real_analysisii_homework4

# real_analysisii_homework4 - Homework 4 Hctor Guillermo...

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Homework 4 ector Guillermo Cu´ ellar R´ ıos February 21, 2006 6.2 (a) If s n 0, then for every ε > 0 there exist N R such that n > N implies s n < ε . True. According to Definition 16.2, if s n 0, then fore each ε > 0 there exists a real number N such that for all n N , n > N implies that | s n - 0 | < ε , then | s n | < ε , clearly s n < ε . (b) If for every ε > 0 there exists N R such that n > N impies s n < ε , then s n 0. False, the sequence s n = - n is always less than ε and it does not converge to any number. (c) Given sequences ( s n ) and ( a n ), if for some s R , k > 0 and m N we have | s n - s | ≤ k | a n | for all n > m , then lim s n = s . False. Consider s n = ( - 1) n , a n = n k and s = 0, none of the sequences converge, furthermore | s n - 0 | ≤ 1 k n k = n for all n . (d) If s n t and s n t , then s = t . True by Theorem 16.14. 16.6 . Using only Definition 16.2, prove the following. (a) For an real number k , lim n →∞ ( k n ) = 0. Choose N = | k | ε . Then by Definition 16.2 k n - 0 = k n = | k | n < | k | N = ε (b) For any real number k > 0, lim n →∞ ( 1 n k ) = 0 Let N = 1 ε 1 /k , then 1 n k < 1 N

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