Limit of a Sequence
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
2
Calculus I
Chapter 2
Limit of a Sequence
2.1
Introduction
2
2.2
Sequences
2
2.3
Convergent Sequences
6
2.4
Divergent Sequences and Oscillating Sequences
7
2.5
Operations on Limits of Sequences
8
2.6
Sandwich Theorem for Sequences
13
2.7
Monotonic Sequences
17
2.8
The Number e
23
2.9
Some Worked Examples
24
Prepared by. K. F. Ngai
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1
Limit of a Sequence
2.1
Advanced Level Pure Mathematics
INTRODUCTION
Some examples of sequences:
2.2
1 1 1
1
, , ,  , 1 , 
2
2 4 8
1.
1,
2.
1, 1, 1, 1,  , (1)r+1 , 
3.
cos x , cos2x , cos3x ,  , cos rx , 
SEQUENCES
Definition 2.1
A sequence {xn} is a function on the set of real numbers, and is usually written as
x1 , x2 , x3 ,  , xn , .
1.
The term xn is called the general term of the sequence.
Example {1, 2, 4, 8, ...} is a sequence of positive integers with general term xn  2 n 1 .
2.
If the sequence has infinite number of terms, it is called an infinite sequence.
3.
If the sequence has finite number of terms, it is called a finite sequence.
4.
Sn = x1 + x2 + x3 +  + xn +  is said to form a series.
5.
Sn =  x r = x1 + x2 + x3 +  + xn is a finite series.
n
r 1

6.
Sn =  x r = x1 + x2 + x3 +  + xn +  is an infinite series.
r 1
n
How to find the series sum , S n   x r ? By using
r 1
(M1)
Mathematical induction.
(M2)
Method of difference where xr  f (r  1)  f (r ) such that
n
n
r 1
r 1
S n   x r   [ f (r  1)  f (r )]  f (n  1)  f (1)
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2
Limit of a Sequence
Advanced Level Pure Mathematics
(M3)
Partial fractions and method of difference.
(M4)
Standard formulae:
(i)
A.P. : a , a+d , a+2d ,  , a+( n1)d ,  .
Sn =
1
[2a  (n  1)d ] .
2
(ii) G.P. : a , a r , a r2 ,  , a rn1 ,  .
Sn =
a (1  r n )
;
1 r
(iii) 1 + 2 + 3 +  + n =
S =
n
r=
r 1
12 + 22 + 32 +  + n2 =
a
where | r| < 1 .
1 r
1
n( n  1)
2
n
r
2
=
1
n(n  1)( 2n  1)
6
3
=
1 2
n (n  1) 2
4
r 1
13 + 23 + 33 +  + n3 =
n
r
r 1
Example 1
If xn denotes the n th term of the series which begins
12
22
32


 ...
1 1(1  2  2 2 ) 1(1  2  2 2 )(1  2  32 )
x
Prove that the sum of the series to n terms is 1 (3  n2 ).
2
n
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Limit of a Sequence
Advanced Level Pure Mathematics
Example 2
Show that (sin2 +sin4 +  +sin2n ) sin = sin n sin (n+1) .
Example 3
(a) Find
n
1
 r (r  1) .
r 1
(b) If ur = r(r+1)(r+2) for positive integer r , show that 3r(r+1) = ur  ur1 .
n
Hence, or otherwise, find
 r (r  1) .
r 1
(M5)
Arithmetic-Geometric Series (A.G.P.)
Sequence :
a , (a+d)R , (a+2d)R2 ,  , [a+( n1)d)Rn1 ,  .
n
n1
Sn = a  [a  (n  1)d ]R  dRa (1  R2 ) .
1 R
(1  R)
In particular, 1 , 2x , 3x2 ,  , nxn1 ,  .
n
Sn =  rx r 1 
r 1
(M6)
1  (n  1) x n  nx n 1
.
(1  x) 2
Harmonic Series
Hn = 1 
1 1 1
   ...
2 3 4
There is no simple formula for the sum of the first n terms.
(M7)
Difference equation / Recurrence relation
Recurrent sequence :
(1) a n+1 = Aan
for n = 1 , 2 , 3 , ... .
(2) an+2 = Aan+1 + Ban for n = 1 , 2 , 3 , ... .
where the coefficients A and B are constants.
For case (1)
, we have
an+1 = Aan = A(Aan1) = A2 an1 = A3 an2 == Ar a1
If a1 = Aa , then an+1 = An+1 a.
For case (2) , we have
an+2  Aan+1  Ban = 0 .
Let  ,  be the roots of the auxiliary equation t 2  At  B = 0 .
(i)
If    , then an = c1n + c2n.
(ii) If  =  , then an = (nc1 + c2 )n.
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Limit of a Sequence
Example 4
Advanced Level Pure Mathematics
A sequence of real numbers an is defined by
a0 = 0 , a1 = 1 and an+2 = an+1 + an for all n = 0, 1, ... .
1
Show that for all non-negative integers n , an =
(n  n ), where  ,  are roots of the
5
equation x2 + x  1 = 0 with  > .
an+2 = an+1 + an
Soln.

an+2 + an+1  an = 0
Consider the auxiliary equation t 2 + t  1 = 0 , we have t =
=
1 5
.
2
1 5
1 5
and  =
2
2
Let an = c1 (
1
1 5 n
) + c2 (
2
2
5
)n , where c1 , c2  R.
Since a0 = 0 , we have
0 = c1 + c2 .................(1)
Since a1 = 1 , we have
1 = c1 (
1
1 5
) + c2 (
2
2
5
) ...................(2)
On solving (1) and (2) , we obtain c1 =
Hence, an =
1
1
and c2 = 
.
5
5
1 n n
(   ) , where  ,  are roots of the equation x2 + x  1 = 0
5
with  > .
Example 5
(a) The general term of a sequence ur satisfies ur  2kur1 + k2 u r2 = 0 for r  2
and if u1 = 2k , u2 = 3k2 , show that ur = (r+1)kr , r  1 .
(b) Find the general term for the Fibonacci sequence
x1 = x2 = 1 and xn+1 = xn + xn 1 .
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Limit of a Sequence
2.3
Advanced Level Pure Mathematics
CONVERGENT SEQUENCES
A CONVERGENT SEQUENCE {xn} is a sequence whose terms will approach a
Definition 2.2
finite value a as n tends to infinite.
We say that xn  a as n  .
Symbolically, lim x n  a , which is called the limit of the convergent sequence.
n 
1
2
n
If xn = ( ) , the sequence {xn} is convergent to 0 .
Example
1
n
Example
The sequence { a } is convergent to 1.
Example
The sequence {
Definition 2.3
sin r
} is also convergent to 0.
r
A sequence {xn} is said to converge to a if and only if for any  > 0 , there exists a
positive integer N such that when n  N , we have |xn  a| <  .
a is called the limit value of {xn} and we write lim x n  a .
n
Definition 2.4
A sequence {xn} not convergent is called DIVERGENT.
Theorem 2.1
The limit value of a convergent sequence is UNIQUE.
i.e. if xn  a and xn  b as n  , then a = b.
Theorem 2.2
All convergent sequences are BOUNDED. i.e.
| xn |  M for all n  N .
Important Facts:
(1) If lim xn  a , then lim | xn || a | .
n 
n
(2) If lim xn  a , then lim xn  p  a for p = 1, 2, 3, ...
n 
n
(3) {xn } converges to a iff every subsequences of {xn} converges to a.
i.e. odd sequence x1 , x3 , x5 ,  and even sequence x2 , x4 , x6 , 
both converges to a.
(4)
lim xn  0 if | x | < 1.
n 
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Limit of a Sequence
Advanced Level Pure Mathematics
(5)
1
 0.
n  n
(6)
lim x n  1 where x  0 .
(7)
lim
1
n 
lim 1  1n   e .
n
n
 0 if   0
1 
(8) lim    1 if   0
n  n
 if   0

2.4
DIVERGENT SEQUENCES AND OSCILLATING SEQUENCES
Definition2.5
A sequence {xn} is said to diverge to positive infinity if for any positive real number
M, there exists a positive integer N such that when n > N, xn > M.
We write l i m x   .

A sequence {xn} is said to diverge to negative infinity if for any positive real
number M , there exists a positive integer N such that when n > N , xn < M.
We write l i m x   .

Theorem 2.3
Let {xn} be a sequence with xn  0 .
Then lim xn   if and only if lim
n 
n
1
 0.
xn
1
0
n n
lim n    lim
N.B.
n
Definition 2.6
Oscillating sequences are neither convergent nor diverging to infinity.
Example {1, 1, 1, 1, 1, 1, ...} is an oscillating sequence.
Example {0 , 4 , 0 , 8 , ... , n[1+(1)n ] , ...} is a infinitely oscillating sequence.
Example (1) xn = sin n
(2) xn = n cos n
(3) xn = n (1)n
(4) xn =
1
(1)n
n
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Limit of a Sequence
2.5
Advanced Level Pure Mathematics
OPERATIONS ON LIMITS OF SEQUENCES
Theorem 2.4
Let {xn} and {yn} be two convergent sequences.
(a)
(b)
(c)
(d)
lim ( xn  yn )  lim xn  lim yn .
n 
n 
n 
lim ( xn yn )  lim xn  lim yn .
n 
n 
n 
 x  lim xn
lim  n   n  , where lim yn  0 .
n 
n  y
yn
 n  nlim

lim (kxn )  k lim xn , where k is a constant.
n 
n 
(e) For any positive integer m,
(i)
lim ( xn )  ( lim xn )m ,
m
n 
n 
1
1
(ii) lim ( xn m )  ( lim xn ) m ,
n 
n 
(iii) lim xn  m  lim xn .
n 
n2  n  2
n   2n 2  2 n  4
Example 6
Find (a) lim
Example 7
Find. lim ( n  1  n )
*N.B.
n 
2
4n 
(b) lim  
.
n  n
4n  1 

n 
We cannot use these rules of operations if {xn} or {yn} is NOT convergent.
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Limit of a Sequence
1
1
Example lim  n    lim n  lim
n 
 n  n  n  n
Advanced Level Pure Mathematics
( lim n   , i.e. {n} diverges )
n 
1
2
n
Find lim  2  2    2  .
n n
n
n 

Wrong proof :
Example 8
k
= 0 for k  n.
n n2
Since lim
2
n
 1
lim  2  2    2  = 0 + 0 +  = 0
n n
n
n 

It is invalid since the sum of an infinite number of terms, each term tends to
So
zero, may not be zero.
Example 9
Soln.
1 2 3 4
n
Show that the limit value lim     +   (1) n 1  does not exist.
n  n
n n n
n

When n is even, let n = 2k , where k is an integer. Then
1 2 3 4
n
   +   ( 1) n 1
n n n n
n
1
2
3
4
2k  1 2k



+  

=
2k 2k 2k 2k
2k
2k
1
[(1  2)  (3  4)    (2k  1  2k )]
=
2k
1
1
( k ) = 
=
2k
2

1
n
1 2 3 4
lim     +   (1) n 1  = 
n  n
2
n n n
n

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Limit of a Sequence
Theorem 2.5
Advanced Level Pure Mathematics
Let {xn} and {yn} be two sequences.
(a) If lim x n  0 and | yn|  M for all n , then lim x n y n  0 .
n
n 
(b) If lim x n   and | yn|  M for all n , then lim ( x n  y n )   .
n 
n 
(c) If lim x n   and yn  0 for all n > N and yn does not converge to 0 , then
n 
lim x n y n  
n 
sin n
.
n  n
Example 10 Evaluate lim
Soln.
1
sin n
= 0 ,  lim
= 0.
n n
n  n
Since |sinn|  1 for all n and lim
Example 11
(1) n
n 
n
Evaluate (a) lim
(b) lim (n  cos n)
n 
(c)
lim n sin n
n
N.B. For Theorem 2.5 , lim yn does not necessarily exist.
n
Example 12

