Math 132 Exam 3 Fall 2016

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1 Math 3 Exam 3 Fall 06 multiple choice questions worth points each. hand graded questions worth and 3 points each. Exam covers sections.-.6: Sequences, Series, Integral, Comparison, Alternating, Absolute (not Root or Ratio No calculators! For the multiple choice questions, mark your answer on the answer card. Show all your work for the written problems. Your ability to make your solution clear will be part of the grade. Useful Formulas n i= i = n(n+ n i= i = n(n+(n+ 6 ( n i= i3 = n(n+ sin θ + cos θ = + tan θ = sec θ + cot θ = csc θ sin(a ± B = sin A cos B ± sin B cos A cos(a ± B = cos A cos B sin A sin B tan(a ± B = tan A±tan B tan A tan B sin A sin B = [cos(a B cos(a + B] cos A cos B = [cos(a B + cos(a + B] sin A cos B = [sin(a + B + cos(a B] sin x = ( cos x cos x = ( + cos x sin(θ = sin θ cos θ cos(θ = cos θ sin θ csc x dx = ln csc x + cot x + C sec x dx = ln sec x + tan x + C

2 Math 3 Exam 3 Page of 0. Determine whether the sequence defined by a n = ln(n 3 + ln(n 3 + n + 4 converges or diverges. If it converges, find the limit. A. 0 B. ln ( C. ln ( D. E. F. G. Diverges Solution: lim a [ n ln(n 3 + ln(n 3 + n + 4 ] ( n 3 + ln [ = ln lim ( = ln n 3 + n + 4 ( n 3 + n 3 + n + 4 ]. Find ALL possible values of x for which the series 9 + x n converges. n A. It is not possible to find such x because the series diverges. B. x > 0 C. x < D. 3 < x < 3 E. 3 < x < F. < x <

3 Math 3 Exam 3 Page 3 of 0 G. < x < Solution: 9 + x n n = ( n [ = 9 ( 9 + ( x n n ] [ n ] ( x n + These are both geometric series. The first has r = / so it converges. The second has r = x/ so it converges when x <, or < x <.

4 Math 3 Exam 3 Page 4 of 0 3. Determine whether the sequence defined by ( a n = n cos n + π converges or diverges. If it converges, find the limit. A. B. C. 0 D. E. F. π G. Diverges Solution: ( lim a n n cos n + π ( cos + π n Apply L Hopital s Rule /n ( sin + π n ( 4/n 3 sin /n 3 = sin (π/ = ( n + π 4. Which of the following sequences converge? a n = (n +! (n + 4!, b n = πn n 00, c n = ln(n0 n, d n = n 4 (n +! A. {d n } only B. {a n }, {b n } only C. {c n }, {d n } only D. {a n }, {d n } only

5 Math 3 Exam 3 Page of 0 E. {a n }, {b n }, {d n } only F. {a n }, {c n }, {d n } only G. All of them Solution: For a n, write out the factorial and see what cancels. a n = (n +! (n + 4! = (n + (n(n (n + (n + 4(n + 3 (n + 4(n + 3(n + = (n + (n(n (n + (Diverges For b n = πn n 00, you can apply L Hopital s Rule 00 times to get lim b (ln π n π n n 00! For c n, apply L Hopital: = lim c ln(n 0 n n 0/n (/n 0 ln n n / / 0 n / = 0 Apply L Hopital For d n, consider what happens when n > 8: d n = n 4 (n +! = n 4 (n + (n(n (n (n 3 n 4 (n (n (n (n (n 3 ( n 4 4 ( n = (n 4 (n 3! = 0 = 0 n (n 3!

