q10-sol-1042-sp08

q10-sol-1042-sp08 - n x n n ! = 0, for any x so lim n (-3)...

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Calculus II—1042 Solution Quiz # 10 Fri., 4/4/08 Which of the sequences { a n } converge, and which diverge. Find the limit. 1. a n = ( - 1) n + 1 2 n - 1 . Solution: Since lim n →∞ ( - 1) n = DNE and lim n →∞ 1 2 n - 1 = 0 so lim n →∞ ± ( - 1) n + 1 2 n - 1 = DNE and the sequence is divergent. 2. a n = 1 - 3 n 4 2 n 4 + 5 n 3 . Solution: It is form, Either use LH Rule or divide the numerator and denominator by n 4 and then take limit. lim n →∞ ± 1 - 3 n 4 2 n 4 + 5 n 3 = lim n →∞ 1 n 4 - 3 2 + 5 n = 0 - 3 2 + 0 = - 3 2 and the sequence is convergent. 3. a n = cos n n . Solution: Use the Sandwhich Theorem for the limit Since - 1 n cos n n 1 n And lim n →∞ - 1 n = 0 = lim n →∞ 1 n By the sandwhich Theorem lim n →∞ cos n n = 0 and the sequence is convergent. 4. a n = 2 tan - 1 n . Solution: lim n →∞ p 2 tan - 1 n = p 2( π/ 2) = π and the sequence is convergent. 5. a n = ( - 3) n · 2 n n ! . Solution: Since lim
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Unformatted text preview: n x n n ! = 0, for any x so lim n (-3) n 2 n n ! = lim n (-6) n n ! = and the sequence is convergent. 6. a n = n 3 n . Solution: Since lim n n 1 /n = 1 and lim n x 1 /n = 1 so lim n n 3 n = lim n (3 n ) 1 /n = lim n h (3) 1 /n ( n ) 1 /n i = (1) (1) = 1 and the sequence is convergent. 7. a n = ln( n + 2) n 2 . Solution: It is form, use LH Rule here. lim n ln( n + 2) n 2 = lim n 1 n + 2 2 n = lim n 1 2 n ( n + 2 = and the sequence is convergent....
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This homework help was uploaded on 04/18/2008 for the course MATH 1042 taught by Professor Dr.z during the Spring '08 term at Temple.

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