Exam 3. Math Spring 2015 April 8, 2015 Name: } {{ } (from xkcd) Read all of the following information before starting the exam:

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1 Exam 3 Math 2 - Spring 205 April 8, 205 Name: } {{ } by writing my name I pledge to abide by the Emory College Honor Code (from xkcd) Read all of the following information before starting the exam: For full credit, always state which test you are using (unless given in the problem) and check that any necessary conditions are satisfied. Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct). Circle or otherwise indicate your final answers. You will have 50 minutes to complete this exam. You may not use the restroom during the exam. All cell phones, bags, etc must be left at the front of the room during the exam. Good luck!

2 Question Points Score Total 00

3 . (2 points) True or False. You do not need to explain your answers, but no partial credit will be given. a. (2 pts) A convergent series with positive terms is absolutely convergent. True, because a n and a n are the same thing. b. (2 pts) The ratio test can be used to determine whether n 3 False, it s inconclusive for algebraic functions. n 3 ( ) n 3 (n + ) 3 = n + converges c. (2 pts) If a finite number of terms are added to a convergent series, then the new series is still convergent. True, finitely many terms don t affect convergence. d. (2 pts) If the nth term of a series is a n = 2n 3n+, then the series converges. False, because lim n a n = 2 3 0, it diverges by the Test for Divergence. e. (2 pts) If the nth partial sum of a series is s n = 2n 3n+, then the series converges. True, because lim n s n = 2 3, so the series converges to 2 3. f. (2 pts) The sequence {,,,,,,... } converges. False, because the terms aren t approaching anything, and are not getting closer together.

4 2. ( points) Use the ratio test to determine whether the following series converges or diverges. n! a n+ a n = n! (n + )! = n + 0 < So the ratio test shows that the series converges. 3. ( points) Determine whether the following geometric series converges or diverges. If it converges, find its sum. ( 3) n 2 3n ( 3) n 2 3n = ( 3) n 8 n = 8 ( ) 3 n 8 This is a geometric series with first term /8 and ratio 3/8. Since 3/8 <, it converges. Its sum is /8 ( 3/8) = /8 /8 =

5 4. ( points) Use the integral test to determine whether the following series converges or diverges. Hint: The derivative of xe x2 is e x2 ( 2x 2 ). n e n2 First, we need to check that we can apply the integral test. The function f(x) = x e x2 is continuous and positive. It s derivative is e x2 ( 2x 2 ), which is negative for x > / 2, so f(x) is also decreasing. Therefore, we can use the integral test. We will solve the integral using u-substitution, letting u = x 2 and therefore du = 2x dx. x dx = e x2 2 x e x2 dx = e u du = 2e u = ( lim t 2e t2 2e x2 ) + 2e = 2e The improper integral converges, and therefore the series converges too. 5. ( points) Determine whether the following series converges or diverges. n 5 n We will use the limit comparison test, with a n = n 5 n 2 +2 and b n = n n 2 for n > 5 and b n is always positive, so the test applies. = n. Note that a n is positive lim n Since n diverges, this tells us that n 5 n 2 +2 n 5 n n2 n = lim n 3 5n 2 n n 3 + 2n = > 0. does too.

6 6. ( points) Determine whether the following series converges or diverges. (ln n) n We will use the ratio test. lim n (ln n) n /n = lim n ln n = 0 <. This tells us that the series converges. 7. ( points) Determine whether the following series converges absolutely, converges conditionally, or diverges. ( ) n n=2 ln n This is an alternating series. Since ln n is increasing to +, the alternating series test says that the series converges. ln n is decreasing to zero. Hence Next, we need to check if it converges absolutely or conditionally. For n 2, we have that ln n < n and so ln n > n. Both are positive and n diverges, so the Comparison Test tells us that ln n diverges too. Therefore, the original series converges conditionally.

7 8. ( points) Determine whether the following series converges absolutely, converges conditionally, or diverges. Since ( ) n e 2/n lim n e2/n =, the terms do not go to zero, and hence by the Test for Divergence the series diverges. 9. ( points) Determine whether the following series converges absolutely, converges conditionally, or diverges. We have that cos(/n), so cos(/n) n 2 cos(/n) n 2 /n2. We can use the comparison test: Since /n 2 converges, cos(/n) n does too. So cos(/n) 2 n 2 converges absolutely. Note: You might have noticed that for n, 0 < /n, and in this range cos is positive. So absolute values aren t actually needed.

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MATH 153 FIRST MIDTERM EXAM

MATH 153 FIRST MIDTERM EXAM NAME: Solutions MATH 53 FIRST MIDTERM EXAM October 2, 2005. Do not open this exam until you are told to begin. 2. This exam has pages including this cover. There are 8 questions. 3. Write your name on