Lesson 25a - Ratio and Root Test

# Example2 nn

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Unformatted text preview: ( n 3 2) n The highest power is n4, so we divide numerator and denominator by n4 to get: an 1 n a n lim 1 2 n 4 )3 4 n4 n n 0 0 lim n n 1 n3 2 1 ( )( 3 3 ) nnn n ( Since the limit is less than 1, we have the series converges by the ratio test. Example 2 nn Determine if the following series converges or diverges: n 1 ( 2 n )! Here we see a factorial, so we will use the ratio test: an 1 (n 1) n 1 (2n)! lim lim n a n ( 2( n 1))! ( n) n n (2n)! an 1 (n 1)(n 1) n lim lim n a n ( 2n 2)( 2n 1)( 2n)! ( n) n n 1 (n 1)(1 ) n a n lim n 1 lim n a n ( 2 n 2)( 2 n 1) n a (n 1) lim n 1 e( lim ) n a n ( 2 n 2)( 2 n 1) n an 1 n a n lim Thus this series converges by the ratio test. 11 ( 2) nn e( lim ) e ( 0) 0 n 2 1 (2 )(2 ) n n Example 3 Determine if the following series converges or diverges: n2 e (2n 1)3n1 n 1 Here we see f(n)g(n), so we will use root test n lim an 1/ n n lim an 1/ n n 1/ n e n ( 2 n 1) 3 n 1 lim 1/ n lim an n2 e...
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## This note was uploaded on 02/07/2014 for the course MATH 1004 taught by Professor Mark during the Summer '00 term at Carleton CA.

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