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Pre-Calc Homework Solutions 383

# Pre-Calc Homework Solutions 383 - Section 9.4 13 Diverges...

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13. Diverges by the Ratio Test, since lim n } a a n 1 n 1 } 5 lim n } ( n 1 3 1 n ) 1 3 1 2 n 1 1 } ? } n 3 3 2 n n } 5 lim n } ( n 1 3 n 1 3 ) 3 (2) } 5 } 3 2 } . 1. (The n th 5 Term Test can also be used.) 14. Converges by the Ratio Test, since lim n } a a n 1 n 1 } 5 lim n ? } n 2 ln n n } 5 lim n } 1 2 } ? } n 1 n 1 } ? } ln ( l n n 1 n 1) } 5 } 1 2 } , 1. 15. Converges by the Ratio Test, since lim n } a a n 1 n 1 } 5 lim n } ( ( 2 n n 1 1 1 3 ) ) ! ! } ? } (2 n n 1 ! 1)! } 5 lim n 5 lim n } 2(2 n 1 1 3) } 5 0 , 1. 16. Converges by the Ratio Test, since lim n } a a n 1 n 1 } 5 lim n ? } n n ! n } 5 lim n 5 lim n 1 } n 1 n 1 } 2 n 5 lim n } (1 1 1 1/ n ) n } 5 } 1 e } , 1 17. One possible answer: n 5 1 } 1 n } diverges (see Exploration 1 in this section) even though lim n } 1 n } 5 0. 18. One possible answer: Let a n 5 2 2 n and b n 5 3 2 n Then a n and b n are convergent geometric series, but } a b n n } 5 1 } 3 2 } 2 n is a divergent geometric series.
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