Math 76 Practice Problems for Midterm III - Solutions Chapter 11 DISCLAIMER. This collection of practice problems is not guaranteed to be identical, in length or content, to the actual exam. You may expect to see problems on the test that are not exactly like problems you have seen before. Multiple Choice. Circle the letter of the best answer. 1. The nth term of the sequence {−3, 4, 11, 18, 25,...}, counting a1 = −3 as the first term, is 2 (a) an = 5n − 2 (c) an = n − 4 (b) an = 7n − 10 (d) an = −3n + 7 This is an arithmetic sequence with a common difference of 7, so only (b) can be correct. Sure enough, when n = 1, a1 = 7n − 10 = 7 − 10 = −3. ∞ X 1 n 2. 3 2 = n=1 3 (a) 6 (c) 2 (b) 3 (d) ∞ (diverges) 1 This is a geometric series with r = , so it converges. But take care! The sum is 2 ∞ X n 3 3 1 = − 3 = 6 − 3 = 3 since the sum starts from n = 1, not n = 0. 2 1 − 1 n=1 2 ∞ X 2 3. The series 3n+2 n=1 8 (a) converges to (c) converges to 3 9 1 (d) converges to 9 (b) converges to 9 This is a geometric series. There are several ways to get it into a form that fits the formula. Here are two: Solution 1. ∞ ∞ X 2 X 2 We have = . 3n+2 3n n=1 n=3 1 This is exactly in the form we want it, but there are three terms “missing.” (The a formula works when the series starts from n = 0, but this one starts at n = 3.) 1 − r a So we take and subtract off the terms corresponding to n = 0, n = 1, and 1 − r n = 2. We get 2 2 2 2 2 1 1 − 2 − − = 3 − 2 − − = . 1 − 3 3 9 3 9 9 Solution 2. ∞ ∞ ∞ n X 2 X 2 X 2 1 2 1 We have = = = 27 = . 3n+2 3n+3 33 3 1 − 1 9 n=1 n=0 n=0 3 ∞ X 2 2 4. − = n n + 1 n=3 2 (a) 0 (c) 3 1 (b) (d) ∞ (diverges) 6 This is a telescoping series. The n-th partial sum is 2 2 2 2 2 2 2 2 2 2 s = − + − + − + ... + − = − , n 3 4 4 5 5 6 n n + 1 3 n + 1 2 whose limit as n → ∞ is . 3 ∞ X 5n3 5. To determine whether or not the series converges, the limit comparison 1 − 2n + n4 n=2 X test may be used with comparison series bn = X 1 X 5 (a) (c) n n4 X (b) 5n3 (d) none; the limit comparison test cannot be used 5n3 The degree of the denominator of a = is one more than the degree of the n 1 − 2n + n4 1 an numerator. So the best comparison term is bn = . To check, note that the limit of is n bn finite and positive, since 3 5n 3 4 5n n lim 1−2n+n = lim · n→∞ 1 n→∞ 4 n 1 − 2n + n 1 5n4 = lim = 5. n→∞ 1 − 2n + n4 2 ∞ √ X n 6. The series (−1)n−1 √ n2 − 4 n − 1 n=1 (a) converges absolutely (AC) (b) converges conditionally (CC) (c) diverges Since the biggest power on the bottom is n2 and the biggest power on the top is n1/2, the 1 3 difference in the powers is greater than 1 (2 − 2 = 2 ). Thus the series converges absolutely 1 (AC) by the limit comparison test, using b = . n n3/2 Here is a more detailed solution: ∞ √ X n Try for AC: look at the series √ . The terms a of this series are positive, n2 − 4 n − 1 n n=1 1 at least from some point on. So we may use the limit comparison test. Let b = . We n n3/2 have 1/2 3/2 an n n lim = lim 2 1/2 · n→∞ bn n→∞ n − 4n − 1 1 n2 = lim = 1, n→∞ n2 − 4n1/2 − 1 a finite positive limit. Therefore we are using the right bn for the limit comparison test. X 3 Since bn converges (it is a p-series with p = 2 ), our series also converges. In other ∞ √ X n words, the original series (−1)n−1 √ converges absolutely (AC). n2 − 4 n − 1 n=1 ∞ √ X n 7. The series (−1)n−1 √ n − 4 n − 1 n=1 (a) converges absolutely (AC) (b) converges conditionally (CC) (c) diverges 1 1 Here the difference in the powers is less than 1 (1 − 2 = 2 ). Thus the series will not converge absolutely. However, it will still converge (conditionally), by the alternating series test: we have √ x • Let f(x) = √ . Then x − 4 x − 1 √ √ 1 √ 1 √1 √2 √ (x − 4 x − 1) − x(1 − ) 2 x + x f 0(x) = 2 x √ x = − √ < 0 (x − 4 x − 1)2 (x − 4 x − 1)2 (I skipped a lot of algebra here; you can check my work). Therefore the terms are decreasing. 