1. Chapter Three SEQUENCES
Oleh:
Mardiyana
Mathematics Education
Sebelas Maret University

2. 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)

3. 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.

4. 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

5. 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.

6. 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 + .

7. 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, …).

8. 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.

9. 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.

10. Theorem
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.

11. |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).

12. 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.

13. 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.

14. 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.

15. 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.

16. 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  …

17. 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}.

18. Proof:

19. 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!

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

21. 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.

22. 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.

23. 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.

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

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

26. 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| < .

27. 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.

28. 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.

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

30. 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.

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