 1

1
1
Evaluate lim n  (1) n 
 .


n 
1

3
3

5
(2
n

1)(2
n

1)



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Limit of a Sequence
Example
Advanced Level Pure Mathematics
A sequence xn  is defined by
x1  1 , x2  2
and
2 xn  2  xn 1  xn  0
( n  1,2,3, )
Show that for all positive integers n ,
 1
xn  1    
 2
n2
Hence evaluate lim xn .
n
Solution
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Limit of a Sequence
Example
Advanced Level Pure Mathematics
A sequence xn  is defined by
x1  1
and
xn 1  xn 
1
( n  1,2,3, )
3n
Find lim x2 n .
n
Solution
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Limit of a Sequence
2.6
Advanced Level Pure Mathematics
SANDWICH THEOREM FOR SEQUENCES
Theorem
If xn  yn , then lim xn  lim yn .
Theorem 2.6
SANDWICH THEOREM FOR SEQUENCES
n 
n 
Let {xn} , {yn} and {zn} be three sequences such that
xn  yn  zn
lim xn  lim zn  a
and
n 
lim yn  a .
Then
Example13
Soln.
n 
n
sin n
 0.
n
n
Prove that lim
Since
1  sin n  1
So
Since
1

n
lim

1
n
sin n

n
 lim

1
n
1
for n  1
n
0
Hence, by SANDWICH THEOREM for sequence, lim
n 
sin n
 0.
n
1
1
1 
Example 14 Find lim  2 
.

2
2 
n 
n
(
n

1)
(2
n
)


since
1
1
1

 2 for all k = 0 , 1, 2, . . ., n
2
2
n
( 2n )
(n  k )
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Limit of a Sequence
Advanced Level Pure Mathematics
N.B. It is wrong to say
1
1
1 

lim  2 

2
(2n)2 
 n (n  1)
n 
=
=
=
lim
n 
1
1
1
 lim
   lim
2
2
n


n


n
(n  1)
(2n) 2
0 + 0 + ...
0
(totally infinitely many 0)
0  is an indeterminate form.
0 
Other indeterminate forms : ,
,  0 , 1 ,   , etc.
0 
because
Example 15
 1
1
1 
Evaluate lim 
.





2
2
2
n  
n 2
n n 
 n 1
Example 16
Evaluate lim
n 
n
2  (1) n
.
2n
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Limit of a Sequence
Advanced Level Pure Mathematics
Example 17 Let a be a real number greater than 1. Prove that lim
n 
n
 0.
an
Since a > 1, let a = 1+h where h is a positive real number.
Then an = (1+h)n =
Example 18 Let A be a positive real number and {an} be a sequence of real numbers such that a1  A
and an+1 =
1
A2 
 an 
 for n  1.
2
an 
(a) Show that an  A for all positive integers n.
Hence, show that an  A 
1
(an1  A ).
2
(b) Find lim an by using sandwich theroem .
n 
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Limit of a Sequence
Advanced Level Pure Mathematics
(a) If lim xn   and there exists a positive integer N such that xn  yn as n > N ,
Theorem 2.7
n 
then lim y n   .
n 
(b) If lim xn   and there exists a positive integer N such that xn  yn as n > N , then
n 
lim yn   .
n
Example 19
Since
1
1
1 
Find lim 


