JhalyssaWilliams.Analysis.HW5 - Homework 5 Section 4.4 2a False By theorem 4.4.8 Every unbounded sequence contains a monotone subsequence that has

JhalyssaWilliams.Analysis.HW5 - Homework 5 Section 4.4 2a...

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Homework 5 Section 4.4 2a) False; By theorem 4.4.8, “ Every unbounded sequence contains a monotone subsequence that has either +∞ ?? − ∞ as a limit.” Every sequence does not have a convergent subsequence. 2b) True; By theorem 4.4.7, “ Every bounded sequence has a convergent subsequence .” So by definitions 4.4.9, the subsequential limit must not be empty. 2c) False; By definition 4.4.9, if S n converges to s, then the lim inf S n = lim sup S n . This is true only if s ∈ ℝ 2d) True 2e)False: S n =n for every natural number n. It is not bounded about but the lim Inf S n = 1. 3a) Since S 1 = 0, S 2 =2, S 3 =0, S 4 =2, and S 5 =0, then the subsequential limit is {0,2}. The Lim Sup S n =2 and the Lim Inf S n =0 3b) t n = ( 0, ½, 2/3, ¼, 4/5, 1/6, 6/7, …) The subsequential limit is {0,1},thus the Lim Sup t n =1 and the Lim Inf t n =0 3c) Since u 1 =-2, u 2 =0, u 3 =-18, u 4 =0, u 5 = -50, u 6 =0 The subsequential limit is { −∞ , 0}. The lim Sup u n =0 and the Lim Inf u n = −∞ . 3d) Let V n = 𝑛𝑠𝑖𝑛(𝜋𝑛) 2 . Then V 1 = 1, V 2 =0, V 3 = -3, V 4 =0, V 5 =5, V 6 =0, and V 8 = -8. When n is even, the sequence is 0, for all n ∈ ℕ . When n is odd, the sequence diverges to +∞ ?? − ∞ The subsequential limit is { +∞, 0, −∞ } with +∞ as is the Lim Sup V n and −∞ is the limit inf V n. 4a) Since w 1 = -1, w 2 =1/2, w 3 = -1/3, w 4 = ¼, then w n converges to 0 ∀? ∈ ℕ lim sup w n = lim inf w n = 0 4b) The subsequence can be written as X 3n-2 =0 ∀? , X 3n-1 =1 ∀? , and X 3n converges to +∞ The subsequential limit is {0, 1, +∞} and the Lim Sup X n = +∞ ; Lim Inf X n = 0 4c) The subsequence can be written as Y 2n =6n ∀?, and Y 2n-1 =n. Both subsequences converge to +∞
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  • Fall '08
  • Staff
  • Limits, Limit of a sequence, Limit superior and limit inferior, lim sup, subsequence, Lim Sup Vn

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