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Unformatted text preview: moseley (cmm3869) – HW06 – Gilbert – (56380) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points For the series ∞ summationdisplay n = 1 ( 1) n n + 5 x n , (i) determine its radius of convergence, R . 1. R = 1 5 2. R = 5 3. R = 0 4. R = 1 correct 5. R = (∞ , ∞ ) Explanation: The given series has the form ∞ summationdisplay n = 1 a n with a n = ( 1) n x n n + 5 . Now for this series, (i) R = 0 if it converges only at x = 0, (ii) R = ∞ if it converges for all x , while 0 < R < ∞ , (iii) if it converges when  x  < R , and (iv) diverges when  x  > R . But lim n →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n →∞  x  n + 5 n + 6 =  x  . By the Ratio Test, therefore, the given series converges when  x  < 1 and diverges when  x  > 1. Consequently, R = 1 . 002 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. interval convergence = ( 1 , 1] correct 2. converges only at x = 0 3. interval convergence = ( 1 , 1) 4. interval convergence = ( 5 , 5) 5. interval convergence = [ 5 , 5) 6. interval convergence = ( 5 , 5] 7. interval convergence = [ 1 , 1) Explanation: Since R = 1, the given series (i) converges when  x  < 1, and (ii) diverges when  x  > 1. On the other hand, at the point x = 1 and x = 1, the series reduces to ∞ summationdisplay n = 1 ( 1) n n + 5 , ∞ summationdisplay n = 1 1 n + 5 respectively. But by the Alternating Series Test, the first of these series converges. On the other hand, if we set a n = 1 n + 5 , b n = 1 n , then lim n →∞ a n b n = lim n →∞ n n + 5 = 1 . By the pseries Test with p = 1, however, the series ∑ n b n diverges. Thus by the Limit Comparison test, the series ∑ n a n also di verges. Consequently, the given series has interval convergence = ( 1 , 1] . moseley (cmm3869) – HW06 – Gilbert – (56380) 2 keywords: 003 (part 1 of 2) 10.0 points For the series ∞ summationdisplay n =1 ( 1) n n 2 n ( x + 1) n , (i) determine its radius of convergence, R . 1. R = 0 2. R = ∞ 3. R = 1 2 4. R = 1 5. R = 2 correct Explanation: The given series has the form summationdisplay n = 1 a n ( x + 1) n with a n = ( 1) n n 2 n . Now for this series (i) R = 0 if it converges only at x = 1, (ii) R = ∞ if it converges for all x , while if R > 0, (iii) it coverges when  x + 1  < R , and (iv) diverges when  x + 1  > R . But lim n →∞ vextendsingle vextendsingle vextendsingle a n +1 a n vextendsingle vextendsingle vextendsingle = lim n →∞ n 2( n + 1) = 1 2 ....
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This note was uploaded on 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.
 Spring '07
 Sadler
 Multivariable Calculus

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