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Unformatted text preview: Math 131A Analysis Summer Session A Homework 3 Solutions 1. Let ( a n ) be a bounded sequence and let ( b n ) be a sequence such that lim n b n = 0. Prove that lim n a n b n = 0. Proof. Since ( a n ) is bounded, there exists M > 0 such that  a n  < M for all n N . For all n N , we have  a n b n  =  a n  b n  < M  b n  , or equivalently, M  b n  < a n b n < M  b n  . Since b n 0, we have  b n  0 by the limit theorems. Thus M  b n  0 by the limit theorems. By the Squeeze Theorem, a n b n 0. 2. Let p > 0 be a positive real number. Then prove lim n a n n p = if  a  1 + if a > 1 does not exist if a < 1 Proof. If  a  1, the sequence a n is bounded. Indeed,  a n  =  a  n 1 for all n N . Also 1 n p 0 for all p > 0. Thus by Problem 1, lim n a n n p = 0. If a > 1, then by the limit theorems, we have a n +1 ( n +1) p a n n p = a n +1 a n n p ( n + 1) p = a n n + 1 p = a 1 1 + 1 n p a > 1 Thus by Homework 2, problem 7b, we have lim n a n n p = lim n a n n p = . If a < 1, then to show the limit does not exist, it suffices to prove lim n a 2 n (2 n ) p = and lim n a 2 n 1 (2 n 1) p = since that implies liminf a n n p = < = limsup a n n p Since a 2( n +1) 1 (2( n +1) 1) p a 2 n 1 (2 n 1) p = a 2 n +1 (2 n +1) p a 2 n 1 (2 n 1) p = a 2 2 n 1 2 n + 1 p a 2 > 1 1 by Homework 2, problem 7b we have lim n a 2 n 1 (2 n 1) p = lim n  a 2 n 1 (2 n 1) p = + Thus lim n a 2 n 1 (2 n 1) p = Similarly, a 2( n +1) (2( n +1)) p a 2 n (2 n ) p = a 2 n +2 (2 n +2) p a 2 n (2 n ) p = a 2 2 n + 2 2 n p a 2 > 1 by Homework 2, problem 7b we have lim n a 2 n (2 n ) p = lim n a 2 n (2 n ) p = + 3. If lim n s n = + and k < 0, prove lim n ks n = . Proof. Define a sequence k n = k for all n N . Clearly, k n k < 0. Thus k n  k > 0. By Theorem 9 . 9, lim n  s n k n = lim n s n ( k n ) = + , which implies lim n ks n = lim n  ( s n k n ) = . 4. If lim n s n = + and ( t n ) is a bounded sequence, prove lim n s n + t n = + . Proof. Since ( t n ) is bounded, there exists C > 0 such that C < t n < C for all n N . Let M > 0. Since lim n s n = + , there exists N N so that s n > M + C for all n > N . Thus s n + t n > ( M + C ) C = M for all n > N ....
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 Summer '08
 hitrik
 Math

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