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Gilbert_Hmwk06sol

# Gilbert_Hmwk06sol - moseley(cmm3869 HW06 Gilbert(56380 This...

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moseley (cmm3869) – HW06 – Gilbert – (56380) 1 This print-out should have 21 questions. Multiple-choice 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 p -series 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] .

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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 . By the Ratio Test, therefore, the given series converges when | x +1 | < 2 and diverges when | x + 1 | > 2. Consequently, R = 2 . 004 (part 2 of 2) 10.0 points (ii) Determine the interval of convergence of the series. 1. converges only at x = - 1 2. interval convergence = [ - 1 , 3) 3. interval convergence = ( - 3 , 1] correct 4. interval convergence = ( - 3 , 1) 5. interval convergence = ( -∞ , ) 6. interval convergence = [ - 1 , 3] Explanation: Since R = 2, the given series (i) converges when | x + 1 | < 2, and (ii) diverges when | x + 1 | > 2.
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Gilbert_Hmwk06sol - moseley(cmm3869 HW06 Gilbert(56380 This...

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