2022W2_MATH_101C_ALL_2022W2_Integral_Cal.56O9DSPQEY00.WW09

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Daniel Truong 2022W2 MATH 101C ALL 2022W2 Integral Cal Assignment WW09 due 03/30/2023 at 11:59pm PDT Problem 1. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Classify each of the following statements as either true or false. a) If a n b n for all n and the series n = 1 b n converges, then the series n = 1 a n converges. ? b) If S n = n k = 1 a k and lim n S n + 1 S n = 1 3 , then n = 1 a n converges. ? c) If a n 0 for all n and the series n = 1 a n converges, then n = 1 ( a n ) 2 converges. ? d) If lim n S n = 0 where S n = n k = 1 a k , then the series k = 1 a k con- verges. ? e) If lim n a n = 0, then the series n = 1 a n converges. ? f) If the series n = 1 a n converges, then lim n S n = 0 where S n = n k = 1 a k . ? Answer(s) submitted: False True True True False False submitted: (correct) recorded: (correct) 1
Problem 2. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Analyze each of the following attempted proofs, and decide if the given argument is correct or incorrect. Each of the given statements appeals to the Comparison Test (and NOT the Limit Comparison Test.) Note : If the conclusion is true but the justification provided is flawed, the appropriate answer is ”Incorrect”. a) For all n > 2, 1 n 2 - 6 < 1 n 2 , and the series n = 1 1 n 2 converges, so by the Comparison Test, the series n = 1 1 n 2 - 6 converges. ? b) For all n > 2, log ( n ) n 2 > 1 n 2 , and the series n = 1 1 n 2 converges, so by the Comparison Test, the series n = 2 log ( n ) n 2 converges. ? c) For all n > 1, n 4 - n 3 < 1 n 2 , and the series n = 1 1 n 2 converges, so by the Comparison Test, the series n = 1 n 4 - n 3 converges. ? d) For all n > 1, 1 n log ( n ) < 2 n , and the series 2 n = 1 1 n diverges, so by the Comparison Test, the series n = 2 1 n log ( n ) diverges. ? e) For all n > 2, n n 3 - 7 < 2 n 2 , and the series 2 n = 1 1 n 2 converges, so by the Comparison Test, the series n = 1 n n 3 - 7 converges. ? f) For all n > 1, arctan ( n ) n 3 < π 2 n 3 , and the series π 2 n = 1 1 n 3 con- verges, so by the Comparison Test, the series n = 1 arctan ( n ) n 3 con- verges. ? Answer(s) submitted: Incorrect Incorrect Incorrect Incorrect Correct Correct submitted: (correct) recorded: (correct) 2
Problem 3. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. For each of the following series decide if it converges or diverges and choose the most appropriate test listed, which you could use to make the judgement. a) n = 1 ( - 1 ) n n ( n + 1 ) 2 ? by the ? b) Suppose that lim n a n + 1 a n = 1 2 . The series n = 1 n 3 a n ? by the ? c) n = 1 ( n ! ) 3 ( 3 n ) ! ? by the ? d) n = 2 1 n p ln ( n ) ? by the ? e) Suppose that n = 1 a n converges and that a n > 0 for all n . The series n = 1 a n 1 + a n ? by the ? f) n = 1 arctan ( n ) 2 n + n 2 ? by the ? Answer(s) submitted: diverges divergence test converges ratio test converges ratio test diverges integral test converges comparison test converges comparison test submitted: (correct) recorded: (correct) Problem 4. (1 point) Prologue: The series n = 1 r n n p converges when - 1 < r < 1 and di- verges when | r | > 1. This is true regardless of the value of the constant p . When r = 1 the series is a p -series. It converges if p > 1 and diverges otherwise. Problem: Each of the series below can be compared to a series of the form n = 1 r n n p . In each case, determine the best value of r and decide whether the series converges. A. n = 1 ( 7 + n ( 5 ) n ) - 4 r = . This series ? . B. n = 1 n π 4 2 n 4 n + n 9 r = . This series ? . C. n = 1 n + 7 n 4 + 5 r = . This series ? . D. n = 1 7 n 2 + 4 n + 7 - 9 n 7 n + 8 + 4 n + 5 n 9 r = . This series ? . Answer(s) submitted: 1 5 4 CONVERGES 4 DIVERGES 1 CONVERGES 1 7 9 CONVERGES submitted: (correct) recorded: (correct) 3
