# WA5 Clark.docx - Name Diana Clark University ID 0644492...

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Name: Diana Clark University ID: 0644492 Thomas Edison State University Calculus II (MAT-232) Section no.: WA5 Semester and year: September 2020 Written Assignment 5 Answer all assigned exercises, and show all work. Each exercise is worth 5 points. *Submitting a graph is not required; however, you are encouraged to create one for your own benefit and to include (or describe) one if possible. Section 8.6 4. Determine the radius and interval of convergence. 0 2 k k k k x lim k →∞ | a k + 1 a k | = lim k →∞ | ( k + 1 ) x k + 1 ( 2 k ) 2 k + 1 ( x k ) | = lim k →∞ | x 2 ( k + 1 k ) | = | x | 2 < 1 x = 2 , k = 1 k 2 k 2 k = k = 1 k = k = 1 a k lim k →∞ a k 0 , diverges. x =− 2 , diverges Interval convergence ( 2,2 ) , radius of convergence is r = 2 10. Determine the radius and interval of convergence. 2 4 1 (3 2) k k x k WA 5, p. 1
lim k →∞ | ( 3 x + 2 ) k + 1 ( k + 1 ) 2 k 2 ( 3 x + 2 ) k | ¿ lim k →∞ | k 2 ( 3 x + 2 ) ( k 2 + 2 k + 1 ) | ¿ lim k →∞ ( 3 x + 2 ) 1 < 3 x + 2 < 1 ¿ 1 < x 1 3 Interval of convergence is ( 1, 1 3 ) and radius is R = 1 3 12. Determine the radius and interval of convergence. 1 ( 1) (3 1) k k k x k u k = 1 k ( 3 x 1 ) k ,u k + 1 = 1 k + 1 ( 3 x 1 ) k + 1 lim k →∞ | u k + 1 u k | = lim k→ ∞ | 1 k + 1 ( 3 x 1 ) k + 1 1 k ( 3 x 1 ) k | ¿ lim k →∞ | ( 3 x 1 ) k k + 1 | ¿ | 3 x 1 | lim k →∞ 1 1 + 1 k ¿ | 3 x 1 | ( 1 ) 1 < ( 3 x 1 ) < 1 = 0 < x < 2 3 Interval of convergence is ( 0, 2 3 ) , radius is R = 2 3 16. Determine the radius and interval of convergence. 2 2 1 2 ( !) (2 )! k k k x k WA 5, p. 2
lim k →∞ | [ ( k + 1 ) ! ] 2 ( 2 k + 2 ) ! x 2 k + 2 ( x ( 2 k + 1 ) ! ( k! ) 2 x 2 k + 1 ) | = lim k →∞ | ( k + 1 ) 2 2 k + 2 ( 1 ) | | x | = lim k →∞ | ( k + 1 ) 2 ( 2 k + 2 ) | | x | lim k→ ∞ 1 2 | 1 ( k + 1 ) | | x | = 0 Converges for every x Radius of convergence will be R = , interval of convergence is ( ∞ ,∞ ) 24. Determine the interval of convergence and the function to which the given power series converges. 0 3 4 k k x k = 0 3 ( x 4 ) k = 3 ( 1 + x 4 + ( x 4 ) 2 + ) For each fixed x , this is a geometric series with common ratio r = x 4 This converges whenever | r | = | x 4 | < 1 and diverges | x | > 4 .
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