131a4 - Problems graded: 10.1 (15 points), 10.7 (10...

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Unformatted text preview: Problems graded: 10.1 (15 points), 10.7 (10 points), 11.1 (10 points). The remaining problems were graded based on completion to be worth five points each, except for the long problem 11.3 which was worth ten. To rescale this problem set to be worth 100 points (instead of 65), we multiplied the number of points missed by 1.5 and rounded the number of missed points DOWN to the nearest integer (note that this effectively gives everyone 2.5 to 3 points for free). 10.1. (b) In { ( 1) n /n 2 } , the first term, 1, is less than the second term, 1 / 4, so the sequence is NOT nonincreasing. As the second term is greater than the third term, 1 / 9, the sequence is NOT nondecreasing. However, as | 1 n /n 2 | = | n- 2 | 1 for all n , the sequence is bounded. (d) In { sin( n/ 7) } , the first term, sin( / 7), is positive, although the seventh term, sin(7 / 7), is zero, so the sequence is NOT nondecreasing. The second term, sin(2 / 7), is greater than the first term (as the sine function is strictly increasing on [0 ,/ 2]), so the sequence is NOT nonincreasing. However, as | sin( x ) | 1 for all x , the sequence is bounded. (f) In { n/ 3 n } , one shows the sequence is nonincreasing by noting that if s n is the n th term and s n +1 is the ( n + 1)th term, both are positive but s n +1 s n = n + 1 n 1 3 = (1 + n- 1 ) 3 2 3 < 1 . In other words, each term is strictly positive but at most two-thirds of the preceding term. Clearly the sequence is NOT nondecreasing as the first term, 1 / 3, is greater than the second term, 2 / 9. Finally, the sequence is bounded from below by zero (all terms are positive) and from above (because the sequence is decreasing, an obvious induction shows each term is bounded above by 1 / 3) so it is bounded. 10.2. (NOTE: This proof was made as similar to the books style as possible; a slicker proof can be found by noting { s n } is bounded nondecreasing.) Let ( s n ) be a bounded nonincreasing sequence. Let S denote the set { s n : n N } and let v = inf S . Since S is bounded, v represents a real numbeer. We show that lim s n = u . Let > 0. Since v + is not an upper bound for S , there exists N such that s N < v + . Since ( s n ) is nonincreasing, we have s N s n for all n > N . Of course, s n v for all n and so n > N implies v + > s n v , which implies | s n v | < . This shows that lim s n = v ....
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131a4 - Problems graded: 10.1 (15 points), 10.7 (10...

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