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Problem 5. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing “Submit”. Test each series below for convergence. a) n = 1 n ( n 6 + 5 ) 1 / 2 [?/Converges/Diverges] b) n = 1 3 n 2 + 4 n [?/Converges/Diverges] c) n = 1 8 n 9 n [?/Converges/Diverges] d) n = 1 sin 6 n [?/Converges/Diverges] e) n = 1 n 5 + 4 n 1 + 9 [?/Converges/Diverges] Answer(s) submitted: Converges Converges Converges Diverges Diverges submitted: (correct) recorded: (correct) Problem 6. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing “Submit”. Test each series below for convergence. a) n = 1 ( - 1 ) n sin 2 n [?/Converges/Diverges] b) n = 1 cos ( n π ) n 4 / 7 [?/Converges/Diverges] c) n = 1 ( - 1 ) n 6 n 9 n + 9 [?/Converges/Diverges] d) n = 1 ( - 1 ) n - 1 n + 6 • ? • Converges Absolutely • Converges Conditionally • Diverges e) n = 1 ( - 8 ) n n 4 • ? • Converges Absolutely • Converges Conditionally • Diverges Answer(s) submitted: Converges Converges Diverges Converges Conditionally Diverges submitted: (correct) recorded: (correct) 4
Problem 7. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Each of the series given below converges. Enter the one-letter label of the FIRST correct reason for convergence. A. Convergent geometric series B. Convergent p series C. Comparison (or Limit Comparison) with a geometric or p series D. Alternating Series Test E. None of the above 1. n = 1 ( n + 1 )( 3 ) n 2 2 n 2. n = 1 4 ( 7 ) n 9 2 n 3. n = 1 n 2 + n n 4 - 1 4. n = 1 cos ( n π ) log ( 3 n ) 5. n = 1 sin 2 ( 5 n ) n 2 6. n = 1 ( - 1 ) n 4 n + 2 Answer(s) submitted: C A C D C D submitted: (correct) recorded: (correct) Problem 8. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. The three series A = a n , B = b n , and C = c n have terms a n = 1 n 10 , b n = 1 n 2 , c n = 1 n . Use the Limit Comparison Test to compare each of the follow- ing series to one of the series above. Enter one of the following two-letter codes for each answer: AC , BC , CC , AD , BD , CD . The first letter is the name of the series above used to make a valid comparison; the second letter is C if the given series converges, or D if it diverges. 1. n = 1 5 n 2 + 8 n 9 2 n 10 + 7 n 3 - 2 2. n = 1 2 n 2 + n 10 1309 n 12 + 7 n 2 + 5 3. n = 1 8 n 6 + n 2 - 8 n 7 n 16 - 5 n 12 + 7 Answer(s) submitted: CD BC AC submitted: (correct) recorded: (correct) 5
Problem 9. (1 point) For each sequence a n below, find a number p such that n p a n has a finite non-zero limit as n . (According to the limit comparison test, the convergence or diver- gence of the p -series n = 1 1 n p then predicts the same behaviour for the series n = 1 a n .) A. a n = ( 4 + 2 n ) - 2 p = B. a n = 2 n 7 + n p = C. a n = 6 n 2 + 2 n + 6 7 n 2 + 7 n + 2 p = D. a n = 6 n 2 + 2 n + 4 7 n 2 + 7 n + 2 n 3 p = Answer(s) submitted: 2 7 0 0 submitted: (correct) recorded: (correct) Problem 10. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Consider the series S = n = 1 a n , with a n = ( - 1 ) n - 1 3 n n 3 . (a) Evaluate the following limit. (Enter a number or one of the 3-letter codes ”inf” or ”div”.) Answer: lim n a n + 1 a n = (b) What does the Ratio Test say about the series S ? Choose from ”Convergent”, ”Divergent”, or ”Inconclusive”, below. Answer: • choose one • Convergent • Divergent • Inconclusive (c) Decide whether the series is absolutely convergent, condition- ally convergent, or divergent. Use the menu below to present your findings. Answer: • choose one • Absolutely Convergent • Conditionally Convergent • Divergent Answer(s) submitted: 3 Divergent Divergent submitted: (correct) recorded: (correct) 6
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Problem 11. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Consider the series S = n = 1 a n , with a n = ( - 1 ) n n 5 . (a) Evaluate the following limit. (Enter a number or one of the 3-letter codes ”inf” or ”div”.) Answer: lim n a n + 1 a n = (b) What does the Ratio Test say about the series S ? Choose from ”Convergent”, ”Divergent”, or ”Inconclusive”, below. Answer: • choose one • Convergent • Divergent • Inconclusive (c) Decide whether the series is absolutely convergent, condition- ally convergent, or divergent. Use the menu below to present your findings. Answer: • choose one • Absolutely Convergent • Conditionally Convergent • Divergent Answer(s) submitted: 1 Inconclusive Absolutely Convergent submitted: (correct) recorded: (correct) Problem 12. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Consider the series S = n = 1 a n , with a n = e - n n !. (a) Evaluate the following limit. (Enter a number or one of the 3-letter codes ”inf” or ”div”.) Answer: lim n a n + 1 a n = (b) What does the Ratio Test say about the series S ? Choose from ”Convergent”, ”Divergent”, or ”Inconclusive”, below. Answer: • choose one • Convergent • Divergent • Inconclusive (c) Decide whether the series is absolutely convergent, condition- ally convergent, or divergent. Use the menu below to present your findings. Answer: • choose one • Absolutely Convergent • Conditionally Convergent • Divergent Answer(s) submitted: INF Divergent Divergent submitted: (correct) recorded: (correct) 7
