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Advanced Calculus Unit XIX
SEQUENCES OF REAL NUMBERS – PART III
Objectives
From this session a learner is expected to achieve the following

Introduce the concept of monotone sequences

Study that a bounded monotonic sequence is convergent

Learn that monotone convergence theorem helps in finding a sequence of real
numbers that converges to the positive square root of a positive real number.

Construct a sequence that converges to the transcendental number e.
Contents
1. Introduction
2. Montone Sequences
3. The Calculation of Square Roots
4. Euler Number
1. Introduction
In the previous session (Sequences of Real Numbers -Part II) we have seen that every
convergent sequence is bounded, but the converse is not true. In this session, we
search for a condition which ensures the convergence of a bounded sequence; and end
up with the Monotone Convergence Theorem that establishes the existence of the limit
of a bounded monotone sequence. We begin with the definition of monotone sequence
and discuss associated results and examples. We will see that Monotone Convergence
Theorem is helpful in finding limit of certain seqeunces. It helps us in finding a
sequence of real numbers that converges to the positive square root of a positive real
number. A discussion on Euler Number will be made in this session.
2. Montone Sequences
1
Definition Let X  ( xn ) be a sequence of real numbers. We say that the sequence X is
increasing (or monotonically increasing) if it satisfies the inequalities
x1  x2 
 xn  xn 1 
We say that X is decreasing (or monotonically decreasing)if it satisfies the inequalities
x1  x2 
 xn  xn 1 
We say that X is monotone if it is either increasing, or it is decreasing.
The following sequences are increasing:
(1, 2, 3, 4,
, n,
),
( a, a 2 , a 3 ,
, an ,
(1, 3, 5, 7,
),
if a  1 .
)
The following sequences are decreasing:
1, 1 , 1 ,
 2 3

(b, b2 , b3 ,
 1 1
1, 3 , 2 ,
3

,


1
, ,
3
, bn
,
1
n1
3
,

,

) if 0  b  1.
The following sequences are not monotone:
(1, 1,  1,
, (1)n ,
(1,  2,  3,
),
, (1)n n,
).
The following sequences are not monotone, but they are ultimately monotone:
(5, 3, 2, 4, 1, 2, 3, 4,
),
 3, 8,  1, 1 , 1 , 1 ,

2 3 4

.


Example 1 Show that the sequence ( xn ) where
xn 
1
1
1



n n 1 n  2

1
, n ,
2n  1
is monotonically decreasing.
Solution
We note that for n  ,
xn1 
1
1


n 1 n  2

1
2(n  1)  1

1
1


n 1 n  2

1
1
1


2n  1 2 n 2 n  1
Hence
xn1  xn 
1
1
1
1

 
 0.
2n 2n  1 n
2n(2n  1)
Therefore,
xn 1  xn for n 
and hence the given sequence ( xn ) is monotonically decreasing.
Theorem 1 (Monotone Convergence Theorem) A monotone sequence of real numbers
is convergent if and only if it is bounded. Also,
(a) If X  ( xn ) is a bounded increasing sequence, then
2
lim( xn )  sup{xn : n  }
(b) If Y  ( yn ) is a bounded decreasing sequence, then
lim( yn )  inf{ yn : n  } .
Proof. First of all we note that a convergent sequence is bounded. Hence, in particular,
a bounded monotone sequence of real numbers is also convergent.
Conversely, let X be a bounded monotone sequence. Then X is either increasing or
decreasing.
We first treat the case where X  ( xn ) is a bounded increasing sequence. By
(a)
hypothesis, there exists a real number M such that xn  M for all n  . Thus the
subset {xn : n  } of the set of real numbers is bounded above. Hence by the Supremum
Property of real numbers, supremum of the set {xn : n  } exists. Let the supremum be
x*  sup{xn : n  } .
Claim x*  lim( xn ).
If   0 is given, then x *  is not an upper bound for the set {xn : n  } , and hence
there exists a natural number K  K ( ) such that x K is a member of the set  xn : n  }
and x *   xK . But since ( xn ) is an increasing sequence it follows that
x *   xK  xn  x *
for all n  K .
Therefore it follows that x    xn  x   for all n  K ,
i.e.,
  xn  x   for all n  K
i.e.,
| xn  x*|  
for all n  K .
Since   0 is arbitrary, we have ( xn ) converges to x * .
(b) If Y  ( yn ) is a bounded decreasing sequence, then it is clear that X  Y  ( yn ) is a
bounded increasing sequence. We have seen in part (a) that lim X  sup{ yn : n  } . On
the other hand, lim X   limY ; and also, we have
sup{ yn : n  }   inf{ yn : n  } .
Therefore
lim Y   lim X   sup{ yn : n  }  inf{ yn : n  } .
This completes the proof.
Remark The above theorem gives us a way of calculating the limit of the sequence
provided we can evaluate the supremum in case (a), or the infimum in case (b).


