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Unformatted text preview: momin (rrm497) – Homework 12 – cheng – (58520) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine whether the series ∞ summationdisplay n = 0 1 √ n + 2 cos nπ converges or diverges. 1. series converges correct 2. series diverges Explanation: Since cos nπ = ( − 1) n , the given series can be rewritten as the alternating series ∞ summationdisplay n =0 ( − 1) n 1 √ n + 2 = ∞ summationdisplay n = 0 ( − 1) n f ( n ) with f ( x ) = 1 √ x + 2 . Now f ( n ) = 1 √ n + 2 > 1 √ n + 3 = f ( n + 1) for all n , while lim n →∞ f ( n ) = lim n →∞ 1 √ n + 2 = 0 . Consequently, by the Alternating Series Test, the given series converges . 002 10.0 points Find the smallest number of terms of the series ∞ summationdisplay k =1 ( − 1) k k 2 k we need to add in order to estimate the sum of the series with error less than 7 / 2 7 . 1. #terms = 10 2. #terms = 9 3. #terms = 6 correct 4. #terms = 8 5. #terms = 7 Explanation: For an infinite series ∞ summationdisplay k =1 a k with sum s the error in using the n th partial sum s n = a 1 + a 2 + . . . + a n instead of s is the difference s − s n = ∞ summationdisplay k = n +1 a k . For an alternating series summationdisplay k =1 ( − 1) k b k , b k > in which the b k satisfy the hypotheses ( ∗ ) b n +1 ≤ b n lim n →∞ b n = 0 we know that the error can be estimated:  s − s n  ≤ b n +1 (text page 774). To be able to apply this to the given series ∞ summationdisplay k =1 ( − 1) k b k , b k = k 2 k we first have to check that the terms b n satisfy hypotheses ( ∗ ). But (i) b n +1 = n + 1 2 n +1 < 2 n 2 · 2 n = n 2 n = b n , while momin (rrm497) – Homework 12 – cheng – (58520) 2 (ii) lim n →∞ n 2 n = 0 . Consequently, the error estimate:  s − s n  ≤ n + 1 2 n +1 . applies, so we have to find the smallest value of n for which n + 1 2 n +1 = b n +1 ≤ 7 2 7 < b n = n 2 n . Thus the smallest value is n = 6 . 003 10.0 points Determine all values of p for which the series ∞ summationdisplay n =2 ( − 1) n − 1 (ln n ) p 7 n is convergent, expressing your answer in in terval notation. 1. [0 , ∞ ) 2. ( −∞ , 0) 3. p = { } 4. ( −∞ , ∞ ) correct 5. (0 , ∞ ) Explanation: The given series can be written in the form ∞ summationdisplay n =2 ( − 1) n − 1 (ln n ) p 7 n = ∞ summationdisplay n =1 ( − 1) n f ( n ) with f defined by f ( x ) = (ln x ) p 7 x . Then by the Quotient Rule, f ′ ( x ) = (ln x ) p − 1 ( p − ln x ) 7 x 2 < if x > e p , so f is eventually decreasing for every p . On the other hand, lim x →∞ (ln x ) p 7 x = 0 if p ≤ 0, while if p > 0 we can apply l’Hospital’s Rule sufficiently many times to get a limit of 0 as well. Consequently the series ∞ summationdisplay n = 2 ( − 1) n − 1 (ln n ) p 7 n converges for all p in the interval ( −∞ , ∞ ) ....
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 Fall '08
 RAdin
 Calculus, Mathematical Series, lim

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