Calculus II Sample Exam 3

Calculus II Sample Exam 3: Chapter 11 1. Does the series described by ∞ X n=0 (−1)n n2 converge? 2n2 − 5 n2 1 = ̸= 0, the series does not converge. 2 n→∞ 2n − 5 2 Answer. No. Since lim 2. To what number does ∞ X (−2)n converge? 7n+1 n=1 Answer. This is a geometric series missing the first term:  n n ∞ ∞ ∞  X X (−2)n 1 −2 1 1 X −2 1 1 1 1 1 2 = =− + =− + =− + =− n+1 7 7 7 7 7 n=0 7 7 7 1 − (−2/7) 7 9 63 n=1 n=1 3. Does ∞ X 1 converge? 4 n(ln n) n=2 Answer. Use the integral test, and the substitution u = ln x: Z lim t→∞ 2 t 1 dx = lim t→∞ x(ln x)4 Z ln t −1 t→∞ 3u3 u−4 du = lim ln 2 ln t −1 1 1 + = , 3 3 t→∞ 3(ln t) 3(ln 2) 3(ln 2)3 = lim ln 2 so the series converges. 4. Does ∞ X n2 − 2 converge? n3 + n + 2 n=0 Answer. This “looks like” the harmonic series, since n2 /n3 = 1/n, so we guess that it diverges. We’d like to see that the terms are larger than terms that look like 1/n. It is true that for large enough n: n2 − 2 n2 /2 1 n2 /2 = = ≥ , 3 3 3 3 3 n +n+2 n +n +n 3n 6n because eventually n2 − 2 > n2 − n2 /2 = n2 /2 and n3 + n + 2 < n3 + n3 + n3 = 3n3 . So P the series diverges because 1/(6n) diverges. Calculus II—Sample Exam 3: Chapter 11 5. Does 2 ∞ X (−1)n converge absolutely, converge conditionally, or diverge? n ln n n=2 Answer. It converges conditionally. It converges by the alternating series test, but it ∞ X 1 doesn’t converge absolutely because diverges by the integral test. n ln n n=2 1 1 1 = 0, and since < , meaning the terms are n→∞ n ln n (n + 1) ln(n + 1) n ln n decreasing, the series converges by the alternating series test. First: Since lim Second: Z lim t→∞ 2 t 1 dx = lim t→∞ x ln x Z ln t ln 2 du = lim ln |u| t→∞ u ln t ln 2 = lim ln | ln t| − ln | ln 2| = ∞. t→∞ ∞ X n2 6. Does converge? n! n=1 Answer. Yes, by the ratio test: (n + 1)2 n! lim = lim n→∞ (n + 1)! n2 n→∞  n+1 n 2 1 = 1 · 0 = 0. n+1 ∞ X 3n n 7. Find the interval of convergence and radius of convergence for x . n n=1 Answer. Using the ratio test: 3n+1 n |x| = 3|x|, n→∞ n + 1 3n so the radius of convergence is 1/3. When x = 1/3 the series is the harmonic series, so lim it diverges; when x = −1/3 the series is the alternating harmonic series, so it converges; thus the interval of convergence is [−1/3, 1/3). ∞ X n3 8. Find the interval of convergence and radius of convergence for (x − 2)n . n 2 n=0 Answer. Using the ratio test: (n + 1)3 2n lim |x − 2| = lim n→∞ 2n+1 n→∞ n3  n+1 n 3 1 1 |x − 2| = |x − 2|, 2 2 so the radius of convergence is 2. When x = 0, ∞ ∞ X X n3 n (0 − 2) = (−1)n n3 , n 2 n=0 n=0 which diverges, by the divergence test. When x = 4, ∞ ∞ X X n3 n (4 − 2) = n3 , n 2 n=0 n=0 which diverges, by the divergence test. Thus, the interval of convergence is (0, 4). Calculus II—Sample Exam 3: Chapter 11 3 9. Find a power series representation for x2 ex ; find the radius of convergence for your series. Answer. Start with the series for ex and multiply by x2 : x2 ∞ ∞ X X xn xn+2 = . n! n! n=0 n=0 Using the ratio test: n! 1 |x| = |x| = 0, n→∞ (n + 1)! n+1 lim so the radius of convergence is ∞. 10. Find a power series representation for cos(x3 ); find the radius of convergence for your series. Answer. Starting with the series for the cosine: cos x = ∞ X (−1)n 2n x , (2n)! n=0 and substituting x3 for x: cos x3 = ∞ ∞ X (−1)n 3 2n X (−1)n 6n (x ) = x . (2n)! (2n)! n=0 n=0 Since the series for cosine converges for all numbers, so does the second series, so the radius of convergence is infinity.