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Unformatted text preview: SOLUTIONS FOR HOMEWORK 6 11.2. (i) a n = ( − 1) n . a 2 n = 1 and a 2 n 1 = − 1 for every n ∈ N , hence the sequences ( a 2 n ) and ( a 2 n 1 ) are monotone (in fact, they are constant). Subsequential limits: lim a 2 n = lim sup a n = 1, lim a 2 n 1 = lim inf a n = − 1. These are the only subsequential limits: indeed, if s is a subsequen tial limit, then lim inf a n ≤ s ≤ lim sup a n . Moreover, suppose, for the sake of contradiction, that s ∈ ( − 1 , 1). Let ε = min { 1 − s, s − ( − 1) } . If lim s n k = s , then s n k ∈ ( s − ε, s + ε ) ⊂ ( − 1 , 1) for k > N , which is impossible. − 1 = lim sup a n negationslash = lim inf a n = 1, hence the sequence is bounded, does not converge, and does not diverge to + ∞ or −∞ . (ii) b n = 1 /n . This sequence is monotone nonincreasing, hence so is any subsequence of ( b n ). lim b n = 0, hence, by Theorem 11.7, 0 is the only subsequential limit of ( b n ). Moreover, lim sup b n = lim inf b n = 0. The convergence of ( b n ) implies that this sequence is bounded, and cannot diverge to + ∞ or −∞ . (iii) c n = n 2 . This sequence is monotone nondecreasing, hence so is any subsequence of ( c n ). lim c n = + ∞ , hence ( c n ) is not bounded, and cannot converge. More over, lim sup c n = lim inf c n = + ∞ (Theorem 10.7), hence the set of subsequential limits consists of only one element  + ∞ . (iv) d n = (6 n + 4) / (7 n − 3). Note first that, for every n , d n − d n +1 = 46 (7 n − 3)(7 n + 4) > , hence the sequence ( d n ) is monotone nonincreasing. Thus, any subse quence of ( d n ) is monotone nonincreasing. Furthermore,) is monotone nonincreasing....
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This note was uploaded on 05/08/2010 for the course MATH 140a taught by Professor Staff during the Winter '08 term at UC Irvine.
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 Limits

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