tapia jat4858 HW09 clark 52990 This print out should have 22 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 X To check for conditional convergence consider the series X 1 n 1 n 1 e1 n 4n is absolutely convergent conditionally convergent or divergent where e1 x 4x Then f x 0 on 0 On the other hand f x f x 2 conditionally convergent correct lim e1 x 1 x 3 absolutely convergent we see that f n 0 as n By the Alternating Series Test therefore the series Explanation Since n 1 1 n 1 1 1 x e1 x 1 x 1 x e e 4x3 4x2 4x3 Thus f x 0 on 0 so f n f n 1 for all n Finally since 1 divergent X 1 n f n n 1 10 0 points Determine whether the series 1 1 X e1 n e1 n 1 n 4n 4 n X n 1 we have to decide if the series X 1 n n 1 e1 n n is absolutely convergent conditionally convergent or divergent First we check for absolute convergence Now since e1 n 1 for all n 1 e1 n 1 0 4n 4n But by the p series test with p 1 the series X 1 4n n 1 diverges and so by the Comparison Test the series X e1 n n 1 4n too diverges in other words the given series is not absolutely convergent 1 n f n n 1 is convergent Consequently the given series is conditionally convergent keywords 002 10 0 points XTo apply the ratio test to the infinite series an the value of n lim n an 1 an has to be determined Compute for the series X 9n n nn n 1 tapia jat4858 HW09 clark 52990 1 9 2e 2 conditionally convergent correct 3 divergent 2 0 Explanation By the Alternating Series test the series 3 9 4 2 9 correct e X 1 n 2 4 n n 1 9 5 2 converges On the other hand by the p series 1 test with p 1 the series 4 Explanation Since X 1 4 n n 1 1 2 n n 1 n 1 n 1 2 n while n 1 is divergent Consequently the series is 1 n 1 n nn n n 1 n 1 n n 1 n 1 1 n we see that an 1 9 an 9 n 1 n n 1 1 n 1 1 1 n n But lim n 1 1 n n e Consequently for the given series 003 9 e 10 0 points Determine whether the series X 1 n 2 4 n n 1 is absolutely convergent conditionally convergent or divergent 1 absolutely convergent conditionally convergent 004 10 0 points Determine whether the series X 1 n 1 n 3 ln n 6n 1 converges conditionally converges absolutely or diverges 1 diverges 2 converges conditionally correct 3 converges absolutely Explanation The given series can be written as X 1 n 1 n 3 X ln n 1 n f n 6n 1 n 3 where f x ln x 6x 1 tapia jat4858 HW09 clark 52990 But xf x 1 lim x ln x 6 so by the Limit Comparison Test and the Integral Test applied to the series Determine whether the series I X n 2 n 3 2n n 1 II X ln n n 3 2 X 1 n 1 n 2n n n 1 n 3 the given series is not absolutely convergent On the other hand by the Alternating Series Test it will converge conditionally if 1 only series II converges 2 both series converge i f n f n 1 for n 3 ii lim f x 0 3 only series I converges correct x 4 both series diverge Now Explanation I Since the series has the form 6x 1 6 ln x x f x 6x 1 2 6 1 ln x 6x 1 2 1 6x ln x 1 0 f x x 6x 1 2 so f x is decreasing for all x 3 Thus n 3 f n f n 1 To determine the limit of f x as x we use L Hospital s Rule for then lim x 1 ln x lim 0 x 6x 6x 1 lim f x 0 x Consequently the given series is conditionally convergent X 1 x When x 3 therefore so converge or diverge an an n 1 n 2 n 3 2n we apply the Ratio Test For then 3 n 1 n 1 2n an 1 an n n 2n 2 3 n 1 n 1 3 n2 2n 1 2n 1 2n 2 4n2 6n 2 in which case lim n 3 an 1 1 an 4 By the Ratio Test therefore the given series converges II Since the series has the form X 2 an n 1 an 1 n 1 n 2n n we apply the Root Test For then 005 10 0 points 1 n an 1 n 1 n 2 n tapia jat4858 HW09 clark 52990 But lim n n 1 n n 1 n lim 1 e n n and so an 1 n 1 an n in which case lim an 1 n n e 1 2 By the Root Test therefore the given series diverges 4 1 3 5 2n 1 1 3 5 2n 1 2 n 1 1 n 1 2n 1 In this case lim n an 1 n 1 1 lim 1 n 2n 1 an 2 Consequently the given series is keywords tExam RatioRootTest RatioRootTes 006 007 10 0 points Determine whether the series 1 absolutely convergent 2 3 4 1 3 1 3 5 1 3 5 7 1 n 1 n 1 3 5 2n 1 is absolutely convergent conditionally convergent or divergent Determine which if any of the series n n X 4 3n A 5n 5 5 B n 1 X n 1 5n 1 n are divergent 1 divergent 1 both of them 2 conditionally convergent 2 B only 3 absolutely convergent correct 3 A only Explanation The given series is of the form X an n 1 where an 1 n 1 n 1 3 5 2n 1 But then an 1 1 n n 1 1 3 5 2 n 1 1 10 0 points 4 neither of them correct Explanation To check for divergence we shall use either the Ratio test or the Root test which means computing one or other of an 1 lim lim an 1 n n n an for each of the given series A The root test is the better one to apply because of the nth powers For then 12 3n 4 1 n 1 an 5 5n 5 25 tapia jat4858 HW09 clark 52990 as n so the given series converges B The ratio test is the better one to use because 5n 6 an 1 n an 5n 1 n 1 Thus the alternating series test applies since 5 5 8k ln …