f11mt1sol - MAT 127, Lecture 01, Spring 2011 Solutions to...

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Unformatted text preview: MAT 127, Lecture 01, Spring 2011 Solutions to Midterm 1 Test the following series for convergence. 1. n =1 ne n . Answer: Divergent. Solution: Let us use the Ratio Test: a n = ne n , a n +1 = ( n + 1) e n +1 , L = lim n a n +1 a n = lim n ( n + 1) e n +1 ne n = e lim n n + 1 n = e. Since L > 1, the series is divergent. 2. n =1 3+(- 2) n 5 n . Answer: Convergent. Solution: Remark that 3 + (- 2) n 5 n = 3 5 n + (- 2) n 5 n . The first series is a geometric series with a 1 = 3 5 and r 1 = 1 5 . The second series is a geometric series with a 2 =- 2 5 and r 2 =- 2 5 . Since | r 1 | < 1 , | r 2 | < 1, both series are convergent, and their sum is convergent too. 3. n =1 ( n !) n (2 n )! . Answer: Convergent. Solution: Let us use the Ratio Test: a n = ( n !) n (2 n )! , a n +1 = (( n + 1)!)( n + 1) (2 n + 2)! , 1 L = lim n a n +1 a n = lim n ( n + 1)! ( n + 1) (2 n )!...
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f11mt1sol - MAT 127, Lecture 01, Spring 2011 Solutions to...

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