Chapter ThreeSEQUENCES

Oleh:Mardiyana

Mathematics EducationSebelas Maret University

Definition 3.1.1A 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 3.1.3If X = (xn) and Y = (yn) are sequences of real numbers, then we definea. X + Y = (xn + yn)

b. X – Y = (xn - yn)

c. X.Y = (xnyn)

d. cX = (cxn)

e.X/Y = (xn/yn), if yn 0 for all n N.

Definition 3.1.4Let 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 notationlim 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 3.1.6Let X = (xn) be a sequence of real numbers, and let x R. The

following statements are equivalent:1. X convergent to x.2. For every -neighborhood V(x), there is a natural number

K() such that for all n K() the terms xn belong to V(x).

3. For every > 0, there is a natural number K() such that for all n K(), the terms xn satisfy |xn – x | <

4. For every > 0, there is a natural number K() such that for all n K(), the terms xn satisfy x - < xn < x + .

Tails of Sequences

Definition 3.1.8If 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, …).

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 3.1.10Let 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.

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

1. Show that if a > 0, then lim (1/(1 + na)) = 0.2. Show that lim (1/2n) = 0.3. Show that if 0 < b < 1, then lim (bn) = 0.4. Show that if c > 0, then lim (c1/n) = 1.5. Show that lim (n1/n) = 1.

Limit Theorems

Definition 3.2.1A 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 3.2.2A 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

DefinitionLet 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 inequalitiesx1 x2 … xn-1 xn …

Monotone Convergence Theorem

A monotone sequence of real numbers is convergent if and only if it is bounded. Furthera). 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

2221

21

11

nxn

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

DefinitionLet 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

),,,,(21

nrrr xxx

is called a subsequence of X.

TheoremIf 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

DefinitionA 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.

LemmaA 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 CriterionA sequence of real numbers is convergent if and only if it is

a Cauchy sequence.

Proof:

Definition

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.

TheoremEvery contractive sequence is a Cauchy sequence, and therefore is convergent

Proof: