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calc2exam3

# calc2exam3 - Math 1700 Exam 3 1-Use the Integral Test to...

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Math 1700 Exam 3 1.-Use the Integral Test to determine whether the series converges or diverges X n =1 ne - n First of all check that the function f ( x ) = xe - x is positive, continuous and decreasing. f 0 ( x ) = e - x - xe - x = e - x (1 - x ) 0 if x 1 R 1 xe - x dx = u = x du = dx dv = e - x v = - e - x = lim t →∞ (( - xe - x ] t 1 + R t 1 e - x dx ) = lim t →∞ (( - 1 - x ) e - x ] t 1 = 2 e - 1 < The indefinite integral is convergent so the series n =1 ne - n is convergent by Integral Test 2).- Explain why the series converges. X n =1 ( - 1) n - 1 2 n + 1 Since it’s an alternating series we are going to apply the alternating series test. b n = 1 2 n + 1 0

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lim n →∞ b n = 0 f ( x ) = 1 2 x + 1 , f 0 ( x ) = - 2 (2 x + 1) 2 0 or 2 n + 3 = 2( n + 1) + 1 > 2 n + 1 0 b n +1 = 1 2( n + 1) + 1 1 2 n + 1 = b n Then by Alternating series Test, the series converges. 3.- Determine whether the series converges or diverges (using the Comparison Test or the Limit Comparison Test) (a) n =1 2 + ( - 1) n n n n =1 1 n n X n =1 2 + ( - 1) n n n X n =1 3 n n 2 + ( - 1) n n
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calc2exam3 - Math 1700 Exam 3 1-Use the Integral Test to...

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