Math 21, Winter 2018 Schaeffer/Solis Midterm Exam 2 (February 28th, 2018) Last/Family Name First/Given Name Seat # Exam # Failure to follow the instructions below will constitute a breach of the Honor Code: You may not use consult any book or notes during the exam.* You may not use a calculator or the calculator function on any electronic device during the exam.* You may not access any internet-enabled electronic device during the exam,* including smartphones, for any reason except to check the time. You may only check the time on your device if and only if you (1) Do not access any other information, and (2) Have turned off all network/internet capabilities before the start of the exam (i.e., it is in airplane mode with wifi off). You must sit in your assigned seat. You may not communicate with anyone other than the course staff during the exam,* or look at anyone else s solutions. * During the exam is defined as: After you start the exam, and before you turn in the exam and leave the testing site. You have 80 minutes to complete this exam. If the course staff must ask you to stop writing or to turn in your exam more than once after time is called, you may receive a score of zero. You may not discuss the contents of this exam with ANYONE other than the course staff until 10:00 PM tonight. I understand and accept these instructions. Signature: Remember to show your work and justify your answer if required. Present all solutions in as organized a manner as possible. GOOD LUCK! 1
Unless otherwise specified, in problems 1 8 you do not need to justify your answer. 1. In a. c. below, either write down an example of a sequence (not a series) with the given properties, or state that such a sequence does not exist (DNE). A formula in n or k suffices, e.g. + 1 would be a valid answer. a. Bounded and not convergent. {( 1) n } = {1, 1, 1, 1,...}is one example. b. Convergent and not bounded. This is not possible. Every convergent sequence is bounded. (This is one of two important relationships between sequence properties we learned. The other is that every bounded monotone sequence is convergent.) c. Convergent and not monotonic. {( 1/2) n } = {1, 1/2, 1/, 1/8,...} is not monotone but it is convergent (with limit = 0). 2. For which values of s does the series converge? Describe all such values. n3s The series is = 1 n 3s so it converges if 3s > 1 so if s > 1/3. 3. Write down an example of a series that converges conditionally. ( 1) n (the alternating harmonic series) is one example.. Let {a k } = {a 1, a 2, a 3,...} be a sequence, and let {S n } be the associated sequence of partial sums: S n = n a k = a 1 + + a n a. Which of the following statements below must be true about these sequences? Circle all true statements. i. If {a k } converges to zero, then {S n } must converge. ii. If {a k } diverges, then {S n } must also diverge. iii. If {S n } converges, then {a k } must also converge. iv. If {a k } is bounded, then {S n } must also be bounded. v. If the terms of {a k } are all positive, then {S n } must be increasing. vi. None of the above statements are true. 2
b. Suppose that S n = + 1 n. Does the series It converges: a k = lim n so the series converges (to ). 5. Let f(x) = ln, graphed below: x(ln x) 2 n a k converge or diverge? ( a k = lim S n = lim + 1 ) = n n n Consider the series f(n) = n= n= Some partial sums of this series are given below: ln n(ln n) = ln 2 (ln ) + ln 2 5(ln 5) + ln 2 6(ln 6) + 2 n 5 6 7 8 9 S n 0.28737 0.3593 0.116 0.5172 0.8362 a. What three properties of f(x) on [, ) should we verify before we can apply the integral test to the series f(n) above? List them. Positive, decreasing, and continuous. b. Using u = ln x and du = dx x, we have What can we conclude about the series f(x) dx = 1 lim b ( ) ln. ln b f(n)? Circle the correct answer. n= i. It converges to a value < 1. ii. It converges to 1. iii. It converges to a value > 1. iv. It converges, but there is not enough information to decide between i. iii. v. It diverges. vi. Nothing. The integral test does not apply to the series in question, the integral test is inconclusive, or there is not enough information to determine convergence/divergence. Hint: If you re stumped, it may help you to draw a diagram using the graph above. 3
6. Partario has taken a sacred Vow of Honour to never use a calculator again. One day, he decides to compute ln and does the obvious thing : He plugs x = into the power series representation ( 1) n (x 1) n+1 ln x = ( 1) n 3 n+1 and concludes that ln =. Believing that the partial sums will estimate ln, Partario makes a table: n 5 6 7 8 9 S n 85.65000 226.77857 593.363 1593.65357 311.263 These partial sums are out of control! What is the deal? Partario wonders, aloud. a. Explain briefly why Partario s approach did not work. (Your reason should be more sophisticated than these are bad estimates for ln. ) Partario s approach did not work because x = is not in the interval of convergence for the power series for ln x. The interval of convergence is (0, 2] and > 2. You can also see that the series diverges by the ratio test (with L = 3). ( 1) n ( 3/) n+1 Quartario suggests that Partario plug in x = 1/, and estimate The partial sums of this new series are instead. n 5 6 7 8 9 S n 1.32810 1.3717 1.35968 1.36803 1.37366 b. To determine whether this series converges, Partario uses the ratio test. Is this a good choice? (That is, will the test be conclusive?) Circle your answer. A GOOD CHOICE, PARTARIO! NO, PARTARIO, NOOO! c. What is the relationship between the value of Q = ( 1) n ( 3/) n+1 and ln? We have Q = ln(1/) = ln. This means the partial sums in the second table above converge to ln : in fact, ln 1.38630.
