3.2. LIMIT THEOREMS 51Theorem (3.2.10). Suppose lim(x n ) = x and x nlim( p x n ) = p x.Proof. [Using the conjugate].First, x0 by Theorem 3.2.4. Let ✏ > 0 be given.Case x = 0 9 K 2 N 3 8 n K, |x n 0| < ✏ 2 ()0 apple x n < ✏ 2 () 0 apple p x n < ✏ () |x n 0| < ✏.0 8 n 2 N. ThenCase x > 0 Then p x > 0. 9 K 2 N 3 8 n K, |x n x| < p x✏ =)| p x np x| =( p x np x) · (pxn + p x)pxn + p x=|x n x|pxn + p x apple |x n x|p < xp x✏p x= ✏.HomeworkPages 69-70 #1d, 5b, 6bd (both find and prove)⇣ xn⌘Extra: If lim = x 6= 0, then (x n ) is not bounded.n⇣ xn⌘ ⌘bounded =) lim = 0 .n⇤⇣Hint: Prove (x n )
52 3. SEQUENCES AND SERIES3.3. Monotone SequencesDefinition (3.3.1). Let X = (x n ) be a sequence.X is increasing if x 1 apple x 2 apple · · · apple x n apple x n+1 apple · · · .X is decreasing if x 1 x 2 · · · x n x n+1 · · · .X is monotone if it is either increasing or decreasing.Example.(1) (1, 1, 2, 3, 5, 8, . . . ) is increasing.(2) For 0 < b < 1, (b, b 2 , b 3 , . . . ) is decreasing.(3) (2, 0, 2, 0, . . . ) is not monotone.(4) (4, 2, 1, 3, 3, 5, 5, 7, 7, . . . ) is ultimately increasing.Theorem (3.3.2 — Monotone Convergence Theorem (MCT)). A monotonesequence converges () it is bounded. Further:(a) if X = (x n ) is a bounded, increasing sequence, then lim(x n ) = sup{x n }.(b) if X = (x n ) is a bounded, decreasing sequence, then lim(x n ) = inf{x n }.Proof. (=)) Follows from Theorem 3.2.2.((=) (a) Since (x n ) is bounded,9 M > 0 3 |x n | apple M 8 n 2 N =) x n apple M 8 n 2 N.Thus, by Completeness, x ? = sup{x n } exists. Let ✏ > 0 be given.By Property S, 9 K 2 N 3 x ? ✏ < x K (x K = s ✏ ).Since (x n ) is increasing, x K apple x n 8 n K. Thus, 8 n K,| {z } ✏ < x K apple |{z}x n apple x ? < x | ? {z + } ✏ =) |x n x ? | < ✏.Thus lim(x n ) = x ? .x ?(b) Similar to (a), except uses Property I.⇤