3 √ n • lim √ = 0 since the power on the bottom is bigger than the power on n→∞ n − 4 n − 1 the top. ∞ X 10n 8. The series (−1)n 7n! n=0 (a) converges absolutely (AC) (b) converges conditionally (CC) (c) diverges Using the ratio test we get n+1 an+1 10 7n! lim = lim · n n→∞ an n→∞ 7(n + 1)! 10 10 = lim = 0 < 1. n→∞ n + 1 Therefore the series is AC. ∞ n X 2n2 + 1 9. The series n2 + 5n − 6 n=2 (a) converges absolutely (AC) (b) converges conditionally (CC) (c) diverges Using the root test we get s 2 n pn n 2n + 1 lim |an| = lim n→∞ n→∞ n2 + 5n − 6 2n2 + 1 = lim = 2 > 1. n→∞ n2 + 5n − 6 Therefore the series diverges. ∞ X 1 10. The interval of convergence of the power series (x − 1)n is n n=1 (a) [0, 1] (c) (0, 2] (b) (0, 1) (d) [0, 2) Since the power series is centered at x = 1, we can see immediately that the answer must be either (c) or (d) (You can also check this using the ratio test). 4 ∞ X 1 Now we check the endpoints: at x = 0 we have (−1)n which converges by the n n=1 ∞ X 1 alternating series test. At x = 2 we have which diverges by the p-series test. n n=1 3 11. A power series representation for the function f(x) = is 4 − x ∞ ∞ X 3 X 3 (a) xn (c) xn 4 4n+1 n=0 n=0 ∞ n+1 ∞ X 3 X (b) xn (d) 3(4 − x)n 4 n=0 n=0 We have 3 3 3 = = 4 4 − x 1 1 4 1 − 4 x 1 − 4 x ∞ n ∞ X 3 1 X 3 = x = xn. 4 4 4n+1 n=0 n=0 12. The Maclaurin series for the function f(x) = x3 cos(4x2) is ∞ ∞ X (−16)n X (−1)n (a) x4n+3 (c) (4x2)n (2n)! n! n=0 n=0 ∞ n ∞ n X (−1) X (−1) 2 (b) x2n+3 (d) x4n +3 (2n)! (2n + 1)! n=0 n=0 ∞ X (−1)n Recall that the Maclaurin series for cos(x) is x2n, for all x. Therefore for cos(4x2) (2n)! n=0 ∞ X (−1)n it is (4x2)2n. Finally, for x3 cos(4x2) it is (2n)! n=0 ∞ ∞ ∞ X (−1)n X (−1)n42nx4n X (−16)n x3 (4x2)2n = x3 = x4n+3. (2n)! (2n)! (2n)! n=0 n=0 n=0 √ 13. The Maclaurin series for the function f(x) = 3 − x is ∞ ∞ X − 1 1 X 1 (a) 2 xn (c) 3 xn n 3n n n=0 n=0 ∞ n ∞ X 1 √ 1 X n(−1)n (b) 2 3 − xn (d) xn n 3 1 3n n=0 n=0 2 5 We have √ √ r x √ x1/2 3 − x = 3 1 − = 3 1 − 3 3 ∞ ∞ n √ X 1 xn X 1 √ 1 = 3 2 − = 2 3 − xn. n 3 n 3 n=0 n=0 Fill-In. 1. Circle the best answer. On the line, indicate one valid test that can be applied to get your answer. You may choose from the following list: • divergence test • integral test • alternating series test • p-series test • direct comparison test • ratio test • geometric series test • limit comparison test • root test ∞ X 500 (a) ( converges | diverges ) n2 n=1 Test: p-series test, integral test, limit comparison test ∞ X n − 5 (b) (−1)n ( converges | diverges ) Test: divergence test 3n n=1 ∞ √ X 3 n (c) ( converges | diverges ) n2 − 3n + 1 n=1 Test: limit comparison test, integral test ∞ X cos(nπ) (d) ( converges | diverges ) Test: divergence test tan−1(n) n=1 ∞ X 10n (e) ( converges | diverges ) Test: ratio test (5n)! n=1 ∞ X 5n 2. (x + 3)n−1 is a power series centered at −3 . n2 + 1 n=1 ∞ X n A power series of the form cn(x − a) is centered at a. Here we have a = −3. n=? ∞ X (−1)n 3. The radius of convergence of the power series √ xn is 1 . n n=1 Using the ratio test, we have n+1 √ √ x n n lim √ · = lim √ |x| n→∞ n + 1 xn n→∞ n + 1 set = |x| < 1. 6 Therefore the radius of convergence is R = 1. ∞ X 6(−1)n 4. A power series representation of 6 tan−1(x2) is x4n+2 .