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Unformatted text preview: Final Solutions 1. Calculations. Find the following quantities. No proof required, but show any sidework for partial credit. (a) (3 pts) X n =2 1 2 n . Solution: X n =2 1 2 n = X n =0 1 2 n 1 1 2 = 1 1 1 2 1 1 2 = 1 2 . (b) (4 pts) The limit of the sequence 2 , p 2 2 , q 2 p 2 2 ,... . Solution: The sequence can be written recursively as a n +1 = 2 a n . So, assuming ( a n ) converges, letting a = lim a n , we get the equation a = 2 a . The solutions are a = 0 or a = 2. Because the sequence is increasing and starts out positive, the limit must be a = 2. (c) (3 pts) The value lim sup s n where s n = ( 1) n 4 n 3 3 n +4 . Solution: The limit of 4 n 3 3 n +4 is 4 3 , so the sequence ( s n ) eventually oscillates between values close to 4 3 and values close to 4 3 . Therefore, limsup s n = 4 3 . (d) (2 pts) The set of limit points of S = [0 , 1) { 3 } . The set of limit points is [0 , 1]. Note that 3 is not a limit point, its an isolated point. (e) (3 pts) X n =1 1 n ( n + 1) . This can be written as a telescoping series whose limit is 1. (see homework 5 solutions.) 1 2. Suppose ( a n ) is a sequence of real numbers so that limsup a n = . Assume R . (a) (5 pts) Let > 0 be arbitrary. Prove by contradiction that there exist infinitely many sequence elements in the neighborhood (  , + )....
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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

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