Explanation Given: The power series is ∑ k = 1 ∞ ( 3 k ) ! x k ( k ! ) 3 . Result used: Ratio test Let ∑ k = 1 ∞ a k be an infinite series and let r = lim k → ∞ | a k + 1 a k | . If r < 1 , then the series ∑ k = 1 ∞ a k converges. Convergence of Power Series : A power series ∑ k = 0 ∞ c k ( x − a ) k centered at a converges if there is a real number R > 0 such that the series converges for | x − a | < R and diverges for | x − a | > R in which case the radius of convergence is R . Calculation: The radius of convergence is computed as follows, Let a k = ( 3 k ) ! x k ( k ! ) 3 . Then, a k + 1 = ( 3 k + 3 ) ! x k + 1 ( k + 1 ! ) 3 . Use the Ratio test to compute the value of | a k + 1 a k | as follows. | a k + 1 a k | = | ( 3 k + 3 ) ! x k + 1 ( k + 1 ! ) 3 ( 3 k ) ! x k ( k ! ) 3 | Take lim k → ∞ on both sides. lim k → ∞ | a k + 1 a k | = lim k → ∞ | ( 3 k + 3 ) ! x k + 1 ( ( k + 1 ) ! ) 3 ( 3 k ) ! x k ( k ! ) 3

Calculus Early Transcendentals 3rd.edition I.r.c.

3rd Edition

ISBN: 9780134766843

Author: Briggs

Publisher: PEARSON

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Chapter 11, Problem 27RE

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To find: The radius of convergence for the given power series.

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Chapter 11, Problem 26RE

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Chapter 11 Solutions

Calculus Early Transcendentals 3rd.edition I.r.c.

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Solution Summary: The author calculates the radius of convergence for a given power series using the Ratio test.

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