Case 2 suppose there are only finitely many dominant terms and we only focus on

# Case 2 suppose there are only finitely many dominant

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Case 2: suppose there are only finitely many dominant terms, and we only focus on this case in the following. Select n 1 , such that s n 1 is larger than all dominant terms. Then (11.1) For any N n 1 , there exists m > N , such that s m s N . We are going to use “inductive construction” to find a monotone sequence. (1) Applying (11.1) with N = n 1 gives n 2 > n 1 , such that s n 2 s n 1 ; (2) Suppose n 1 , · · · , n k have been selected such that n 1 < n 2 < · · · < n k , and s n 1 s n 2 ≤ · · · ≤ s n k . Applying (11.1) with N = n k gives n k +1 > n k , such that s n k +1 s n k . By Mathematical Induction, ( s n k ) is a nondecreasing subsequence. Corollary 11.9. Let ( s n ) be a sequence, then there exists a monotone subsequence with limit lim sup s n , and a monotone subsequence with limit lim inf s n . Proof. We only prove the first statement for lim sup s n . Let v N = sup { s n : s > N } , N N , then v = lim v N = lim sup s n . If v = -∞ , then lim s n = -∞ , so any monotone sequence will converge to lim sup s n , and we finished. If v 6 = -∞ , select an arbitrary monotone increasing sequence ( t N ), such that lim t N = v : if v is finite, let t N = v - 1 N ; if v = + , let t N = N . Now we discuss the two cases as in the proof of the above theorem. Case 1: by assumption, s n k = sup { s m : m n k } = v n k - 1 , so { s n k } is decreasing and lim k →∞ s n k = lim v N = lim sup s n . Case 2: Given N > n 1 as above, by (11.1), there exists m 1 > N , such that s m 1 s N . Since t N < v v N = sup { s m : m > N } ,
38 XIN ZHOU there exists m 2 > N , such that s m 2 > t N . either s m 1 s m 2 , = s m 1 s N and s m 1 > t N ; or s m 2 s m 1 , = s m 2 s N and s m 2 > t N . Then we have (11.2) Given N n 1 , there exists m > N , such that s m s N and s m > t N . We are going to use “inductive construction” to find the monotone subse- quence. (1) Applying (11.2) with N = n 1 gives n 2 > n 1 , such that s n 2 s n 1 and s n 2 > t n 1 ; (2) Suppose n 1 , · · · , n k have been selected such that n 1 < n 2 < · · · < n k , s n 1 s n 2 ≤ · · · ≤ s n k , and s n j +1 > t n j , for all 1 j k - 1 . Applying (11.2) with N = n k gives n k +1 > n k , such that s n k +1 s n k and s n k +1 > t n k . Moreover, s n k 1 sup { s m : m > n k } = v n k , so t n k < s n k +1 v n k . By Mathematical Induction, ( s n k ) is a nondecreasing sequence and lim s n k = lim sup s n . Definition 11.10. Let ( s n ) be a sequence. A subsequential limit is any real number, or + , or -∞ , that is the limit of some subsequence. Example 11.11. If lim s n = s , then the set of subsequential limit is { s } . Example 11.12. Let ( r n ) be the list of rational numbers, then the set of subsequential limits are R ∪ { + , -∞} . Theorem 11.13. Let ( s n ) be a sequence, and S be the set of subsequential limits of ( s n ) . Then (i) S 6 = ; (ii) sup S = lim sup s n , inf S = lim inf s n ; (iii) lim s n exists if and only if S consists of only one element, i.e. S = { s } .
MATH 117 FALL 2017 UCSB 39 Proof. (i) follows from the above corollary since lim sup s n , lim inf s n S , and (iii) follows from (ii).

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