# mt1s - MATH 153 Midterm 1 Solutions Name Mrs Potter Points...

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MATH 153 Midterm 1 Solutions Name: Mrs. Potter # Points Possible 1 15 2 15 3 15 4 15 5 15 6 25 Total 100 1

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2 1. [15 points] Use an appropriate test to show whether the series converges or diverges . a. [5] X n =0 n n 2 + 1 Limit Comparison Test: We know the series 1 n diverges ( p -series with p 1). lim n →∞ n n 2 +1 1 n = lim n →∞ n 2 n 2 + 1 = 1 Since this limit is positive and ﬁnite, by the Limit Comparison Test our series diverges with 1 /n . b. [5] X n =1 n 3 n - 1 4 n Ratio Test: lim n →∞ ± ± ± ± ± ( n +1)3 n 4 n +1 n 3 n - 1 4 n ± ± ± ± ± = lim n →∞ ( n + 1)3 n 4 n n 3 n - 1 4 n +1 = lim n →∞ 3( n + 1) 4 n = 3 4 . Since this limit is less than 1, the series converges. c. [5] X n =2 n ln n Test for Divergence: lim n →∞ n ln n = lim n →∞ 1 1 /n = lim n →∞ n = . Thus, since the terms of the series don’t go to 0, by the Test for Diver- gence the series diverges.
3 2. [15 Points] Find the interval of convergence of the following power series. X n =0 ( x + 1) n n 2 + 1 First, we ﬁnd the radius via Ratio Test. lim n →∞ ± ± ± ± ± ± ( x +1) n +1 ( n +1) 2 +1 ( x +1) n n 2 +1 ± ± ± ± ± ± = lim n →∞ | x + 1 | n +1 ( n 2 + 1) (( n + 1) 2 + 1) | x + 1 | n = lim n →∞ | x + 1 | ( n 2 + 1) (( n + 1) 2 + 1) = | x + 1 | . The Ratio Test says this series converges if

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## This note was uploaded on 07/26/2011 for the course MATH 153 taught by Professor Rempe during the Spring '08 term at Ohio State.

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mt1s - MATH 153 Midterm 1 Solutions Name Mrs Potter Points...

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