Example 2 lim  1   0 .
 n
Solution
Obviously the sequence
 1   1 1
,
,

  1,
2 3
 n 

 is decreasing and bounded (1 is a

bound for the sequence). Hence by part (b) of Theorem 1,
3
 1 
 1

lim 
: n  .
  inf 
 n
 n

…(1)


Clearly 0 is a lower bound for the set  1 : n   , and also 0 is the infimum of the set
 n

 1 
 1

: n   ; hence (1) implies lim 

  0.
 n
 n

Example 3 Examine the convergence of the sequence ( xn ) , where
1
1


n 1 n  2
xn 
1
nn

for n  .
Solution
It can be seen that
xn1  xn 
1
0
2(2n  1)(n  1)
and hence ( xn ) is monotonically increasing.
Obviously the sequence ( xn ) is bounded below by
1
.
2
Also,
1
1


n 1 n  2
xn 

1

n


1
nn
1
n
n times

n
 1.
n
Hence the sequence ( xn ) is bounded above by 1.
We conclude that the sequence is a bounded monotone sequence and hence is
convergent.
Example 4 Examine the convergence of the sequence ( xn ) , where
xn  1 
1 1
 
2 3

1
for n 
n
.
Solution
Since xn1  xn  1  xn , we see that ( xn ) is an increasing sequence. Hence by the
n 1
Monotone convergence Theorem the question of whether the sequence is convergent
or not is reduced to the question of whether the sequence is bounded or not.
We note that
x2n  1 
1 1 1
 

2  3 4 
 1
1 1 1



2  4 4 
 1
  n1 
 2 1
 1
 n 
2


1 

2n 
1 

2n 
4
 1
1 1
 
2 2
 1
n
.
2

1
2
Hence the sequence ( xn ) is unbounded, and therefore by Monotone convergence
Theorem ( xn ) is divergent.
Example 5 Let X  ( xn ) be defined inductively by x1  1, xn 1  14 (2 xn  3) for n  1 . Show
that lim X  3 .
2
Solution
Direct calculation shows that x2  5 . Hence we have x1  x2  2 . We show, by induction,
4
that xn  2 for all n  . We have already noted that this is true for n  1, 2 . If xk  2
holds for some k  , then
xk 1  14 (2 xk  3)  14 (4  3)  74  2 ,
so that xk 1  2 . Therefore xn  2 for all n  .
We now show, by induction, that xn  xn 1 for all n  . The truth of this assertion has
been verified for n  1 . Now suppose that xk  xk 1 for some k ; then
2 xk  3  2 xk 1  3 ,
and hence it follows that
xk 1  14 (2 xk  3)  14 (2 xk 1  3)  xk  2 .
Thus xk  xk 1 implies that xk 1  xk  2 . Therefore xn  xn 1 for all n  .
We have shown that the sequence X  ( xn ) is increasing and bounded above by 2. It
follows from the Monotone convergence Theorem that the sequence X converges to a
limit that is at most 2. In this case it is not so easy to evaluate lim( xn ) by calculating
sup{xn : n  } . Hence we proceed as follows:
Let x  lim( xn ). Then limit of the 1-tail ( xn 1 ) of X also has the limit x .
Since xn 1  14 (2 xn  3) for all n  , we have the limit of the sequence ( xn 1 ) and the limit
of the sequence
 14 (2xn  3)  are the same.
Hence, we obtain
x  14 (2 x  3) ,
Hence it follows that x  3 . .
2
Example 6 Let a and b be such that a  0 and b  a. Let ( xn ) be the sequence of real
numbers defined by
x1  a, xn 1 
ab 2  xn2
a 1
for n  .
Show that lim( xn )  b.
5
Solution
We first prove that
( xn ) is bounded above by b.
i.e.,
we have to show that
xn  b for n  . We prove this by Principle of Mathemtical Induction.
The result is true for n  1, becuase x1  a  b.
As induction argument, assume the result is true for n  k . i.e., suppose xk  b.
Then
ab2  xk2
xk2  b2
2
xk 1  b 
b 
 0, since xk2  b2  0.
a 1
a 1
2
2
Hence
x k 1  b,
and this shows that
xn  b for n  . Hence ( xn ) is bounded above by b.
Also, for n 
xn1 
2
xn2
ab2  xn2
a(b2  xn2 )
2

