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Unformatted text preview: Homework 4 Solutions 1. For each of the following sequences ( s n ), find the set S of subsequential limits, and use this to find liminf s n and lim sup s n . (a) (11.1) s n = 3 + 2( 1) n . Solution: ( s n ) = (5 , 1 , 5 , 1 ,... ), so S = { 1 , 5 } , and therefore limsup s n = 5 and liminf s n = 1. (b) s n = n 2 ( 1 + ( 1) n ). Solution: ( s n ) = (0 , 2 , , 18 , , 50 ,... ), so S = { , } , and there fore lim sup s n = 0 and lim inf s n = . (c) ( s n ) = (0 , 1 , 2 , , 1 , 3 , , 1 , 4 ,... ). Solution: S = { , 1 , } , so limsup s n = and lim inf s n = 0. 2. Prove that every unbounded sequence contains a monotone subsequence that diverges to either or . Solution: If ( s n ) is unbounded above, for each choice of K , there exists a sequence element s n k so that s n k > K . The subsequence ( s n k ) clearly diverges to . Proof is similar if ( s n ) is unbounded below. 3. Suppose x > 1. Prove that lim x 1 /n = 1. Solution: First of all, x 1 /n is bounded below by 1 since x > 1. Note the quantity x 1 n ( n +1) > 1, so x 1 n +1 < x 1 n ( n +1) x 1 n +1 = x 1 n , which means the sequence is decreasing. Since it is decreasing and bounded below by 1, a limit exists for the sequence, call it...
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This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.
 Winter '11
 Wiley
 Differential Equations, Equations, Limits

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