1 Dr. Bradley Quoc Phung 230A Theorem Limit of Sequences Theorem: Suppose sn and lim k .sn k .s lim( sn tn ) s t lim( sn .tn ) s.t lim tn are real sequences with lim sn s , lim tn t x x x x x x 1 1 sn s Sn 0 and Provide S 0 Prove: lim( sn .tn ) s.t x Let > 0. sntn st sntn snt snt st sn tn t t sn s Since lim tn t , x N1 such that n > N1 tn t 2M when | Sn | < M for all n. Since lim sn s , x So. N2 such that n > N1 sntn st sn tn t t sn s M Therefore: sn s 2 2M lim( sn .tn ) s.t x 2 t 2 t 1 Q.E.D. 2 t 1 2 Dr. Bradley Quoc Phung 230A Theorem 1: If sequence { an }is convergent , then { an } is bounded. Solution: * We need to prove * | a n| = | a n Assume { an }is convergent to a 0, | a n| < M , for all n N. - a+ a| |an -a|+|a| lim an a n N* such that Let = 1, then N* such that Therefore, | a n| n > N* | an - a | < 1 |an |an -a|+|a| 1 Let M = Max { a1 , a2 , … , a n-1 , Then | a n | Therefore: | an - a | < = = n > N* - a+ a| + |a| 1+|a|} M for n > N* If sequence { an }is convergent , then { an } is bounded. Q.E.D. 3 Dr. Bradley Quoc Phung 230A Theorem 2: If a sequence { an }is bounded, monotonic , then { an } is convergent. Solution: * We need to prove: lim an L n 0, N* such that n > N* | an - L | < Suppose a sequence { an }is bounded and monotone increasing. Let A = { an | n N } be a nonempty bounded set. Then by the Completeness axiom: If every nonempty subset A of R is bounded above, then subset A has a least upper bound. Therefore A = { an | n N } has a least upper bound Sup A = L. Sup A - = L - Sup A = L aM Since L= Sup A, then L- is not Sup A. So there exist M such that L - aM L A sequence { an }is monotone increasing From ( 1 ) and ( 2 ) i e: 0, that L - a n M such that L - aM aM a n L (1) for n > M. aM a n L L + ( 2) L+ | an - L | < n>M | an - L | < Therefore: If a sequence { an }is bounded, monotonic , then { an } is convergent Q.E.D. 4 Dr. Bradley Quoc Phung 230A