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Unformatted text preview: Homework 3 Solutions 1. Prove the following using only the definition of convergence or divergence: (a) (8.2(e)) sin( n ) n → 0. Solution: Note that  sin( n )  ≤ 1 for all n . Let > 0 be given. Choose N = 1 / . Then for n > N , sin( n ) n ≤ 1 n < . (b) 4 n 2 3 5 n 2 2 n → 4 5 . Solution: Let > 0 be given. Choose N = 8 3 . Then for n > N : 4 n 2 3 5 n 2 2 n 4 5 = 8 n 15 5(5 n 2 2 n ) ≤ 8 n 3 n 2 = 8 3 n < . (c) x n = 1 + ( 1) n does not converge to any x ∈ R . Solution: We need to prove: ∃ > 0 so that ∀ N , ∃ n > N so that  x n x  > . When n is even, x n = 2, and when n is odd, x n = 0. Case 1: x ≤ 1. Then for = 1 2 , given any N , choose n to be the next largest even number. Then x n = 2, so  x n x  > . Case 2: x ≥ 1. Then for = 1 2 , given any N , choose n to be the next largest odd number. Then x n = 0, so  x n x  > . (d) y n = n ! diverges to ∞ . Solution: We need to prove: ∀ M > 0, ∃ N so that ∀ n > N , y n > M . Given M > 0, choose N = M . Then for n > N = M , y n = n ! > n > M . 2. (8.4) Let ( t n ) be a bounded sequence and let ( s n ) be a sequence that converges to 0. Prove that s n t n → 0 using only the definition of convergence (no theorems)....
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This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.
 Winter '11
 Wiley
 Differential Equations, Equations

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