is an interval, enter your answer 3 using interval notation. If the interval of convergence is a finite set, enter your answer using set notation.)
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