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Would you like this assignment to be marked? YES NO MA104 Lab Report 9 - Strategies for Testing Series, Power Series Name: Student Number: Winter 2019 1. [10 marks ] Consider the series n =2 ( - 1) n (2 x + 3) n n ln n . For each of the following values of x , determine whether the resulting series converges absolutely, converges conditionally, or diverges. (a) x = - 3 n =2 ( - 1) n (2 x + 3) n n ln n = n =2 ( - 1) n ( - 3) n n ln n = n =2 3 n n ln n which diverges by the Test for Divergence: lim n →∞ a n = lim n →∞ 3 n n ln n I.F. = H lim n →∞ 3 n ln 3 ln n + 1 I.F. = H lim n →∞ 3 n (ln 3) 2 1 /n = lim n →∞ n 3 n (ln 3) 2 = (b) x = - 1 [You may use Maple to find any derivatives or integrals: diff(???,x); simplify(%); int(???,x=?..?); ] n =2 ( - 1) n (2 x + 3) n n ln n = n =2 ( - 1) n (1) n n ln n = n =2 ( - 1) n n ln n 1. n =2 | a n | = n =2 1 n ln n Using the integral test: Let f ( x ) = 1 x ln x . Then: (i) f is continuous for all x 2. (ii) f 0 ( x ) = - ln x + 1 x 2 (ln x ) 2 < 0 for all x 2 f is decreasing on [2 , ). (iii) f ( x ) > 0 for all x > 0 f is positive on [2 , ). Therefore, the integral test may be applied. Z 2 f ( x ) dx = lim t →∞ Z t 2 1 x ln x dx =

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