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9. the series diverges by the 2. Thus the series converges when r + 3| < 1, or - < a < -. At z = - Divergence Test. At r = - 3 the series diverges by the Divergence Test.. Thus the interval of convergence 2'

9. the series diverges by the 2. Thus the series converges when r + 3| < 1, or - < a < -. At z = - Divergence Test. At r = - 3 the series diverges by the Divergence Test.. Thus the interval of convergence 2'

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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For number 34 for the part circled in pink , can you show the Algebra for how we solved the inequality to get -9/2 and -3/2?  

x = 0, the series diverges as well by the Divergence Test. Thus the interval of convergence is (0, 2). 11.2.34 Using the Ratio Test: (-2)사1(z + 3)서1 3k+2 3k+1 |ak+1l |ak| lim k00 (-2)*(x+ 3)k +3|. -을 <u< -흙. At z= - 9. the series diverges by the 2' Thus the series converges when x + 3| < 1, or < x < 3. Divergence Test. At r = - the series diverges by the Divergence Test. Thus the interval of convergence 21 (을 9. is
Transcribed Image Text:x = 0, the series diverges as well by the Divergence Test. Thus the interval of convergence is (0, 2). 11.2.34 Using the Ratio Test: (-2)사1(z + 3)서1 3k+2 3k+1 |ak+1l |ak| lim k00 (-2)*(x+ 3)k +3|. -을 <u< -흙. At z= - 9. the series diverges by the 2' Thus the series converges when x + 3| < 1, or < x < 3. Divergence Test. At r = - the series diverges by the Divergence Test. Thus the interval of convergence 21 (을 9. is
(-2)* (x +3)* 34. 3k+1 k=0 ,20 xk 00
Transcribed Image Text:(-2)* (x +3)* 34. 3k+1 k=0 ,20 xk 00
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