6 Math 3 Exam 3 Page 6 of 0. Assume the terms of a sequence {a n } are given by the following formula a n = 3n 3 + 3n n n 3n 3 Find the limit of the sequence or conclude that it diverges. A. 0 B. C. D. 3 E. 6 F. 9 G. Diverges Solution: We use the formula on the formula sheet: ( lim a n 3n + 3 3n n + + n 3 3n 3 ( n 3n 3 3n 3 ( n i= i 3n 3 n(n + (n + 8n 3 = 9 ( n(n + (n Determine the value of the series n= 6 n(n + 3 or conclude that it diverges. A. 0 B. 3

7 Math 3 Exam 3 Page 7 of 0 C. 3 D. 3 6 E. 3 3 F. 0 G. Diverges Solution: This is telescoping so you need to use partial fractions to write out the series. So 6 n(n + 3 = n n + 3 n= 6 n(n + 3 = ( n = n= ( n ( 3 6 S N = N + N + 3 N + 4 and lim S N = = ( 4 ( + 7 (

8 Math 3 Exam 3 Page 8 of 0 7. Determine the value of the series n + 3 n+ 4 n or conclude that it diverges. A. 4 B. 9 C. 3 D E. F G. Diverges Solution: This is two geometric series: ( n + 3 n+ ( n 3 n+ = + 4 n 4 n 4 n ( ( ( n = ( ( /4 3 = + / 3/4 = + = ( n Assume a n is an infinite series with partial sums given by S N = 4 +. What is a N? A. B. C. 0 D. 3 0 E. - 9

9 Math 3 Exam 3 Page 9 of 0 F. - G. - 0 H Solution: Note that a = S S 4 = ( 4 + ( 4 + = 4 = 0

10 Math 3 Exam 3 Page 0 of 0 9. Which of the following series converge? I. ( n + n II. n= ln(n 4 III. n n + A. None of them B. I only C. II only D. III only E. I and II F. I and III G. II and III H. All of them Solution: For Series I note that lim a n = 0 so the series diverges by the test for divergence. Similarly, for series III, lim a n = 0 so diverges by the test for divergence. For series II, note that too. = ln n 4 4 ln 4 4n Since 4n diverges, series II must diverge 0. The series A. B. 8π 4 4 8π 4 4 n = π Find the value of the series ( 4. n ( π 4 C D. 8π 4 E. 90 n=

11 Math 3 Exam 3 Page of 0 8π 4 F G. Diverges Solution: Must take into account that our series starts at n = instead of at n =. ( 4 4 ( = n n = 6 4 n = 6 + = 6 ( + π4 4 n 4 90 n= n= n=

12 Math 3 Exam 3 Page of 0. Which of the following series converge? I. n + n 4 4 n + n II. 4 n n + n III. n! 3 n A. None of them B. I only C. II only D. III only E. I and II F. I and III G. II and III H. All of them Solution: For I you can do a limit comparison with b n = n 4 n. Note that lim n4 lim ( n +n 4 4 n +n n /4 n This series I converges. For II 4 n n + n 4n n = ( ( n +n 4 + n4 ( n 4 ( n = n +n 4 + n n 4 n ( n 4 And so the series converges by comparison to a geometric series with r = 4/. For III, lim n! 3 n = 0 so the series diverges. n = 0. Which of the following series converge? I. ln n e n + II. n= sin (n n + n 3/ III. ( n n /3 A. None of them

13 Math 3 Exam 3 Page 3 of 0 B. I only C. II only D. III only E. I and II F. I and III G. II and III H. All of them Solution: For I, ln n e n + n e n and the series n e n For II, sin n n + n3/ n 3/ converges so series I converges by comparison. so series II converge by comparison to a p-series with p = 3/. For III, this is an alternating series with ( n = ( n b n 3/ k with b k = /n /3. Note that the sequence {b k } is positive, lim b k = 0 and is decreasing. Thus, series III converges by the alternating series test.

14 Math 3 Exam 3 Page 4 of 0 3. Which of the following alternating series converge conditionally, but not absolutely? I. n= ( n n ln n II. n= ( n n(ln n III. cos(πn n 3 A. None of them B. I only C. II only D. III only E. I and II F. I and III G. II and III H. All of them Solution: All the series are alternating and all converge by the alternating series test. For I, note that n > ln n and so n ln n > n n = n and the series diverges by comparision to the harmonic series. For II, the series n(ln n converges by the integral test. For III, n 3 is a geometric series with r = / and thus converges. 4. For which values of p does the series A. All values of p B. < p < C. p > e n ( + e n p converge? D. p E. p >

15 Math 3 Exam 3 Page of 0 F. p G. No values of p Solution: Do a limit comparison to the series e n. A little work should show e np you that ( lim e n (+e n p ( e n e np = and thus both series either diverge or converge. The series e n e np = ( e e p n = ( e p n is geometric and thus converges when e p <. Solving gives p < 0 or p > /.