7. Evaluate the series below (that is, find its value) or state that it diverges. ( 1) n 5 2n+1 = 5 5 3 + 55 2! 57 3! + 59! 511 + 5! You do not need to show your work or simplify your answer. Draw a box around your final answer. ( 1) n 5 2n+1 ( 1) n 5 2n = 5 ( 25) n = 5 = 5e 25 8. Consider the power series below with unknown coefficients {c n }: F (x) = c n (x + 3) n = c 0 + c 1 (x + 3) + c 2 (x + 3) 2 + c 3 (x + 3) 3 + c (x + 3) + Suppose we know that F (x) CONVERGES at x = 2 and at x = 3 and that it DIVERGES at x =. For each of the x-values below, determine whether F (x) converges or diverges there, or that there is no enough information to make a decision for sure. Circle your answers. a. x = 3.5: CONVERGES DIVERGES NOT ENOUGH INFORMATION b. x = 2.5: CONVERGES DIVERGES NOT ENOUGH INFORMATION c. x = 3: CONVERGES DIVERGES NOT ENOUGH INFORMATION d. x = 9: CONVERGES DIVERGES NOT ENOUGH INFORMATION e. x = 10.5 CONVERGES DIVERGES NOT ENOUGH INFORMATION Hint: It may help you to draw a number line and plot points where you know the power series converges/diverges. Explanation: This question relies on remembering (1) that the interval of convergence (IoC) is an interval, centered at the center of the power series (in this case at a = 3), (2) that the endpoints are equidistant from the center, and (3) that the series can either converge or diverge at each of the endpoints. Let s call our endpoints p (left endpoint) and q (right endpoint). We know from the given information that 3 q. Since the center is 3 and the endpoints are equidistant from the center, we have 10 p 9. So: We don t have enough info for (a) because we could have 3.5 < q, 3.5 = q, or 3.5 > q. We have convergence for (b) because 3 < 2.5 < q (so it lies strictly between the center and the right endpoint). (c) We have convergence at 3 because it is the center (in fact, we will have F ( 3) = c 0 ). We don t have enough info for (d) because we could have p = 9 (in which case it could diverge) or p < 9 (in which case it would definitely converge). Finally, we have divergence in (e) because 10.5 < 10 p, so 10.5 lies strictly to the left of the left endpoint. 5
9. Find the endpoints of the interval of convergence of the power series (x 1) 2n+ n () = (x (x 1)6 (x 1)8 (x 1)10 (x 1)12 2 1) + + + + + 16 1 102 600 Show all work. You do not need to test convergence at the endpoints. Draw a box around your final answer. We have L(x) = lim (x 1) 2n+6 n n+1 (n + 2) n () 2 2 (x 1) 2n+ = lim (x 1) 2 ()2 n (n + 2) 2 = x 1 2 To find the endpoints, we solve L(x) = 1, so x 1 2 =, so x 1 = 2, so the endpoints are x = 1 and x = 3. For the six series 10 15 below, fill in the required information: (a) Whether the series converges or diverges; (b) Which test(s) you ultimately used to determine convergence/divergence; and (c) Additional info: If you used the integral test, write down the value of the relevant integral (or that it diverges); if you used a direct or limit comparison test, state what series you compared the given series to; if you used the ratio test, state the value of L you found. (Leave blank for other tests.) The next page can be used for any extra work, but it will not be graded. 10. ( + 1)( + 2)( + 3)( + ) (n 3 + 1)(n 3 + 3)(n 3 + 5) Diverges Limit comparison With 1 n 11. n 7 n () Converges Ratio test L = 1/7 6
12. cos n Converges Direct comparison With 1 13. 3! n 3 + 1 Diverges Divergence test 1. n=2 ( 1) n ln n Converges Alternating series test 15. ( 1 5n 1 ) = 5n + 5n(5n + ) = 25 + 20n Converges Limit comparison With 1 7