HW12 - momin (rrm497) Homework 12 cheng (58520) 1 This...

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Unformatted text preview: momin (rrm497) Homework 12 cheng (58520) 1 This print-out should have 18 questions. Multiple-choice 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 n converges or diverges. 1. series converges correct 2. series diverges Explanation: Since cos n = ( 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 lHospitals 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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HW12 - momin (rrm497) Homework 12 cheng (58520) 1 This...

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