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MATH 4389 (17)
Almus (17)
Final

# MATH 4389 Final: Real_Analysis(part_2) Premium

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School
Department
Mathematics
Course
MATH 4389
Professor
Almus
Semester
Spring

Description
PART II. SEQUENCES OF REAL NUMBERS II.1. CONVERGENCE Deﬁnition 1. A sequence is a real-valued function f whose domain is the set positive integers (N). The numbers f(1),f (2), ··· are called the terms of the sequence. Notation Function notation vs subscript notation: f(1) ≡ 1 ,f (2) ≡2s ,··· ,f (n) n s , ··· . In discussing sequences the subscript notation is much more common than functional notation. We’ll use subscript notation throughout our treatment of analysis. Specifying a sequence There are several ways to specify a sequence. 1. By giving the function. For example: 1 1 1 1 1 1 (a) sn= n or {sn} = n . This is the sequence {12 ,3 ,4 ,...,n,... }. (b) s = n − 1. This is the sequence {0, ,2 , 3,...,n − 1,... }. n n 2 3 4 n (c) sn=( −1) n . This is the sequence {−1,4,−9,16,..., (−1) n ,... }. 2. By giving the ﬁrst few terms to establish a pattern, leaving it to you to ﬁnd the function. This is risky – it might not be easy to recognize the pattern and/or you can be misled. (a) {sn} = {0,1,0,1,0,1,... }( The pattern here is obvious; can you devise the function? It’s 1 − (−1)n) 0, n odd sn= or s n= 2 1, n even 2 (b) {s } = 2, 5,10,17 ,6 ,... ,s = n +1 . n 2 3 4 5 n n (c) {sn} = {2,4,8,16,32,... }. What is6s ? What is the function? While you might say 64 n and s n2 , the function I have in mind gives s6= π/6: n π 64 sn=2 +( n − 1)(n − 2)(n − 3)(n − 4)(n − 5) − 720 120 3. By a recursion formula. For example: (a) s = 1 s ,s = 1. The ﬁrst 5 terms are 1,1, , 1 , 1 ,.... Assuming that n+1 n +1 n 1 2 6 24 120 1 the pattern continues n = . n! 1 (b) sn+1 = (n +1) ,s 1 = 1. The ﬁrst 5 terms are {1,1,1,1,1,... }. Assuming that the 2 pattern continues n = 1 for all n; {n } is a “constant” sequence. 13 Deﬁnition 2. A sequence {s }nconverges to the number s if to each > 0 there corresponds a positive integer N such that |s − s| for all n>N. n The number s is called the limit of the sequence. Notation “{s } converges to s” is denoted by n n→∞ sn= s, or by limsn= s, or by n → s. A sequence that does not converge is said to diverge. Examples Which of the sequences given above converge and which diverge; give the limits of the convergent sequences. THEOREM 1. If s → s and s → t, then s = t. That is, the limit of a convergent sequence is unique. Proof: Suppose s 6= t. Assume t>s and let = t − s. Since sn→ s, there exists a positive integer N such that |s − s | / 2 for all n>N . Since s → t, there exists a positive integer 1 n 1 n N 2 such that |t−s n / 2 for all n>N 2. Let N = max{N ,N1} a2d choose a positive integer k>N . Then t − s = |t − s| = |tk− s k s − s|≤| t k s | + |sk− s |+< = = t − s, 2 2 a contradiction. Therefore, s = t. THEOREM 2. If {s } connerges, then {s } is bnunded. Proof: Suppose s → s. There exists a positive integer N such that |s−s | < 1 for all n>N . n n Therefore, it follows that |sn| = |n − s + s|≤| sn− s| + |s| < 1+ |s| for alln>N. Let M = max{|s |, |s |, ..., |s |, 1+ |s|}. Then |s | N. If an→ 0, then s →n0. Proof: Note ﬁrst that a ≥n0 for all n>N . Since a → 0,nthere exists a positive integer N 1 such that |an| < /k. Without loss of generality, assume that1N ≥ N. Then, for all n>N 1, |sn− 0| = |sn|≤ ka n
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