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Unformatted text preview: MATH 214: SEQUENCES AND SERIES 1. Sequences 1.1. Definitions. Definition. An (infinite) sequence is a list of numbers. We can also define a sequence to be a function from N to R , where n 7 a n . We often write { a n } n =1 for a sequence. Example 1. The following sequences will come up often: (1) { 1 , 1 2 , 1 3 ,... } = { 1 n } n =1 (2) { ( 1) n } n =1 (3) { n } n =1 The order of a sequence matters, for instance the first sequence is not equal to { 1 2 , 1 , 1 3 ,... } . Note that the first sequence gets closer and closer to 0, but the second bounces between 1 and 1 and the third gets larger and larger as we move down the sequence. We formalize each of these concepts in the following way: Definition. A sequence converges to L if for every > there exists an N N such that  a n L  < for all n > N . A sequence diverges if there is no L such that a n converges to L A sequence diverges to (negative) infinity if for every number M there is an integer N M such that a n > M ( a n < M ) for all n > N M So the first sequence converges to 0, the second sequence diverges, and the third diverges to infinity. Exercise. Prove all of the above statements. 1.2. Limits. The only way that we can calculate the limits of sequences at this point is directly. The following theorems allow us to use some limits to calculate others: Theorem 1.1. Let { a n } n =1 and { b n } n =1 be convergent sequences of real numbers with lim n a n = A and lim n b n = B (1) lim n ( c 1 a n + c 2 b n ) = c 1 A + c 2 B (2) lim n ( a n b n ) = AB (3) If B 6 = 0 lim n a n b n = A B Example 2. Let { a n } n =1 = { 1 } n =1 and { b n } n =1 = { 1 n } n =1 , then lim n 1 1 n = lim n 1 lim n 1 n = 1 Note that the sequences a n and b n both have to be convergent for the theorem to apply. Example 3. If we let a n = b n = n , then a n b n = 1 converges, but each of the two sequences diverges to infinity. Theorem 1.2 (Sandwich Theorem) . Given sequences { a n } n =1 , { b n } n =1 , { c n } n =1 be sequences of real numbers with a n b n c n for all n beyond some index N and lim n a n = lim n c n = L . Then lim n b n = L . 1 2 MATH 214: SEQUENCES AND SERIES Exercise. Let b n = n ! n n , a n = 0 and c n = 1 n . Use the above theorem to show that lim n b n = 0 . In single variable calculus, we learned that continuous functions take limits to limits. The same holds true for sequences: Theorem 1.3. If { a n } n =1 is an infinite sequence and f ( x ) is a function that is continuous at L and defined at all a n , then lim n f ( a n ) = f ( L ) Exercise. Let a n = p 1 n for any p > . Use the above theorem and f ( x ) = p x to show that lim n p 1 n = 1 ....
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 Spring '11
 AlexOndrus
 Calculus, Sequences And Series

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