Explanation The Radio Test for series is very useful test for determining whether or not a series converges or diverges. The Ratio Test says that: Let the following limit exists, L = lim n → ∞ | a n + 1 a n | i. If L < 1 , then the original series ∑ n = 1 ∞ a n will be absolutely convergent. ii. If L > 1 or L = ∞ , then the original series ∑ n = 1 ∞ a n will be divergent. iii. If L = 1 , then no conclusion can be made (the series may diverge or converge). Given: (a) ∑ n = 1 ∞ ( x − 5 ) n 9 n (b) ∑ n = 1 ∞ ( x − 5 ) n n 9 n (c) ∑ n = 1 ∞ ( x − 5 ) n n 2 9 n Formula Used: L = lim n → ∞ | a n + 1 a n | Proof: (a) Consider the power series: ∑ n = 1 ∞ ( x − 5 ) n 9 n So, let a n = ( x − 5 ) n 9 n . Then L = lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( x − 5 ) n + 1 9 n + 1 ( x − 5 ) n 9 n | = lim n → ∞ | x − 5 9 | = 1 9 lim n → ∞ | x − 5 | L = 1 9 | x − 5 | Here, r = 1 9 Since radius of convergence, R = r − 1 Then, R = ( 1 9 ) − 1 R = 9 Now, the interval of convergence can be calculated as: | x − 5 | < 9 − 9 < | x − 5 | < 9 − 4 < x < 14 Now, we have to prove that the power series diverges at the endpoints. So, L ( − 4 ) = ∑ n = 1 ∞ ( − 4 − 5 ) n 9 n = ∑ n = 1 ∞ ( − 9 ) n 9 n = ∑ n = 1 ∞ ( − 1 ) n 9 n 9 n = ∑ n = 1 ∞ ( − 1 ) n By the geometric series test, this series diverges. Thus, divergent at x = − 4 And L ( 14 ) = ∑ n = 1 ∞ ( 14 − 5 ) n 9 n = ∑ n = 1 ∞ 9 n 9 n = ∑ n = 1 ∞ 1 By the limit test, this series diverges. Thus, divergent at x = 14 Thus, it is proved that the power series diverges at the endpoints x = − 4 and x = 14 . (b) Consider the power series: ∑ n = 1 ∞ ( x − 5 ) n n 9 n So, let a n = ( x − 5 ) n n 9 n . Then L = lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( x − 5 ) n + 1 ( n + 1 ) 9 n + 1 ( x − 5 ) n n 9 n | = lim n → ∞ | n ( n + 1 ) ⋅ ( x − 5 ) n + 1 9 n ( x − 5 ) n 9 n + 1 | = lim n → ∞ | 1 ( 1 + 1 n ) ⋅ x − 5 9 | = 1 9 lim n → ∞ | x − 5 | L = 1 9 | x − 5 | Here, r = 1 9 Since radius of convergence, R = r − 1 Then, R = ( 1 9 ) − 1 R = 9 Now, the interval of convergence can be calculated as: | x − 5 | < 9 − 9 < | x − 5 | < 9 − 4 < x < 14 Now, we have to prove that the power series converges at one endpoint but diverges at the other endpoint...

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.6, Problem 4E

To determine

To prove:

The power series (a)-(c) have the same radius of convergence.

(a) Diverges at both the endpoints.

(b) Converges at one endpoint but diverges at the other endpoint.

(c) Converges at both the endpoints.

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Chapter 11.6, Problem 3E

Chapter 11.6, Problem 5E

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Chapter 11 Solutions

Calculus - Standalone book

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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To prove: The power series (a)-(c) have the same radius of convergence. (a) Diverges at both the endpoints. (b) Converges at one endpoint but diverges at the other endpoint. (c) Converges at both the endpoints.

Solution Summary: The author explains that the Radio Test for series is useful test for determining whether or not a series converges or diverges.

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To prove: The power series (a)-(c) have the same radius of convergence. (a) Diverges at both the endpoints. (b) Converges at one endpoint but diverges at the other endpoint. (c) Converges at both the endpoints.

Want to see the full answer?

Chapter 11 Solutions

To prove:

The power series (a)-(c) have the same radius of convergence.

(a) Diverges at both the endpoints.

(b) Converges at one endpoint but diverges at the other endpoint.

(c) Converges at both the endpoints.