 xn 
 0,
a 1
a 1
since b  xn , and a  0.
Hence
x n21  xn2
so that
x n 1  xn .
Hence ( xn ) is monotonic increasing.
Being monotonic increasing and bounded, the sequence ( xn ) is convergent. Let
l  lim( xn ). Then, also, lim( xn 1 )  l.
Since
x n21 
ab2  xn2
,
a 1
we have
lim xn 1 
2
ab2  lim xn2
n
n
a 1
,
implies
l2 
ab 2  l 2
,
a 1
implies
al 2  l 2  ab 2  l 2 ,
implies
6
l 2  b2 ,
implies
l  b.
i.e., lim( xn )  b.
Example 7 Let Y  ( yn ) be the sequence of real numbers defined by y1  1, yn 1  2 yn
for n  . Show that lim( yn )  2.
Solution
Note that y1  1 and y2  2 ; hence 1  y1  y2  2 . We claim that the sequence Y is
increasing and bounded above by 2. To show this we will show, by induction, that
1  yn  yn 1  2 for all n  . This fact has been verified for n  1 . Suppose that it is true
for n  k ; then 2  2 yk  2 yk 1  4 , and hence it follows that
1  2  yk 1  2 yk  2 yk 1  4  2.
Noting that yk  2  2 yk 1 , the above implies
1  yk 1  yk  2  2.
Hence the validity of the inequality 1  yk  yk 1  2 , implies the validity of
1  yk 1  yk  2  2 . Therefore 1  yn  yn 1  2 for all n 
.
Since Y  ( yn ) is a bounded increasing sequence, it follows from the Monotone
convergence Theorem that it converges. Let the limit of Y be y .
The relation yn 1  2 yn gives lim  yn1   lim


2 yn which implies
y  2y .
Hence y 2  2 y which has the roots y  0, 2 . Since the terms of y  ( yn ) all satisfy
1  yn  2 , it follows that we must have 1  y  2. Therefore y  2 .
M2
3. The Calculation of Square Roots
Now we show that Monotone Convergence Theorem helps us in finding a sequence of
real numbers that converges to the positive square root of a positive real number.
Let a  0 ; we will construct a sequence ( sn ) of real numbers that converges to
a.


Let s1  0 be arbitrary and define sn1  12  sn  a  for n  . We now show that the
s

n

sequence ( sn ) converges to a .
7
We first show that sn2  a for n  2 . Since sn satisfies the quadratic equation
sn2  2sn1sn  a  0 , this equation has a real root. Hence the discriminant 4sn21  4a must
be nonnegative; that is sn21  a for n  1 .
To see that ( sn ) is ultimately decreasing, we note that for n  2 we have
2
1
a  1 ( s  a)
sn  sn1  sn   sn     n
 0.
2
sn  2
sn
Hence, sn 1  sn for all n  2 . It follows from the Monotone Convergence Theorem that
s  lim( sn ) exists. Moreover, it follows that the limit s must satisfy the relation
1
a
s   s   .
2
s
whence it follows that
1
1a
s
2
2s
s
or
a
s
or
s2  a .
s a .
Thus
For the purposes of calculation, it is often important to have an estimate of how rapidly
the sequence ( sn ) converges to
a . As above, we have
a  sn for all n  2 , whence it
follows that a  a  sn . Thus we have
sn
0  sn  a  sn 
2
a ( sn  a)

sn
sn
Using this inequality we can calculate
for n  2 .
a to any desired degree of accuracy.
Mod 3
4. Euler’s Number
We now construct a sequence that converges to one of the most important
transcendental numbers in mathematics.
For this, let en  1  1 

n
n
for n  . We will now show that the sequence (en ) is
bounded and increasing; hence it is convergent.
If we apply the Binomial Theorem, we have
1
n 1 n(n  1) 1 n(n  1)(n  2) 1
en  1    1   
 2
 3
1 n
2!
3!
 n
n
n
n


n(n  1)
n!
2 1

1
.
nn
If we divide the powers of n into the terms in the numerators of the binomial
coefficients, we get
8
1  1  1  1  2 
1

1
1 
2!  n  3!  n 
 n 
en  1  1 

1  n  1  .

n 

1  1  2 
1
1 
n!  n 
 n 

Similarly, we have
en 1  1  1 
1
1  1
1 
2 
1
 1 
1



2!  n  1  3!  n  1  n  1 



1  n  1 
 n 1


1
1 
2 
1
1
n!  n  1 
n
 1 

1 
1 
2 
1
1


(n  1)!  n  1  n  1 
1  n  .
 n 1


Note that the expression for en contains n  1 terms, while that for en 1 contains n  2
terms. Moreover, each term appearing in en is less than or equal to the corresponding
term in
en 1 , and
2  e1  e2 
en 1
 en  en 1 
has one more positive term. Therefore we have
, so that the terms of E are increasing.
To show that the terms of the sequence (en ) are bounded above, we note that if
p  1, 2,
, n, then 1 
p
1
1
 1 . Moreover 2 p 1  p ! so that
. Therefore, if n  1 ,