.
n 
n

1
n

2
2
n


[compare with Example14 and 15]
1
1
1
1
1
1

  


  
n 1
n2
2n
nn
nn
2n
 0 if | a | 1
.
Example 20 Prove that lim a n  
n 


if
|
a
|

1

What happens when |a| = 1?
Case 1 : Let a > 1,
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Limit of a Sequence
2.7
Advanced Level Pure Mathematics
MONOTONIC SEQUENCES
Definition 2.7
(1) A sequence {xn} is said to be monotonic increasing if and only if xn  xn+1
for n = 1, 2, 3, ... .
(2) A sequence {xn} is said to be monotonic decreasing if and only if xn  xn+1 for n
= 1, 2, 3, ... .
(3) A sequence {xn} is said to be monotonic if and only if it is either increasing or
decreasing.
(4) A sequence {xn} is said to be strictly increasing or strictly decreasing if
and only if xn < xn+1 or xn > xn+1 for all n.
Example 21 Show that the sequence {
Soln.
1
} is strictly decreasing.
n
1
1
>
for all n  N
n n 1
1
So the sequence { } is strictly decreasing.
n
Since
n
Example 22 Let a sequence {xn} be defined by xn =
1
 r . Show that the sequence is
r 1
monotonic increasing.
Definition 2.8
(a) A sequence {xn} is said to be BOUNDED ABOVE if and only if
there exists a constant M such that xn  M for n = 1, 2, 3, ... .
(b) A sequence {xn} is said to be BOUNDED BELOW if and only if
there exists a constant M such that xn  M for n = 1, 2, 3, ... .
Prepared by. K. F. Ngai
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Limit of a Sequence
Theorem 2.8
Advanced Level Pure Mathematics
(a) If a monotonic increasing sequence is bounded above, then
the sequence is convergent and limit of {xn} exists.
(b) If a monotonic decreasing sequence is bounded below, then
the sequence is convergent and limit of {xn} exists.
decreasing and bounded below
increasing and bounded above
n
1
for n  1 .
r 1 3  1
Example 23 Let the sequence {xn} be defined by xn = 
r
(a) Show that {xn} is an increasing sequence and bounded above.
(b) Hence, show that {xn} is convergent.
N.B.
(1) {xn} converges and xn < L , then lim xn  L.
n
(2) {xn} converges and xn > L , then lim xn  L.
n
Prepared by. K. F. Ngai
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Limit of a Sequence
Example 24
Advanced Level Pure Mathematics
Let a and b be two positive real numbers. A sequence {xn} is defined by
x n 1 
ab 2  x n
a 1
2
for n  1. It is given that 0 < x1 < b.
(a) Show that xn  b for all positive integer n.
(b) Show that {xn} is monotonic increasing. Hence, show that {xn} is convergent.
(c) Find lim xn .
n
Prepared by. K. F. Ngai
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Limit of a Sequence
Advanced Level Pure Mathematics
Example 25 Let a and b be two real numbers such that a > b > 0 . Two sequences {an} and {bn} are
defined by
an 
a n 1  bn 1
ab
, bn  a n 1bn 1 for n > 1 and a1 
, b1  ab .
2
2
(a) Prove that {an} is monotonic decreasing. Hence deduce that {bn} is monotonic
increasing.
(b) Prove that {an} and {bn} converge to the same limit.
Prepared by. K. F. Ngai
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Limit of a Sequence
Example26
Let x1 > x2 > 0 and x n 1 
Advanced Level Pure Mathematics
1
( x n  x n 1 ) .
2
(a) Show that {x2n1} is a decreasing sequence and {x2n} is an increasing sequence .
(b) Show that x2n1 > x2m for all positive integers n and m .
(c) Hence, show that {x2n1} and {x2n} have a common limit and find it .
Prepared by. K. F. Ngai
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Limit of a Sequence
Advanced Level Pure Mathematics
Example 27 Given two positive numbers a and b where a > b and {an}, {bn} are two sequences defined
by
a1 
2a n 1bn 1
a  bn 1
ab
ab
, b1 
and a n  n 1
, bn 
for all n  2,
2
ab
a n 1  bn 1
2
(a) Show that bn  bn+1  an+1  an .
(b) Show that {an} and {bn} have the same limit. Find this common limit.
Prepared by. K. F. Ngai
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Limit of a Sequence
2.8
Advanced Level Pure Mathematics
THE NUMBER e
Consider the sequence of numbers defined by:
1
(1  ) n , n  1, 2, 3, 
n
The following table give the value of the sequence corresponding to different values of n.
n
(1 
1
)
n
1
2
3
10
100
1000
10000
100000
2
2.25
2.3704
2.5937
2.7048 2.71692 2.71875 2.71827
1000000
2.718281
1
As the value of n increases without bounds, the value of (1  ) increases steadily, but it seems to
n
increase slower and slower. We can see that it would stop somewhere around 2.7182….
This number plays an important role in advanced mathematics and is denoted by e.
n
Definition
e = lim 1  1  , where n takes positive integral values.
n
 n
Or
lim ( 1 
n 
1 1
1
  )  e
1! 2!
n!
e is an important irrational number in calculus.
N.B.
Theorem 2.9
2e3.
Example 28
1
Find lim 1   .
n 
 3n 
n
1
1
3n
3n
1

