This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View Full
Document
 Spring '09
 WEISBART

Click to edit the document details