Homework SourseUse Lagrange multipliers to find the point on the plane x 2
Solution
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Monotone sequences have particularly simple convergence properties. Definition 3.27. A sequence (xn) of real numbers is increasing if xn+1 xn for all n N, decreasing if xn+1 xn for all n N, and monotone if it is increasing or decreasing. A sequence is strictly increasing if xn+1 > xn, strictly decreasing if xn+1 < xn, and strictly monotone if it is strictly increasing or strictly decreasing. We don’t require a monotone sequence to be strictly monotone, but this usage isn’t universal. In some places, “increasing” or “decreasing” is used to mean “strictly increasing” or “strictly decreasing.” In that case, what we call an increasing sequence is called a nondecreasing sequence and a decreasing sequence is called nonincreasing sequence. We’ll use the more easily understood direct terminology. Example 3.28. The sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, . . . is monotone increasing but not strictly monotone increasing; the sequence (n 3 ) is strictly monotone increasing; the sequence (1/n) is strictly monotone decreasing; and the sequence ((1)n+1) is not monotone. Bounded monotone sequences always converge, and unbounded monotone sequences diverge to ±. Theorem 3.29. A monotone sequence of real numbers converges if and only if it is bounded. If (xn) is monotone increasing and bounded, then limn xn = sup{xn : n N}, 46 3. Sequences and if (xn) is monotone decreasing and bounded, then limn xn = inf{xn : n N}. Furthermore, if (xn) is monotone increasing and unbounded, then limn xn = , and if (xn) is monotone decreasing and unbounded, then limn xn = . Proof. If the sequence converges, then by Proposition 3.19 it is bounded. Conversely, suppose that (xn) is a bounded, monotone increasing sequence. The set of terms {xn : n N} is bounded from above, so by Axiom 2.17 it has a supremum x = sup{xn : n N}. Let > 0. From the definition of the supremum, there exists an N N such that xN > x . Since the sequence is increasing, we have xn xN for all n > N, and therefore x < xn x. It follows that |xn x| < for all n > N, which proves that xn x as n . If (xn) is an unbounded monotone increasing sequence, then it is not bounded from above, since it is bounded from below by its first term x1. Hence, for every M R there exists N N such that xN > M. Since the sequence is increasing, we have xn xN > M for all n > N, which proves that xn as n . The result for a monotone decreasing sequence (xn) follows similarly, or by applying the previous result to the monotone increasing sequence (xn). The fact that every bounded monotone sequence has a limit is another way to express the completeness of R. For example, this is not true in Q: an increasing sequence of rational numbers that converges to 2 is bounded from above in Q (for example, by 2) but has no limit in Q. We sometimes use the notation xn x to indicate that (xn) is a monotone increasing sequence that converges to x, and xn x to indicate that (xn) is a monotone decreasing sequence that converges to x, with a similar notation for monotone sequences that diverge to ±. For example, 1/n 0 and n 3 as n . The following propositions give some examples of monotone sequences. In the proofs, we use the binomial theorem, which we state without proof. Theorem 3.30 (Binomial). If x, y R and n N, then (x + y) n = Xn k=0 n k x nk y k , n k = n! k!(n k)!. Here, n! = 1 · 2 · 3 · · · · · n and, by convention, 0! = 1. The binomial coefficients n k = n · (n 1) · (n 2)· · · · (n k + 1) 1 · 2 · 3 · · · · k , 3.5. Monotone sequences 47 read “n choose k,” give the number of ways of choosing k objects from n objects, order not counting. For example, (x + y) 2 = x 2 + 2xy + y 2 , (x + y) 3 = x 3 + 3x 2 y + 3xy2 + y 3 , (x + y) 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy3 + y 4 . We also recall the sum of a geometric series: if a 6= 1, then Xn k=0 a k = 1 a n+1 1 a . Proposition 3.31. The geometric sequence (a n) n=0, 1, a, a2 , a3 , . . . , is strictly monotone decreasing if 0 < a < 1, with limn a n = 0, and strictly monotone increasing if 1 < a < , with limn a n = . Proof. If 0 < a < 1, then 0 < an+1 = a · a n < an, so the sequence (a n) is strictly monotone decreasing and bounded from below by zero. Therefore by Theorem 3.29 it has a limit x R. Theorem 3.26 implies that x = limn a n+1 = limn a · a n = a limn a n = ax. Since a 6= 1, it follows that x = 0. If a > 1, then a n+1 = a · a n > an, so (a n) is strictly increasing. Let a = 1 + where > 0. By the binomial theorem, we have a n = (1 + ) n = Xn k=0 n k k = 1 + n + 1 2 n(n 1) 2 + · · · + n > 1 + n. Given M 0, choose N N such that N > M/. Then for all n > N, we have a n > 1 + n > 1 + N > M, so an as n . The next proposition proves the existence of the limit for e in Example 3.16. Proposition 3.32. The sequence (xn) with xn = 1 + 1 n n is strictly monotone increasing and converges to a limit 2 < e < 3. 48 3. Sequences Proof. By the binomial theorem, 1 + 1 n n = Xn k=0 n k 1 nk = 1 + n · 1 n + n(n 1) 2! · 1 n2 + n(n 1)(n 2) 3! · 1 n3 + · · · + n(n 1)(n 2). . . 2 · 1 n! · 1 nn = 2 + 1 2! 1 1 n + 1 3! 1 1 n 1 2 n + · · · + 1 n! 1 1 n 1 2 n . . . 2 n · 1 n . Each of the terms in the sum on the right hand side is a positive increasing function of n, and the number of terms increases with n. Therefore (xn) is a strictly increasing sequence, and xn > 2 for every n 2. Moreover, since 0 (1k/n) < 1 for 1 k n, we have 1 + 1 n n < 2 + 1 2! + 1 3! + · · · + 1 n! . Since n! 2 n1 for n 1, it follows that 1 + 1 n n < 2 + 1 2 + 1 2 2 + · · · + 1 2 n1 = 2 + 1 2 1 (1/2)n1 1 1/2 < 3, so (xn) is monotone increasing and bounded from above by a number strictly less than 3. By Theorem 3.29 the sequence converges to a limit 2 < e < 3.
