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**Unformatted text preview: **Therefore, S = N ; the equation holds for all positive integers n . (c) s n → 3 2. The sequence ( s n ) is defined by the recursion formula: s n +1 = 1 4 (2 s n + 5) , s 1 = 1. (a) Determine a ”formula” for ( s n ). s 2 = 7 4 , s 3 = 17 8 , s 4 = 37 16 , s 5 = 77 32 , ··· Conjecture: s n = 5 2 n- 1- 3 2 n (b) Let S be the set of positive integers for which the equation holds. 1 ∈ S ? s 1 = 5 2- 3 2 1 = 2 2 = 1. Therefore, 1 ∈ S . Assume that the positive integer k ∈ S . That is, assume s k = 5 2 k- 1- 3 2 k . Prove that k + 1 ∈ S : s k +1 = 1 4 (2 s k + 5) = 1 4 parenleftbigg 2[5 2 k- 1- 3] 2 k + 5 parenrightbigg = 1 4 parenleftbigg 5 2 k- 6 + 5 2 k 2 k parenrightbigg = 5 2 k +1- 6 2 k +2 = 5 2 k- 3 2 k +1 Therefore, k + 1 ∈ S . Therefore, S = N ; the equation holds for all positive integers n . (c) s n → 5 2 3. Prove that lim n →∞ 5 n + 4 3 n- 1 = 5 3 . An epsilon1- N argument: Let epsilon1 > 0. vextendsingle vextendsingle vextendsingle vextendsingle 5 n + 4 3 n- 1- 5 3 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 15 n + 12- (15 n- 5) 3(3 n- 1) vextendsingle vextendsingle vextendsingle vextendsingle = 17 3 1 3 n- 1 = 17 9 1 n- 1 3 . Since 1 n- 1 3 → 0, there is a positive integer N such that 1 N- 1 3 < 9 17 epsilon1 . Therefore, for all n > N , vextendsingle vextendsingle vextendsingle vextendsingle 5 n + 4 3 n- 1- 5 3 vextendsingle vextendsingle vextendsingle vextendsingle = 17 9 1 n- 1 3 < 17 9 1 N- 1 3 < 17 9 9 17 epsilon1 = epsilon1. Therefore lim n →∞ 5 n + 4 3 n- 1 = 5 3 . A proof using Theorem 8, Section 16. vextendsingle vextendsingle vextendsingle vextendsingle 5 n + 4 3 n- 1- 5 3 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 15 n + 12- (15 n- 5) 3(3 n- 1) vextendsingle vextendsingle vextendsingle vextendsingle = 17 3 1 3 n- 1 = 17 9 1 n- 1 3 . 4 Since 1 n- 1 3 → 0, it follows that (by Theorem 8) lim n →∞ 5 n + 4 3 n- 1 = 5 3 . 4. Prove that lim n →∞ radicalbig n 2 + 1- n = 0. A proof using Theorem 8, Section 16. vextendsingle vextendsingle vextendsingle radicalbig n 2 + 1- n- vextendsingle vextendsingle vextendsingle = radicalbig n 2 + 1- n = parenleftBig radicalbig n 2 + 1- n parenrightBig √ n 2 + 1 + n √ n 2 + 1 + n = n 2 + 1- n 2 √ n 2 + 1 + n = 1 √ n 2 + 1 + n < 1 2 n < 1 n . Since 1 n → 0, it follows that (by Theorem 8) lim n →∞ radicalbig n 2 + 1- n = 0. 5...

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- Math, Topology, 1 k, Metric space, 1 k, 1 5 2k