# 5 - Math 310 - hw 5 solutions Wednesday, 14 Oct 2009 10.2,...

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Math 310 - hw 5 solutions 10.2, 10.6a, 10.10; Wednesday, 14 Oct 2009 10.2 Prove Theorem 10.2 for bounded nonincreasing sequences. Suppose that ( s n ) n is a bounded nonincreasing sequence and set s := inf { s n : n N } (Note that s R ). We will prove that s n s . Let ε > 0. Since s + ε is not an lower bound, there is N N so that s N < s + ε . Since ( s n ) n is a nonincreasing sequence, for all n N , n N implies s n s N . Let n N and suppose n N , then since s s n we have s - ε < s s n s N < s + ε ; hence, | s n - s | < ε and so s n s . ± 10.6a Let ( s n ) n be a sequence such that | s n +1 - s n | < 2 - n for all n N . Prove that ( s n ) n is a Cauchy sequence and hence a convergent sequence. We will need the following straightforward fact: For a (0 , 1) we have 1 + a + ··· + a k = 1 - a k +1 1 - a < 1 1 - a . Let ε > 0. By Example 9.7(a), 2 - n 0 (note that for all n N , 0 < 2 - n < 1). Hence, there is N 0 such that for
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## This note was uploaded on 03/26/2010 for the course MATHEMARIC 131a taught by Professor Janeday during the Spring '10 term at San Jose State.

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