M413notes10 - 10 Monotone sequences and Cauchy sequences...

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Unformatted text preview: 10 Monotone sequences and Cauchy sequences This section introduces two important ideas that are useful for proving that sequences converge. Definition. a. A sequence ( s n ) is called nondecreasing if s n s n +1 for all n . b. A sequence ( s n ) is called increasing if s n < s n +1 for all n . c. A sequence ( s n ) is called nonincreasing if s n s n +1 for all n . d. A sequence ( s n ) is called decreasing if s n > s n +1 for all n . e. If a sequence obeys any of these, it is called monotone . Some writers use increasing for (a) and strictly increasing for (b); also decreasing for (c) and strictly decreasing for (d). Example. The sequence s n = 2 n n + 1 is increasing. We can verify that s n < s n +1 . This would be the inequality 2 n n + 1 < 2( n + 1) ( n + 1) + 1 or 2 n n + 1 < 2 n + 2 n + 2 . Heres how to verify this inequality: < 2 2 n 2 + 4 n < 2 n 2 + 4 n + 2 2 n ( n + 2) < (2 n + 2)( n + 1) 2 n n + 1 < 2 n + 2 n + 2 Here is the key theorem about monotone sequences: Theorem. Suppose ( s n ) is bounded above. (So there is a number M such that s n < M for all n .) If ( s n ) is nondecreasing, then it is convergent. Proof. This theorem is a consequence of the Completeness Axiom. Consider the set S = { s n : n N } . (Recall, for example, if x n = (- 1) n that the set { x n : n N } is {- 1 , 1 } but the sequence itself is (- 1 , 1 ,- 1 , 1 ,... ).) Now the set S is bounded above: there is a number M such that s n < M for all n . So by the Completeness Axiom, the supremum of the set S exists. Let s = sup S . We claim that lim s n = s . Let > 0. We seek N such that if n > N then | s n- s | < . Now since s = sup S , there is a number s N in S such that s- < s N s . (Otherwise, the number s- would be an upper bound for S .) So now let....
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M413notes10 - 10 Monotone sequences and Cauchy sequences...

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