16 Math 3 Exam 3 Page 6 of 0. Let a n be a series with partial sums S N. always true? Which of the following statements are I. If lim a n = 0, then a n converges. II. If III. If a n = L, then lim a n = L. a n converges, then lim a n = 0. IV. If lim N S N = L, then A. None of them B. I and II C. I and III D. II and III E. III and IV F. I, II, IV G. II, III, IV H. All of them a n = L. Solution: I is false (look at, which diverges. n II is false because if the series converges then the limit of the terms has to actually be 0. III is always true (we talked about this fact in class. IV is always true, this is the definition of what it means for a series to converge.

17 Math 3 Exam 3 Page 7 of 0 Name: ID: Discussion Section Letter: You can find your discussion section on the front of your exam book Written Problem. You will be graded on the readability of your work. Use the back of this sheet, if necessary. 6. Determine whether the following series converge or diverge. State any tests used and show that all conditions of the test are satisfied. ln n (a n n=3 Solution: Limit compare to n 3/. ( a ln n n lim ( n b n n 3/ ln n n / (Apply L Hopital /n (/n / = n / = 0 Since n 3/ converges, so does ln n n by the limit comparison test. (b 3 n n n 3

18 Math 3 Exam 3 Page 8 of 0 Solution: Do a limit comparison to 3 n n, which converges (its a geometric series and converges. a n lim b n ( 3 n n n 3 ( 3 n n n n n 3 = n3 n Note we used the fact that lim n3 = 0, which can be shown by applying L Hopital s n rule. Thus, we must have 3 n converges just like (3/ n. n n 3

19 Math 3 Exam 3 Page 9 of 0 Name: ID: Discussion Section Letter: You can find your discussion section on the front of your exam book Written Problem. You will be graded on the readability of your work. Use the back of this sheet, if necessary. 7. Determine whether the following series converge absolutely, converge conditionally, or diverge. State any tests used and show that all conditions of the test are satisfied. (a ( n arctan(n Solution: Let a n = ( n b n where b n = arctan(n. Note that: lim b n = π which means that lim a n does not exist. Thus, by the Test for Divergence (some students may call this the a n limittest, the series must diverge. The series a n diverges and the sum b n = an both diverge. (b n= ( n n(n + (n 3 Solution: Let a n = ( n b n where b n = = n(n+(n 3 n 3 n 3n Note that b n 0 and lim b n = 0, lim a n = 0, thus the Test for Divergence does not apply. We can do a limit comparison of b n to the series c n = n 3 ( b n lim = c n n 3 n 3n ( n 3 = n 3 n 3 n 3n =

20 Math 3 Exam 3 Page 0 of 0 Thus this is a good comparison. Since c n is a p-series with p = 3, c n converges. By the limit comparison test b n also converges. Thus the series a n converges absolutely. Because a n converges absolutely, an must converge. (c n= ( n n n + Solution: Let a n = ( n b n where b n = n n + Note that b n 0 and lim b n = 0, lim a n = 0, thus the Test for Divergence does not apply. We can do a limit comparison of b n to the series c n = n ( n b n n lim = + = n c n n + = ( n Thus this is a good comparison. Since c n is a p-series with p =, c n diverges. By the limit comparison test b n also diverges. This means that an can not converge absolutely. We not do alternating series test: Test that lim b n = 0: lim b n n n + = 0 Test that b n decreasing. If f(x = x x + then f (x = x (x + < 0 for all x /. Thus f is decreasing and b n decreasing. Thus a n converges and thus a n converges conditionally.

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