Exam1_070207_Sol

# 7 if for every ε 0 there exists n r such that for

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(7). If for every ε > 0 there exists N R such that for every n > N we have s n < ε , then lim n →∞ s n = 0. False. Example: Let s n = - 1 for all n N , then for every ε > 0 there exists N = 0 R such that for every n > N we have s n < ε but lim n →∞ s n = - 1 = 0. (8). If ( s n ) and ( t n ) converge to s and t , respectively, and s n < t n for every n N , then s < t . False. Example: Let s n = 0 and t n = 1 n for n N , then s n < t n but s = t = 0. 1

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(9). A sequence is bounded and monotone if and only if it is a Cauchy sequence. False. Example: Let s n = ( - 1) n n for n N , which is a Cauchy sequence (since it converges to 0) but it is not monotone. (10). Every bounded sequence has a convergent subsequence. True. By Theorem 19.7. 2. (10 points) Let S be a bounded infinite set and let s = sup S . Prove: If s / S , then s S . Proof By the definition of s , ε > 0 a S such that s a > s - ε , which implies a N ( s, ε ). Since s / S , we have a = s and a N * ( s, ε ) S . By the definition of S , we have s S . 3. (14 points) (1). Prove the following statement by using only the definition: lim n →∞ sin( π 2 n ) n = 0 . Proof Since ε > 0, N = 1 ε R , such that n > N , we have sin( π 2 n ) n - 0 1 n < ε.
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