11.8 Power Series-solutions

# 11.8 Power Series-solutions - im(gi768 11.8 Power Series...

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im (gi768) – 11.8 Power Series – sadun – (53088) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind the interval oF convergence oF the se- ries s n =1 x n n + 2 . 1. interval oF cgce = [ 1 , 1] 2. interval oF cgce = ( 1 , 1) 3. interval oF cgce = [ 2 , 2] 4. interval oF cgce = [ 1 , 1) correct 5. interval oF cgce = ( 2 , 2] 6. converges only at x = 0 Explanation: When a n = x n n + 2 , then v v v v a n +1 a n v v v v = v v v v x n +1 n + 3 n + 2 x n v v v v = | x | p n + 2 n + 3 P = | x | r n + 2 n + 3 . But lim n →∞ n + 2 n + 3 = 1 , so lim n →∞ v v v v a n +1 a n v v v v = | x | . By the Ratio Test, thereFore, the given series (i) converges when | x | < 1, (ii) diverges when | x | > 1. We have still to check what happens at the endpoints x = ± 1. At x = 1 the series becomes ( ) s n =1 1 n + 2 . Applying the Integral Test with f ( x ) = 1 x + 2 we see that f is continuous, positive, and de- creasing on [1 , ), but the improper integral I = i 1 f ( x ) dx diverges, so the infnite series ( ) diverges also. On the other hand, at x = 1, the series becomes ( ) s n =1 ( 1) n n + 2 . which is an alternating series s n =1 ( 1) n a n , a n = f ( n ) with f ( x ) = 1 x + 2 the same continuous, positive and decreasing Function as beFore. Since lim x →∞ f ( x ) = lim x →∞ 1 x + 2 = 0 , however, the Alternating Series Test ensures that ( ) converges. Consequently, the interval oF convergence = [ 1 , 1) . 002 10.0 points IF the series s n =0 c n 4 n

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im (gi768) – 11.8 Power Series – sadun – (53088) 2 is convergent, which of the following state- ments must be true without further restric- tions on c n ? 1. s n =0 c n ( 4) n is convergent 2. s n =0 c n ( 3) n is convergent correct 3. s n =0 c n ( 3) n is divergent 4. s n =0 c n ( 4) n is divergent Explanation: The series s n =0 c n x n converges when x = 4, its radius of conver- gence, R , must have the property R 4 since the series (i) converges when | x | < R , and (ii) diverges when | x | > R .
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