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Download SEQUENCES, CONTINUED Definition 3.13. A sequence {sn} of real
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SEQUENCES, CONTINUED De…nition 3.13. A sequence fsn g of real numbers is said to be 1. increasing if sn+1 sn for n 1; 2. decreasing if sn+1 sn for n 1. If fsn g is increasing or decreasing, we say that fsn g is monotonic. Theorem 3.14. Let fsn g be a monotonic sequence. Then fsn g converges if and only if fsn g is bounded. Proof. See p. 55 of Rudin. We shall just prove the following statement: If fsn g is increasing and bounded, then fsn g converges to sup E, where E = fsn : n 2 Ng is the range of fsn g. Let fsn g be an increasing and bounded sequence. Since E is bounded and nonempty, by the least upper bound property we have that sup E 2 R exists. Since sup E is an upper bound for E, we have sn sup E for all n 1. Since sup E is the least upper bound for E, for any " > 0, sup E " is not an upper bound. Hence there exists N 1 such that sN > sup E ". Since fsn g is increasing, this implies sn > sup E " for n N . So we have sup E " < sn sup E for n N: This implies jsn sup Ej < " for n N . We have proved that fsn g converges to sup E. De…nition 3.15. We say that fsn g ! +1 (also written limn!1 sn = +1) if for each M 2 R there exists N 2 N such that sn M for n N . Similarly, we say that fsn g ! 1 (also written limn!1 sn = 1) if for each M 2 R there exists N 2 N such that sn M for n N . If fsn g converges, fsn g ! +1, or fsn g ! 1, then we say that fsn g converges to an extended real number. Remark. The set of extended real numbers is de…ned to be R = R [ f+1; 1g. The usual order on R is extended to R by de…ning 1 < x and x < +1 for all x 2 R and 1 < +1. Lemma. Let fsn g be a sequence of real numbers. Let S R be the set of extended real number subsequential limits of fsn g. Then S is nonempty. Proof. Case 1. fsn g is bounded. Then Theorem 3.6(b) implies that there exists a convergent subsequence (converging to a real number). Case 2. fsn g is unbounded. Then there a subsequence which converges either to +1 or to 1. (Exercise: Prove this.) Again, let S R denote the set of extended real number subsequential limits of a sequence fsn g. De…ne s = sup S; s = inf S to be the upper limit and lower limit of fsn g, respectively. 1 We use the notation: s = lim sup sn ; n!1 s = lim inf sn : n!1 Recall that S R is the set of extended real number subsequential limits of fsn g. Theorem 3.17. (a) s 2 S and s 2 S. (b) For each x > s there exists N 2 N such that sn < x for n N . For each x < s there exists N 2 N such that sn > x for n N . Moreover, s and s are the only numbers with properties (a) and (b). Proof. See p. 56 of Rudin. Example 3.18(a). Let fsn g be a sequence of real numbers whose range E contains all rational numbers. In other words, for each q 2 Q there exists n 1 such that sn = q. Exercise: (1) Prove that for each x 2 R, there exists a subsequence of fsn g converging to x. Solution. Let x 2 R. Choose q1 2 Q such that q1 2 (x; x + 1). Since fsn g contains all rational numbers, there exists n1 2 N such that q1 = sn1 . Now assume that we have chosen n1 ; : : : ; nk 1 2 N. Choose qk 2 Q such that qk 2 (x; x + k1 ). Since fsn g contains all rational numbers and since the interval (x; x + k1 ) contains an in…nite number of rationals, there exists nk > nk 1 such that qk = snk . Then n1 < n2 < and jsnk xj < k1 for each k 1. Hence the subsequence fsnk g converges to x. (2) Prove that lim supn!1 sn = +1 and (similarly) lim inf n!1 sn = 1. Theorem 3.20(c). p lim n n = 1: n!1 Proof. See p. 58 of Rudin. Let e be Euler’s number. Theorem 3.31. lim n!1 1+ Proof. See p. 64 of Rudin. P Here e is de…ned by the series sum e = 1 n=0 1 n 1 ; n! 2 n = e: (in…nite) series is a topic we consider next.