Chapter Three SEQUENCES Oleh: Mardiyana Mathematics Education Sebelas Maret University

Definition 3.1.1 A sequence of real numbers is a function on the set N of natural numbers whose range is contained in the set R of real numbers. X : N  R n  xn We will denote this sequence by the notations: X or (xn) or (xn : n  N)

Definition 3.1.3 If X = (xn) and Y = (yn) are sequences of real numbers, then we define X + Y = (xn + yn) X – Y = (xn - yn) X.Y = (xnyn) cX = (cxn) X/Y = (xn/yn), if yn  0 for all n  N.

Definition 3.1.4 Let X = (xn) be a sequence of real numbers. A real number x is said to be a limit of (xn) if for every  > 0 there exists a natural number K() such that for all n  K(), the terms xn belong to the -neighborhood V(x) = (x - , x + ). If X has a limit, then we say that the sequence is convergent, if it has no limit, we say that the sequence is divergent. We will use the notation lim X = x or lim (xn) = x

Theorem 3.1. 5 A sequence of real numbers can have at most one limit Proof: Suppose, on the contrary, that x and y are both limits of X and that x  y. We choose  > 0 such that the -neighborhoods V(x) and V(y) are disjoint, that is,  < ½ | x – y|. Now let K and L be natural numbers such that if n > K then xn  V(x) and if n > L then xn  V(y). However, this contradicts the assumption that these -neighborhoods are disjoint. Consequently, we must have x = y.

Theorem 3.1.6 Let X = (xn) be a sequence of real numbers, and let x  R. The following statements are equivalent: X convergent to x. For every -neighborhood V(x), there is a natural number K() such that for all n  K() the terms xn belong to V(x). For every  > 0, there is a natural number K() such that for all n  K(), the terms xn satisfy |xn – x | <  For every  > 0, there is a natural number K() such that for all n  K(), the terms xn satisfy x -  < xn < x + .

Xm := (xm+n : n  N) = (xm+1, xm+2, …) Tails of Sequences Definition 3.1.8 If X = (x1, x2, …, xn, …) is a sequence of real numbers and if m is a given natural number, then the m-tail of X is the sequence Xm := (xm+n : n  N) = (xm+1, xm+2, …) Example: The 3-tail of the sequence X = (2, 4, 6, 8, …, 2n, …) is the sequence X3 = (8, 10, 12, …, 2n + 6, …).

In this case lim Xm = lim X. Theorem 3.1.9 If X = (xn) be a sequence of real numbers and let m  N. Then the m-tail Xm = (xm+n) of X converges if and only if X converges. In this case lim Xm = lim X.

Proof: We note that for any p  N, the pth term of Xm is the (p + m)th term of X. Similarly, if q > m, then the qth term of X is the (q – m)th term of Xm. Assume X converges to x. Then given any  > 0, if the terms of X for n  K() satisfy |xn – x| < , then the terms of Xm for k  K() – m satisfy |xk – x| < . Thus we can take Km() = K() – m, so that Xm also converges to x. Conversely, if the terms of Xm for k  Km() satisfy |xk – x| < , then the terms of X for n  Km() + m satisfy |xn – x| < . Thus we can take K() = Km() + m. Therefore, X converges to x if and only if Xm converges to x.

Theorem 3.1.10 Let A = (an) and X = (xn) be sequences of real numbers and let x  R. If for some C > 0 and some m  N we have |xn – x|  C|an| for all n  N such that n  m, and if lim (an) = 0 then it follows that lim (xn) = x.

|xn – x|  C|an| < C (/C) = . Proof: If  > 0 is given, then since lim (an) = 0, it follows that there exists a natural number KA(/C) such that if n  KA(/C) then |an| = |an – 0| < /C. Therefore it follows that if both n  KA(/C) and n  m, then |xn – x|  C|an| < C (/C) = . Since  > 0 is arbitrary, we conclude that x = lim (xn).

Examples Show that if a > 0, then lim (1/(1 + na)) = 0. Show that lim (1/2n) = 0. Show that if 0 < b < 1, then lim (bn) = 0. Show that if c > 0, then lim (c1/n) = 1. Show that lim (n1/n) = 1.

Limit Theorems Definition 3.2.1 A sequence X = (xn) of real numbers is said to be bounded if there exists a real number M > 0 such that |xn|  M for all n  N. Thus, a sequence X = (xn) is bounded if and only if the set {xn : n  N} of its values is bounded in R.

