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
a. 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 3.1.4
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:
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 3.1.8
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, …).
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.
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.
2.
3.
4.
5.
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
xn 
1
2
1

1
2
2

1
n2
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 BolzanoWeierstrass 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
( xr1 , xr2 ,, xrn ,)
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
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: