# 11-12 - SOLUTIONS FOR HOMEWORK 6 11.2(i an =-1)n a2n = 1...

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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 non-increasing, 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 non-decreasing, 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) > 0 , hence the sequence ( d n ) is monotone non-increasing. Thus, any subse- quence of ( d n ) is monotone non-increasing. Furthermore, lim d n = lim 6 + 4 /n 7 3 /n = lim(6 + 4 /n ) lim(7 3 /n ) = 6 7 .

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