# SOLUTIONS - CHAPTER 3 Sequences and Series 3.1 Sequences...

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CHAPTER 3 Sequences and Series 3.1. Sequences and Their Limits Definition (3.1.1) . A sequence of real numbers (or a sequence in R ) is a function from N into R . Notation. (1) The values of X : N ! R are denoted as X ( n ) or x n , where X is the sequence. (2) ( x n : n 2 N ) or simply ( x n ) may denote a sequence — this is not the same as { x n : n 2 N } . (3) ( x 1 , x 2 , . . . , x n , . . . ). Example. (1) (3 n ) = (3 n : n 2 N ) = (3 , 6 , 9 , . . . , 3 n, . . . ) . (2) (1) = (1 : n 2 N ) = (1 , 1 , 1 , . . . , 1 , . . . ) . (3) ( ( - 2) n ) = ( ( - 2) n : n 2 N ) = ( - 2 , 4 , - 8 , . . . , ( - 2) n , . . . ) . (4) 1 2 + 1 2 ( - 1) n = 1 2 + 1 2 ( - 1) n : n 2 N = (0 , 1 , 0 , 1 , . . . , 0 , 1 , . . . ) . 34
3.1. SEQUENCES AND THEIR LIMITS 35 (5) n 2 1 2 + 1 2 ( - 1) n = n 2 1 2 + 1 2 ( - 1) n : n 2 N = 1 , 1 , 1 , 2 , 1 , 3 , 1 , 4 , . . . , 1 , n 2 , . . . . Sequences may also be defined inductively or recursively. Example. (1) x 1 = 5, x n +1 = 2 x n - 3 ( n 1) gives (5 , 7 , 11 , 19 , 35 , . . . ) . (2) Fibonacci sequence : x 1 = x 2 = 1 , x n +1 = x n - 1 + x n ( n 2) gives (1 , 1 , 2 , 3 , 5 , 8 , 13 , . . . ) . Definition (3.1.3) . A sequence X = ( x n ) in R is said to converge to x 2 R , or have limit x , if 8 > 0 9 K ( ) 2 N 3-- 8 n K ( ) , | x n - x | < . We write this as lim X = x , lim( x n ) = x , lim n !1 x n = x , or x n ! x as n ! 1 . A sequence that converges is called convergent , one that does not divergent . Example. lim 1 n = 0. Proof. Let > 0 be given. By the Archimedean property, 9 K ( ) 2 N 3-- 1 K ( ) < . Then, for n K ( ), 1 n 1 K ( ) , and so 1 n - 0 = 1 n 1 K ( ) < . Thus lim 1 n = 0 by definition.
36 3. SEQUENCES AND SERIES Calculator Visualisation lim n !1 1 n = lim x !1 1 x = 0 if 8 > 0, with yMin = 0 - and yMax = 0 + , you can find K ( ) 2 N 3-- if xMin = K ( ) and xMax = 1 E 99, the graph only enters the screen from the left and exits from the right. Theorem (3.1.4 — Uniqueness of Limits) . A sequence in R can have at most one limit. Proof. [The 2 technique.] Suppose lim( x n ) = x 0 and lim( x n ) = x 00 . By Theorem 2.1.9, it su ffi ces to show that | x 0 - x 00 | < 8 > 0, for then | x 0 - x 00 | = 0 = ) x 0 = x 00 . Let > 0 be given. Since lim( x n ) = x 0 , 9 K 0 2 N 3-- 8 n K 0 , | x n - x 0 | < 2 . Since lim( x n ) = x 00 , 9 K 00 2 N 3-- 8 n K 00 , | x n - x 00 | < 2 . Let K = max { K 0 , K 00 } . Then n K = ) n K 0 and n K 00 = ) | x 0 - x 00 | = | x 0 - x n + x n - x 00 | | {z } smuggling | x 0 - x n | + | x n - x 00 | < 2 + 2 = .
3.1. SEQUENCES AND THEIR LIMITS 37 Theorem (3.1.5) . Let X = ( x n ) be a sequence in R , and let x 2 R . The following are equivalent: (a) X converges to x . (b) 8 > 0 , 9 K 2 N 3-- 8 n K, | x n - x | < . (c) 8 > 0 , 9 K 2 N 3-- 8 n K, x - < x n < x + . (d) 8 -nbhd. V ( x ) of x, 9 K 2 N 3-- 8 n K, x n 2 V ( x ) . Proof. (a) () (b) by definition. (b) () (c) () (d) since | x n - x | < () - < x n - x < () x - < x n < x + () x n 2 V ( x ). Technique Given > 0. Produce or verify the existence of an integer K ( ) so that n K ( ) = ) | x n - x | < . Sometimes | x n - x | < can be converted, with reversible steps, to an inequality of the form n > f ( ). Take K ( ) as the first integer greater than f ( ) (by the Archimedean Property), K ( ) = [ f ( )] + 1, for example. Then n K ( ) = ) n > f ( ) = ) | x n - x | < . Example.