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Unformatted text preview: HOMEWORK 4: SOLUTIONS/HINTS 10.1 (d) Let s n := sin( n 7 ), then  s n  1, hence the sequence is bounded. However, it is neither increasing nor decreasing. (f) Let a n := n 3 n . Then a n > 0 for all n , so the sequence is bounded below. Moreover, a n +1 a n = n + 1 3 n = 1 3 + 1 3 n < 1 , therefore the sequence is decreasing and consequently, it is bounded above as well. 10.2 Let s n be a bounded decreasing sequence and set a := inf { s n } . As the sequence is bounded, the infimum a is a real number (and not ). We now want to show that s n converges to a . Let > 0, then a + is not a lower bound of the sequence and hence there is an N such that a + > s N . Since the sequence is decreasing we get that a + > s N s n for all n N . As a is a lower bound, we further have s n a > a and thus a < s n < a + , for all n N and we are done. 10.6 Let > 0 and choose N > 1 ln( ) ln(2) , so we have 2 1 N < . For n m N we then get by the triangle inequality and the formula...
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This note was uploaded on 02/12/2012 for the course MATH 1311a taught by Professor Peter during the Spring '11 term at UCLA.
 Spring '11
 Peter

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