1  3 
1 
1  3


lim 1   = lim 1    =  lim 1    = e 3
3n  
n 
 3n  
 3n  n    3n  

n
Soln.
Example 29
Express the following limits in terms of e.
n
(a)
1 

lim 1 

n 
 n 1
(d)
 3 2
lim 1   2 
n
 n n 
(b)

lim 1 
n

2

n
n
n 
(c) lim 

n n  1


n
n
Prepared by. K. F. Ngai
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Limit of a Sequence
Advanced Level Pure Mathematics
Example 30 Prove that e - ( 1 
2.9
1 1
1
1
  ) 
1! 2!
n!
n!n
SOME WORKED EXAMPLES
Example 31 Let a1 = 2 , b1 =
3
2n
2n  1
and an =
an1 , bn =
bn1 for n  2 .
2n  1
2n
2
(a) Prove that an > bn and an bn = 2n + 1 for n  1 .
(b) Using (a), or otherwise, show that an2 > 2n + 1 for n  1 .
1
.
n  a
n
Hence find lim
[HKAL98]
(7 marks)
Example 32 (a) Let x > 1 and define a sequence {an} by
a n 1  1
a1 = x and an =
for n  2 .
2 a n 1
2
(i)
Show that an > 1 and an > an+1 for all n .
(ii) Show that lim an = 1 .
(8 marks)
n
(b) Let f : [1,]  R be a continuous function satisfying
 x2  1 
 for all x  1 .
f(x) = f 
 2x 
Using (a), show that f(x) = f(1) for all x  1.
[HKAL96]
(7 marks)
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Limit of a Sequence
Example 33
Advanced Level Pure Mathematics
For any  > 0 , define a sequence of real numbers as follows :
a1 =  + 1 , an = an1 +

for n > 1 .
a n 1
(a) Prove that
(i)
an2  an12 + 2 for n  2 ;
(ii) an2  2 + 2n + 1 for n  1
.
(2 marks)
(b) Using (a), show that for n  2 ,
an2  2 + 2n + 1 +
2
n 1

k 1
2
 2k  1
.
(3 marks)
(c) Prove that for k  1 ,
2
  2k  1
2
1
dx .
k 1   2 x  1

k
(2 marks)
2
2
an
exists and find the limit.
n  n
(d) Using the above results, show that lim
State with reasons whether lim
an
n 
2
n
exists.
[HKAL95]
(8 marks)
Example 34 By using the identity sin 2  2sin  cos , prove that
x
x
x
sin x
lim cos cos 2  cos n 1 
n 
2
x
2
2
Example 35 If x1  0 , xn1 
3(1  xn )
, show that xn 1  3  k xn  3 , where 0  k  1 .
3  xn
Hence, show that lim xn  3
n 
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