Unformatted text preview: The sequence converges to 3. lim n →∞ 3 n 2 n 2 + 1 = lim n →∞ 3 1 + 1 n 2 = 3 4. 3 n1 4 n = 1 4 , 3 16 , 9 64 , 27 4 4 , 3 4 4 5 ,... The sequence converges to 0. Use the theorem lim n →∞ x n = 0 for x < 1 . Since 3 4 < 1, then lim n →∞ 1 4 3 4 n1 = 1 4 lim n →∞ 3 4 n1 = 1 4 ⋅ = . 5. 2 n + 1 ( ) ! 2 n1 ( ) ! = 3! 1! , 5! 3! , 7! 5! , 9! 7! , 11! 9! ,... The sequence diverges. Simplify the factorials: 2 n + 1 ( ) ! 2 n1 = 2 n + 1 ( ) 2 n ( ) 2 n1 ( ) ! 2 n1 = 2 n + 1 ( ) 2 n ( ) . Then ....
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 Spring '08
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 Calculus, Limit of a function, Indeterminate form, form lim ln

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