MATH 201 Exam 1, Fall 2001

MATHEMATICS 201 FIRST SEMESTER, 2007-08 QUIZ 1 Solutions Time: 60 Minutes Date: November 3, 2007 Name:————————————- ID Number:—————————— Section:———————————- Mr. Z. Khachadourian Prof. A. Lyzzaik Instructions: The examination consists of two parts: Part I consists of six written questions, and Part II of four multiple-choice questions. Please observe the following: (a) Answers of questions of Part I must be fully justified. (b) Circle only one choice (a), (b), · · · , or (e) for each question of Part II. (c) Each multiple-choice question has exactly one answer. (d) Circling none or more than one answer for a question of Part II results in a zero credit. (e) The grade allocated to each question is set next to it. Part Grade I /68 II /32 TOTAL /100 1 2 PART I: Investigate the following infinite series for conver- gence and divergence providing adequate justification. (1) ∞ � n=1 �ln n n �3 (10 points) Solution. The series converges by the limit comparison test since, by L’Hospital’s rule, lim n→∞ (ln n/n)3 (1/n)2 = (ln n)3 n = 0, and the p−series, p = 2, �∞ n=1 1/n2 converges. (2) ∞ � n=1 3√n sin � 1 3√n � (10 points) Solution. The series diverges by the n−th term test since lim n→∞ sin(1/ 3√n) 1/ 3√n) = 1. (3) ∞ � n=0 �n + 2 n + 3 �n2 (10 points) Solution. The series converges by the n−th root test since lim n→∞ n � � � � �n + 2 n + 3 �n2 = lim n→∞ �n + 2 n + 3 �n = lim n→∞ (1 + 2/n)n (1 + 3/n)n = e2/e3 = 1/e < 1. (4) ∞ � n=1 2n ln n n10 (10 points) Solution. The series diverges by the ratio test since lim n→∞ 2n+1 ln(n + 1) (n + 1)10 �2n ln n n10 � = 2 � n n + 1 �10 ln(n + 1) ln n = 2 > 1. 3 (5) ∞ � n=1 2. 4. 6. · · · .(2n) 1. 3. 5. · · · .(2n − 1).1 n (10 points) Solution. The series diverges by the direct comparison test since 2. 4. 6. · · · .(2n) 1. 3. 5. · · · .(2n − 1).1 n = �2 1 � �4 3 � �6 5 � · · · � 2n 2n − 1 � .1 n > .1 n and the harmonic series diverges. (6) Given the power series ∞ � n=1 (−1)n (x + 2)n 3n n ln n. For what values of x does the series converge (i) absolutely and (ii) conditionally? Find the radius of convergence. (18 points) Solution. Since lim n→∞ ������ (−1)n+1 (x+2)n+1 3n+1 (n+1) ln(n+1) (−1)n (x+2)n 3n n ln n ������ = |x + 2| 3 lim n n + 1 lim ln n ln(n + 1) = |x + 2| 3 , by the ratio test for absolute convergence, the series converges abso- lutely for values of x, |x + 2| < 3, or, −5 < x < 1, and diverges for values of x, |x + 2| > 3, or, x > 1 or x < −5. If x = 1, then the series becomes ∞ � n=1 (−1)n 1 n ln n which converges by the alternating series test since the sequence {1/(n ln n)} of positive terms is strictly decreasing to zero, but diverges absolutely since the series � 1/(nlnn) is a p−logarithmic series with p = 1. Hence, the power series converges conditionally at x = 1. If x = −5, then the series becomes ∞ � n=1 1 n ln n which diverges as argued above. PART II: Circle the correct answer in the following multiple- choice questions: (8 points for each question) (7) If the series ∞ � n=0 (−1)n 2n (n!)2 4 is approximated by the sum of its first three terms, then the magni- tude of the error made (according to the alternating series estimation theorem) is less than (a) 1/9 → (b) 2/9 (c) 3/4 (d) 2/19 (e) none of the above. (8) The sum of the series ∞ � n=0 3n(n! − 1) 4n n! is (a) 5/2 − e3/5 (b) 3 − e3/4 (c) 4 − e−3/4 → (d) 4 − e3/4 (e) none of the above. (9) The approximate value with an error of magnitude less than 10−3 of the integral � 1 0 sin(x3) x2 dx is (a) 15/64 (b) 17/120 → (c) 23/48 5 (d) 23/96 (e) none of the above. (10) If f(x) = x2e−2x, then f 5(0) equals → (a) −160 (b) −120 (c) 80 (d) −540 (e) none of the above. [Hint: It is time consuming to attempt successive derivatives]