Lesson 6.pdf - 6 Convergence of Sequences Sequences sequence as A sequence is a list of real numbers with a particular order We denote a(sn =(s1 s2 s3

Lesson 6.pdf - 6 Convergence of Sequences Sequences...

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6. Convergence of Sequences July 15, 2019 Sequences A sequence is a list of real numbers with a particular order. We denote a sequence as ( s n ) = ( s 1 , s 2 , s 3 , ..., s n , ... ) . Each s n is called a term of this sequence. 1 [Example] There are several ways to describe a sequence as follows. 1. Given the formula for all the terms, write down the sequence. (a) s n = 1 n . This is the sequence ( s n ) = 1 , 1 2 , 1 3 , ... . (b) s n = n - 1 n . This is the sequence ( s n ) = 0 , 1 2 , 2 3 , 3 4 , ... . (c) s n = ( - 1) n n 2 . This is the sequence ( s n ) = ( - 1 , 4 , - 9 , 16 , ..., ( - 1) 2 n 2 , ... ) (d) s n = ( 1 n , n odd. ( - 1) n/ 2 n 2 , n even. This is the sequence ( s n ) = 1 , - 4 , 1 3 , 16 , 1 5 , - 36 , 1 7 , 64 , ... 1 It is also denoted by ( s n ) n =1 , or { s n } , or { s n } n =1 . 50
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2. By giving the first few terms to establish a pattern, leaving it to you to find the function. (a) ( s n ) = (0 , 1 , 0 , 1 , 0 , 1 , .... ). The pattern here is obvious. We can write it as s n = ( 0 , n odd, 1 , n even. or we can write s n = 1 - ( - 1) n 2 . The second expression is better. (b) ( s n ) = 2 , 5 2 , 10 3 , 17 4 , 26 2 , ... . What is s n =?. s n = n 2 + 1 n . (c) ( s n ) = (2 , 4 , 8 , 16 , 32 , .... ). What is s 6 ? You can find s n = 2 n = s 6 = 64 . 3. By a recursion formula. (a) s n +1 = 1 n +1 s n and s 1 = 1. The first 5 terms are 1 , 1 2 , 1 6 , 1 24 , 1 120 , ... Assuming that the pattern continues: s n = 1 n ! , n N . In fact we can prove it: s n +1 = 1 n +1 s n = 1 n +1 · 1 n s n - 1 = ...... = 1 n +1 · 1 n · ... · 1 2 s 1 = 1 ( n +1)! . (b) s n +1 = 1 2 ( s n + 1), s 1 = 1. The first 5 terms are (1 , 1 , 1 , 1 , 1 , ... ) . Assuming that the pattern continues s n = 1 , n N . 51
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In fact, we can prove it. s n +1 = 1 2 ( s n + 1) = 1 2 s n + 1 2 = 1 2 · 1 2 ( s n - 1 + 1) + 1 2 = 1 2 2 s n - 1 + 1 2 + 1 2 2 = 1 2 2 · ( 1 2 s n - 2 + 1) + 1 2 + 1 2 2 = 1 2 3 s n - 2 + 1 2 + 1 2 2 + 1 2 3 = ...... = 1 2 n - 1 s 1 + 1 2 + 1 2 2 + 1 2 3 + ... + 1 2 n - 1 = 1 2 n - 1 + 1 2 (1 - 1 2 n - 1 ) 1 - 1 2 = 1 2 n - 1 + 1 - 1 2 n - 1 = 1 . More simply, we may use mathematical induction to prove it. Let S be the set of positive integers for which this formula holds. We see 1 S . Assume k S , i.e., s k = 1. Then s k +1 = 1 2 ( s k + 1) = 1 2 · 2 = 1 so that k + 1 S . We are done. The induction method is better. Convergence of a sequence Definition 1. A sequence ( s n ) converges to the number s if for any δ > 0, there corresponds a positive integer N such that | s n - s | < for all n > N. (1) The number s is called the limit of this sequence. 2. “( s n ) converges to s ” is denoted by lim n →∞ s n = s, or lim s n = s, or s n s. 3. A sequence that does not converge is said to diverge . [Examples and remarks] 1. After Newton and Leibniz, people did not know how to define “limit”. The first definition of limit was given by Augustin Louis Cauchy (1789-1857) without - N language: 52
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When the values successively assigned to the same variable indefinitely
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