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Unformatted text preview: MATH 1014 3.00 MW – APPLIED CALCULUS II – Prof. Madras Solutions for Practice Test #3 1. (a) Use the Ratio Test: lim n →∞ ( n + 1) 2 / 2 n +1 n 2 / 2 n = lim n →∞ 2 n ( n + 1) 2 2 n +1 n 2 = lim n →∞ 1 2 n + 1 n 2 = lim n →∞ 1 2 1 + 1 n 2 = 1 2 . Since 1 2 < 1, the Ratio Test tells us that this series converges . (The Root Test could also be used: you would show that lim n →∞  n 2 / 2 n  1 /n = 1 / 2 < 1.) (b) This series diverges. To see this, observe that lim n →∞ e n 1 e n + 1 = lim n →∞ 1 e n 1 + e n = 1 . Therefore, lim n →∞ ( 1) n e n 1 e n +1 does not exist, and in particular it does not equal 0. So the Test for Divergence (page 692) tells us that the series diverges. 2. To show the series converges, we can compare it with the series ∑ ∞ j =1 1 j 3 , which we know converges (by the Integral Test—see top of page 700). And 1 j 3 + j ≤ 1 j 3 , so the series ∑ ∞ j =1 1 j 3 + j converges by the Comparison Test (page 705). (The Limit Comparison Test on page 707 will also work.)Comparison Test on page 707 will also work....
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This note was uploaded on 02/26/2011 for the course MATH 1014 taught by Professor Ganong during the Spring '09 term at York University.
 Spring '09
 ganong
 Math, Calculus

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