Problem 13. (1 point) CAUTION: You have five chances to submit an answer for this problem. Make sure you have answered all the parts before press- ing ”Submit”. Consider the series S = n = 1 a n , with a n = ( - 1 ) n n 2 5 n n ! . (a) Evaluate the following limit. (Enter a number or one of the 3-letter codes ”inf” or ”div”.) Answer: lim n a n + 1 a n = (b) What does the Ratio Test say about the series S ? Choose from ”Convergent”, ”Divergent”, or ”Inconclusive”, below. Answer: • choose one • Convergent • Divergent • Inconclusive (c) Decide whether the series is absolutely convergent, condition- ally convergent, or divergent. Use the menu below to present your findings. Answer: • choose one • Absolutely Convergent • Conditionally Convergent • Divergent Answer(s) submitted: 0 Convergent Absolutely Convergent submitted: (correct) recorded: (correct) Problem 14. (1 point) For an alternating series whose summands are decreasing in mag- nitude, the true sum S lies between any two successive partial sums: ( * ) min { S N , S N + 1 } ≤ S max { S N , S N + 1 } . Consider S = n = 1 ( - 1 ) n + 1 n 3 , and write S N = N n = 1 ( - 1 ) n + 1 n 3 . (a) Find the smallest value of N for which the interval bracketing S in line ( * ) above has length at most 10 - 3 . Answer: N min = (b) Using the N found in part (a) , approximate S by the midpoint of the interval implicit in line ( * ) . A spreadsheet may be helpful to calculate the sum S N . Answer: S S N + S N + 1 2 = Answer(s) submitted: 9 0 . 901615 submitted: (correct) recorded: (correct) Problem 15. (1 point) Find all the values of x such that the given series would converge. n = 1 ( n + 7 ) x n Answer: Note: Give your answer in interval notation . Answer(s) submitted: ( - 1 , 1 ) submitted: (correct) recorded: (correct) 8
Problem 16. (1 point) Find all the values of x such that the given series would converge. n = 1 ( - 1 ) n n 2 n x n Answer: Note: Give your answer in interval notation . Answer(s) submitted: - 1 2 , 1 2 submitted: (correct) recorded: (correct) Problem 17. (1 point) For each n 2, let a n = n 3 + 7 n 6 n 2 + 3 n + 1 sin π n . Evaluate the fol- lowing series: S = n = 2 a n + 1 - a n a n a n + 1 = Suggestion : Flex your abstract thinking skills. Work with the given abstract form of S for as long as possible, so that your inter- actions with the detailed definition of a n can be laser-focused and done last. Answer(s) submitted: 31 22 - 6 π submitted: (correct) recorded: (correct) Problem 18. (1 point) Given p ( t ) = 88 e 5 t 95 + 75 e 5 t , consider the series S ( t ) = n = 1 e - p ( t ) log ( n ) . Find the set of real numbers t for which S ( t ) converges. Write your answer in interval notation. Interval for convergence: Answer(s) submitted: ln 95 13 5 , submitted: (correct) recorded: (correct) Problem 19. (1 point) Find the set of real numbers z for which the following series con- verges: S ( z ) = n = 4 30 ( - 5 ) n n ! ( nz ) n . Use interval notation to present your answer. Set of real z values for convergence: Hint : Remember that lim n 1 + 1 n n = e . To treat borderline z -values, Stirling’s approximation of the factorial is accurate enough: n ! 2 π n n e n . Answer(s) submitted: 5 e , - , - 5 e submitted: (correct) recorded: (correct) Problem 20. (1 point) Use interval notation to present the answers to these three ques- tions. (a) Find the set of real numbers t for which the following series converges: A ( t ) = n = 6 ( 2 t + 17 ) n . Answer: (b) Find the set of all real numbers x for which the following series converges: B ( x ) = n = 5 arctan - 61 x 29 n . Answer: (c) Find the largest interval including 0 on which the following series converges: C ( θ ) = n = 16 2 n sin 2 n 39 86 θ . Answer: Answer(s) submitted: ( - 9 , - 8 ) tan ( - 1 ) · 29 61 , tan ( 1 ) · 29 61 ( - 1 . 7319 , 1 . 7319 ) submitted: (correct) recorded: (correct) 9
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