n
p ! 2 p 1
then we have
2  en  1  1 
1 1
 
2 22

1
n 1
2
.
Being the finite sum of geometric progression with common ratio
1 1


2 22

1
2
n 1
1
1
2
n 1
1
, we have
2
1,
and hence we deduce that 2  en  3 for all n  . The Monotone Convergence Theorem
implies that the sequence (en ) converges to a real number. Also, since 2  en  3 for all
n
, the limit of (en ) lies between 2 and 3. We define the number e to be the limit of
the sequence (en ) . The limit of the sequence (en ) is the famous Euler number e, whose
approximate value is 2.718281828459045
, which is taken as the base of the natural
logarithm.
Summary
In this session, we have described Monotone Convergence Theorem and used it for
finding limits of certain sequences. We have seen that it is helpful in finding a sequence
of real numbers that converges to the positive square root of a positive real number. A
discussion on Euler Number have been made in this session.
Assignments
9
1
1. Let a  0 and let z1  0 . Define zn1  (a  zn ) 2 for n  . Show that ( zn ) converges
and find the limit.
2. Let ( xn ) be a bounded sequence, and for each n 
let sn  sup{xk : k  n} and
tn  inf{xk : k  n} . Prove that ( sn ) and (t n ) are convergent. Also prove that if
lim( sn )  lim(tn ) , then ( xn ) is convergent.
that is bounded above and let u  sup A . Show there
3. Let A be an infinite subset of
exists an increasing sequence ( xn ) with xn  A for all n 
4. Let xn : 12  12 
1
2

1
for each n 
n2
such that u  lim( xn ) .
. Prove that ( xn ) is increasing and bounded,
and hence converges.
5. Let x1  8 and xn1  1 xn  2 for n  . Show that ( xn ) is bounded and monotone.
2
Find the limit.
6. Let x1  p , where p  0, and and xn 1  p  xn for n  . Find lim ( xn ).
n
QUIZ
1. Let x1  1 and xn1  2  1 for n  2 . Then pick the true statement.
xn
(a) ( xn ) is monotone, but not bounded.
(b) ( xn ) is bounded, but not monotone. .
(c) ( xn ) is bounded and monotone, but not convergent..
(d) ( xn ) is convergent.
Ans. (d) ( xn ) is convergent.
2. Let x1  a  0 and xn1  xn  1 . Then pick the true statement.
xn
(a) ( xn ) is monotonically decreasing
(b) ( xn ) is monotonically increasing
(c) ( xn ) is monotonically decreasing and bounded below
(d) none of the above.
Ans. (b) ( xn ) is monotonically increasing

1
3. Limit of the sequence  1  

n
n 1 
 is _________

10
(a) e
(b)
1
e
(c) e 2
(d)
1
e2
Ans. (a) e
n

4. Limit of the sequence  1  1   is _________

n 
(a) e
(b)
1
e
(c) e 2
(d)
1
e2
Ans. (b)
1
e
FAQ
1. State Supremum Property of real numbers.
Answer:
The Supremum Property (or Completeness Property) of
is the following: Every
nonempty set of real numbers that has an upper bound has a supremum in .
2. Define transcendental number.
Answer:
A transcendental number is a real or complex number, which is not a root of a nonconstant polynomial equation with rational coefficients.  and e are transcendental
numbers.
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3. A careless assumption that a sequence is ‘convergent’ leads to aburd conclusions.
Comment.
Answer:
The issue of convergence must not be ignored or casually assumed. The following
example illustrates this:
Consider the sequence ( xn ) defined by x1  1, xn 1  2 xn  1. Assuming the
‘convergence’ (actually wrong! The sequence is not convergent) with lim( xn )  x, we
would obtain x  2x  1, so that x  1. Of course, this is absurd as always xn  1 and
hence x must be  1. This absurdity is due to the wrong assumption that the sequence
is convergent. Hence, it is required to examine the convergence of the sequence before
finding its limit.
Glossary
Increasing Sequence:
A sequence X  ( xn ) of real numbers is increasing (or
monotonically increasing) if it satisfies the inequalities
x1  x2 
Decreasing Sequence:
 xn  xn 1 
A sequence X  ( xn ) of real numbers is decreasing (or
monotonically decreasing) if it satisfies the inequalities
x1  x2 
 xn  xn 1 
Monotone Sequence: A sequence X  ( xn ) of real numbers is monotone if it is either
increasing, or it is decreasing.
References:
1. T. M. Apostol, Mathematical Analysis, Narosa Publishing House, New Delhi, 1985.
2. R. R. Goldberg, Real Analysis, Oxford & I.B.H. Publishing Co., New Delhi, 1970.
3. D. Soma Sundaram and B. Choudhary, A First Course in Mathematical Analysis,
Narosa Publishing House, New Delhi, 1997.
4. Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, Wiley India Pvt.
Ltd., New Delhi.
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