5333-Solns-Assignment5-Sp13 - Solutions Assignment#5 Section 19 3(a sn =(1)n(sn =(1 1 1 1 1 1 Subsequential limits S = cfw_1 1 lim sup sn = 1 lim inf sn

# 5333-Solns-Assignment5-Sp13 - Solutions Assignment#5...

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Unformatted text preview: Solutions, Assignment #5 Section 19 3. (a) s n = (- 1) n ; ( s n ) = (- 1 , 1 ,- 1 , 1 ,- 1 , 1 , ··· ). Subsequential limits S = {- 1 , 1 } . lim sup s n = 1, lim inf s n =- 1. (b) t n → 0; S = { } ; lim sup t n = lim inf t n = 0. (c) ( u n ) = (- 1 , ,- 9 , ,- 25 , , ··· ); S = {-∞ , } ; lim sup u n = 0, lim inf u n =-∞ . (d) ( v n ) = (1 , ,- 3 , , 5 , ,- 7 , , ··· ); S = {-∞ , , ∞} ; lim sup v n = ∞ , lim inf v n =-∞ . 4. (a) w n = (- 1) n n ; ( w n ) = parenleftbigg- 1 , 1 2 ,- 1 3 , 1 4 , ··· parenrightbigg ; w n → 0. Therefore, S = { } , lim sup w n = lim inf w n = 0. (b) S = { , 1 , + ∞} ; lim sup x n = + ∞ , lim inf x n = 0. (c) ( y n ) diverges to + ∞ ; lim sup y n = lim inf y n = + ∞ . (d) ( z n ) = (- 1 , 4 ,- 27 , 256 , ··· ); S = {-∞ , ∞} , lim sup z n = + ∞ , lim inf z n =-∞ . 6. (a) True: For example, there is a subsequence which converges to lim sup s n . (b) True: There is a subsequence ( s n k ) which converges to α = lim sup s n and a subsequence ( s n m ) which converges to α = lim inf s n . (c) False: ( s n ) = (1 , 2 , 3 , 4 , 5 , ··· ) diverges and does not oscillate. 7. (a) True: Every bounded sequence has a convergent subsequence; a convergent sequence is a Cauchy sequence. (b) False: The sequence in 6(c) is a counterexample. 11. ( s n ) is a bounded sequence and S is its set of subsequential limits. Let q be an accumulation point of S . To show that S is closed, we must show that q ∈ S . That is, we must show that q is a subsequential limit of ( s n ). Let epsilon1 = 1. There exists an element p 1 ∈ S such that p 1 ∈ N ( q, 1). Since N ( q, 1) is open, there is a neighborhood N ( p 1 , δ ) such that N ( p 1 , δ ) ⊆ N ( q, 1). Since p 1 ∈ S , there is term s n 1 ∈ { s n } such that s n 1 ∈ N ( p 1 , δ ). It now follows that s n 1 ∈ N ( q, 1). Now let epsilon1 = 1 / 2. There exists an element p 2 ∈ S such that p 2 ∈ N ( q, 1 / 2). Since N ( q, 1 / 2) is open, there is a neighborhood N ( p 2 , δ ) such that N ( p 2 , δ ) ⊆ N ( q, 1 / 2). Since p 2 ∈ S , the neighborhood N ( p 2 , δ ) contains infinitely many terms of the sequence (...
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• Fall '08
• Staff
• Limits, lim, Limit of a function, Limit of a sequence, lim sup, subsequence
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