advertisement

Calculus II Sample Exam 3: Chapter 10 1. Does the sequence described by an = (−1)n n2 converge? 2n2 − 5 n2 1 n2 n = , the terms (−1) do not coverge. n→∞ 2n2 − 5 2 2n2 − 5 Answer. No. Since lim ∞ X (−2)n 2. To what number does converge? 7n+1 n=1 Answer. This is a geometric series missing the first term: n n ∞ ∞ ∞ X X (−2)n 1 −2 1 X 1 −2 1 1 1 1 1 2 =− + =− = =− + =− + n+1 7 7 7 7 n=0 7 7 7 7 1 − (−2/7) 7 9 63 n=1 n=1 3. Does ∞ X 1 converge? n(ln n)3 n=2 Answer. Use the integral test: Z 2 ∞ 1 dx = lim t→∞ x(ln x)3 Z ln t u−3 du = ln 2 1 , 2(ln 2)2 so the series converges. 4. Does ∞ X n2 − 2 converge? n3 + n + 2 n=2 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 ≥ = , 3 3 n +n+2 3n 6n because eventually n2 − 2 > n2 /2 and n3 + n + 2 < 3n3 . So the series diverges because P 1/(6n) diverges. Calculus II—Sample Exam 3: Chapter 10 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 6. Does ∞ X n2 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 = 0. n+1 7. Find the interval of convergence and radius of convergence for ∞ X 3n n x . n n=1 Answer. Using the ratio test: 3n+1 n |x| = 3|x|, n→∞ n + 1 3n lim so the radius of convergence is 1/3. When x = 1/3 the series is the harmonic series, so 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). On the real exam, as usual, (−1/3, 1/3) would be acceptable. ∞ 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 and the interval of convergence is (0, 4). It is not hard to see that it does not converge at the endpoints 0 and 4. Calculus II—Sample Exam 3: Chapter 10 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. It’s difficult to do this by computing derivatives of cos(x3 ), but easy to do by starting with the series for the cosine: ∞ X (−1)n 2n cos x = x , (2n)! n=0 and substituting x3 for x: cos x3 = ∞ X (−1)n 6n x . (2n)! n=0 Since the series for cosine converges for all numbers, so does the second series, so the radius of convergence is infinity. 11. Find the first 4 terms (c0 + c1 x + c2 x2 + c3 x3 ) of the Maclaurin series for cos(x + π/2). Answer. We need to compute the first three derivatives and evaluate at x = 0. The derivatives are − sin(x + π/2), − cos(x + π/2), sin(x + π/2), and the values of the function and the derivatives at 0 are 0, −1, 0, 1. The first 4 terms are then 0−x+0+ 1 3 x . 3!