MATH 101 Quiz #5 (v.T2) Last Name: Thursday, March 17 First Name: Grade: Student-No: Section: Very short answer question 1 1 1 1 1 1. 1 mark To what value does the series 1+ + + + + + · · · converge? Simplify 5 25 125 625 3125 your answer completely. Answer: 5/4 Solution: We recognize that this is a geometric series with ratio r = 1/5 and scale factor (first term) a = 1. We know this converges to: a 1 5 = = . 1−r 1 − 1/5 4 Marking scheme: 1 mark for the correct answer. Short answer questions—you must show your work 2. 2 marks Find the solution to the separable initial value problem: 4x3 dy = y , dx e Express your solution explicitly as y = y(x). y(0) = log 5. Solution: Rearranging, we have: ey dy = 4x3 dx. Integrating both sides: ey = x4 + C. Since y = log 5 when x = 0, we have elog 5 = 04 + C 5 = C, and therefore ey = x4 + 5 y = log(x4 + 5). Marking scheme: • 1 mark for integrating the equation. • 1 mark for the correct final answer. Note: y = log(x4 ) + log 5 is incorrect. 3. 2 marks Show that the series ∞ X 2 i=1 2 − 3 i (i + 1)3 converges and find its limit. Simplify your answer completely. Solution: We recognize this as a telescoping series. Successive terms cancel and the nth partial sum is: n X 2 2 2 2 = 3− sn = − . 3 3 3 i (i + 1) 1 (n + 1) i=1 Therefore, we can see directly that the sequence of partial sums {sn } is convergent: 2 2 lim sn = lim = 2. − n→∞ n→∞ 13 (n + 1)3 It follows that the series is also convergent with limit 2. Marking scheme: 2 marks for the correct answer with valid justification; 1 mark for a correct procedure with minor errors. 1 mark for showing that the series converges by some valid method without finding its value. Long answer question—you must show your work 4. 5 marks The nth partial sum of a sequence {an } is known to have the formula sn = 1 + 3n . 3 + 2n (a) Find an expression for an , valid for n ≥ 2. ∞ X an converges. (It will help to make the expression from part (a) (b) Show that the series n=1 a single fraction.) (c) Find the value of the series ∞ X an . n=1 Solution: (a) We can find an by subtracting: an = sn − sn−1 . 1 + 3n 1 + 3(n − 1) 1 + 3n 3n − 2 − = − 3 + 2n 3 + 2(n − 1) 3 + 2n 2n + 1 (1 + 3n)(2n + 1) − (3n − 2)(2n + 3) 7 = = 2 . (2n + 1)(2n + 3) 4n + 8n + 3 an = (b) Define f (x) = 7/(4x2 + 8x + 3), which is a positive and decreasing function for x ≥ 1. Then an = f (n), and by the integral test, the series converges if and only if the following integral converges: Z ∞ Z ∞ 7 f (x) dx = dx. 2 4x + 8x + 3 1 1 We can see that this converges: for example, for x ≥ 1 we have 4x2 + 8x + 3 > 4x2 , and so by the integral comparison test: Z ∞ Z ∞ Z R 7 7 7 7 dx < dx = lim dx = . 2 2 2 R→∞ 1 4x 4x + 8x + 3 4x 4 1 1 P P∞ 1 7 Alternately, we can compare the series ∞ n=1 4n2 +8n+3 . to the series n=1 n2 , which converges by the p-test (2 > 1). (c) To evaluate the limit of the series, we return to the partial sum sn originally given and see that the sequence sn converges: 1/n + 3 3 1 + 3n = lim = . n→∞ 3/n + 2 n→∞ 3 + 2n 2 lim sn = lim n→∞ Marking scheme: • 1 mark for (a), finding an (whether simplified or not). • 2 marks for (b), a correct argument showing that the series converges. 1 mark partial credit for good-faith attempts that include either a comparison test, an integral test, or a p-test. • 2 marks for (c), the correct evaluation of the limit of the series (regardless of whether they proved convergence above).