prob10sol

prob10sol - Math 151 Spring 2008 Problem Set # 10 Solutions...

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Math 151 Spring 2008 Prob lemSet#10 Solutions The Limit of a Sequence) 1. 1 , 3 , 5 , 7 2. 2 3 , 3 8 , 4 15 , 5 24 3. 1 , 1 3 , 1 9 , 1 27 4. 1 , 4 , 9 , 16 5. 5 n 2 +9 2 n 2 +1 = n 2 μ 5+ 9 n 2 n 2 μ 1+ 1 n 2 = 9 n 2 1 n 2 . Therefore, lim n →∞ 5 n 2 2 n 2 =l im n →∞ 9 n 2 1 n 2 =5 . 6. n 2 +10 4 n 3 = n 2 μ 10 n 2 n 3 μ 4+ 1 n 3 = 1 n 10 n 2 1 n 3 . Therefore, lim n →∞ n 2 4 n 3 n →∞ 1 n 10 n 2 1 n 3 =0 . 7. s 3 n 2 9 n 2 2 = v u u u t 3 n 2 n 2 μ 9 2 n 2 = v u u t 3 9 2 n 2 . Therefore, lim n →∞ s 3 n 2 9 n 2 2 = lim n →∞ v u u t 3 9 2 n 2 = r 3 9 = r 1 3 . 8. We have cos ( πn )= ½ 1 if n =1 , 3 , 5 , 7 ,..., 1 if n =2 , 4 , 6 , 8 ,... Thus, the sequence { cos ( ) } n =1 has the subsequences, 1 , 1 , 1 , 1 1

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and 1 , 1 , 1 , 1 ,... that have the limits 1 and 1, respectively. If a sequence has a limit, every subsequence must have the same limit. Therefore, the given sequence does not have limit. 9. lim n →∞ sin μ πn 6 n +2 =s i n μ lim n →∞ 6 n i n lim n →∞ n μ 6+ 2 n i n lim n →∞ π 2 n =sin ³ π 6 ´ = 1 2 . 10. We have ( 1) n n n +4 = (2 k 1) (2 k 1)+4 ,n =2 k 1 ,k =1 , 2 , 3 ,..., 2 k 2 k k, k , 2 , 3 lim k →∞ μ (2 k 1) (2 k 1) + 4 = 1 and lim k →∞ 2 k 2 k . Since the given sequence has two subsequences with di f erent limits, it does not have a limit. 11. We have ¯ ¯ ¯ ¯ ( 1) n n n 2 ¯ ¯ ¯ ¯ = n n 2 , and lim n →∞ n n 2 = lim n →∞ n n 2 μ 1+ 4 n 2 = lim n →∞ 1 n μ 4 n 2 =0 .
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This note was uploaded on 04/21/2008 for the course MATH 151 taught by Professor Geveci during the Winter '08 term at San Diego State.

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prob10sol - Math 151 Spring 2008 Problem Set # 10 Solutions...

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