Theorem 3.2.2 A convergent sequence of real numbers is bounded Proof: Suppose that lim (xn) = x and let  := 1. By Theorem 3.1.6, there is a natural number K := K(1) such that if n  K then |xn – x| < 1. Hence, by the Triangle Inequality, we infer that if n  K, then |xn| < |x| + 1. If we set M := sup {|x1|, |x2|, …, |xK-1|, |x| + 1}, then it follows that |xn|  M, for all n  N.

Theorem 3.2.3 (a). Let X and Y be sequences of real numbers that converge to x and y, respectively, and let c  R. Then the sequences X + Y, X – Y, X.Y and cX converge to x + y, x – y, xy and cx, respectively. (b). If X converges to x and Z is a sequence of nonzero real numbers that converges to z and if z  0, then the quotient sequence X/Z converges to x/z.

3.3 Monotone Sequences Definition Let X = (xn) be a sequence of real numbers. We say that X is increasing if it satisfies the inequalities x1  x2  …  xn-1  xn  … We say that X is decreasing if it satisfies the inequalities x1  x2  …  xn-1  xn  …

Monotone Convergence Theorem A monotone sequence of real numbers is convergent if and only if it is bounded. Further a). If X = (xn) is a bounded increasing sequence, then lim (xn) = sup {xn}. b). If X = (xn) is a bounded decreasing sequence, then lim (xn) = inf {xn}.

Proof:

Example Let for each n  N. a). Prove that X = (xn) is an increasing sequence. b). Prove that X = (xn) is a bounded sequence. c). Is X = (xn) convergent? Explain!

Exercise Let x1 > 1 and xn+1 = 2 – 1/xn for n  2. Show that (xn) is bounded and monotone. Find the limit.

3.4. Subsequences and The Bolzano-Weierstrass Theorem Definition Let X = (xn) be a sequence of real numbers and let r1 < r2 < … < rn < … be a strictly increasing sequence of natural numbers. Then the sequence Y in R given by is called a subsequence of X.

Theorem If a sequence X = (xn) of real numbers converges to a real number x, then any sequence of subsequence of X also converges to x. Proof: Let  > 0 be given and let K() be such that if n  K(), then |xn – x| < . Since r1 < r2 < … < rn < … is a strictly increasing sequence of natural numbers, it is easily proved that rn  n. Hence, if n  K() we also have rn  n  K(). Therefore the subsequence X’ also converges to x.

Divergence Criterion Let X = (xn) be a sequence of real numbers. Then the following statements are equivalent: (i). The sequence X = (xn) does not converge to x  R. (ii). There exists an 0 > 0 such that for any k  N, there exists rk  N such that rk  k and |xrk – x|  0. (iii). There exists an 0 > 0 and a subsequence X’ of X such that |xrk – x|  0 for all k  N.

Monotone Subsequence Theorem If X = (xn) is a sequence of real numbers, then there is a subsequence of X that is monotone.

The Bolzano-Weierstrass Theorem A bounded sequence of real numbers has a convergent subsequence.

3.5 The Cauchy Criterion Definition A sequence X = (xn) of real numbers is said to be a Cauchy Sequence if for every  > 0 there is a natural number H() such that for all natural numbers n, m  H(), the terms xn, xm satisfy |xn – xm| < .

Lemma If X = (xn) is a convergent sequence of real numbers, then X is a Cauchy sequence. Proof: If x := lim X, then given  > 0 there is a natural number K(/2) such that if n  K(/2) then |xn – x| < /2. Thus, if H() := K(/2) and if n, m  H(), then we have |xn – xm| = |(xn – x) + (x – xm)|  |xn – x| + |xm – x| < /2 + /2 = . Since  > 0 is arbitrary, it follows that (xn) is a Cauchy sequence.

Lemma A Cauchy sequence of real numbers is bounded Proof: Let X := (xn) be a Cauchy sequence and let  := 1. If H := H(1) and n  H, then |xn – xH|  1. Hence, by the triangle Inequality we have that |xn|  |xH| + 1 for n  H. If we set M := sup{|x1|, |x2|, …, |xH-1|, |xH| + 1} then it follows that |xn|  M, for all n  N.

Cauchy Convergence Criterion A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Proof:

Definition |xn+2 – xn+1|  C|xn+1 – xn| We say that a sequence X = (xn) of real numbers is contractive if there exists a constant C, 0 < C < 1, such that |xn+2 – xn+1|  C|xn+1 – xn| for all n  N. The number C is called the constant of the contractive sequence.

Theorem Every contractive sequence is a Cauchy sequence, and therefore is convergent Proof: