k For any given sequence s n s n has a convergent subsequence s n k Sometimes

K for any given sequence s n s n has a convergent

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(k) For any given sequence ( s n ), ( s n ) has a convergent subsequence ( s n k ). Sometimes true: The sequence s n = 1 , n odd 1 /n, n even has a con- vergent subsequence s 2 n 0. The sequence s n = n has no convergent subsequences. Similarly, the sequence s n = n 2 has no convergent subsequences. 4
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(l) For any bounded sequence ( s n ), lim sup s n = sup { s n } . Sometimes true: See the sequences in (j). (m) It α is a subsequential limit of a bounded sequence ( s n ), then α lim sup s n Always true: By definition: lim sup s n is the least upper bound of the set of all subsequential limits. (n) If every subsequence of a sequence ( s n ) is convergent, then ( s n ) itself must be convergent. Always true: If every subsequence of ( s n ) is convergent, then ( s n ) must be convergent since ( s n ) is a subsequence of itself. 5
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(o) If ( s n ) is a divergent sequence, then some subsequence of ( s n ) must diverge. Always true: If every subsequence of ( s n ) is convergent, then ( s n ) is convergent as shown immediately above. (p) If ( s n ) is unbounded above, then ( s n ) has an increasing subsequence ( s n k ) which diverges to + . Always true: This is Theorem 3, Section 19. 6
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2. Let ( s n ) be a positive sequence such that lim n →∞ s n +1 s n = L > 1 . Prove that s n + . Hint: lim n →∞ x n = + for any number x such that x > 1. Choose a number c such that 1 < c < L . Let ϵ = L c . Since lim n →∞ s n +1 s n = L There is a positive integer N such that s n +1 s n L < ϵ for all n > N which implies ϵ < s n +1 s n L < ϵ and L ϵ < s n +1 s n < L + ϵ for all n > N.
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  • Fall '08
  • Staff
  • Mathematical analysis, Limit of a sequence, Limit superior and limit inferior, Sn, subsequence

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