Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
Limit of Sequence
A. Sequence
a n 
n  1,2,3,
n must be natural numbers.
 A sequence is usually defined / given by its general term or a difference equation
(recursive relation).
 General term: a n is given by a well-defined function of n.
 Recursive relation: a n is given by a function containing other terms in the
sequence.
Example A1
These two sequences are the same:
an  n
b1  1, bn1  bn  1
B. Limit
 Meaning of lim a n  l
n
 Meaning of lim a n  
n
Adopted from Roy Li’s notes at SPCS
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Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
lim a n   is generally not considered as having a limit.
n
1
0
n a
n
but lim a n    lim
n
 Finding simple limits
Example B1
2n  1
lim
n
2n
Example B2
lim n  1  n  1
n


Example B3
n2  2 n  3
lim 2
n n  8n  6
pn i  lower order terms
In general, given a n 
qn j  lower order terms
Adopted from Roy Li’s notes at SPCS
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Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
if i  j
0


p
lim a n  
n
q



if i  j
if i  j
 Theorem on limits (without proof)
The following theorems can ONLY be applied under the restrictions:
1. ALL the respective limits must exist.
2. They can be applied on a finite number of terms only.
lim  a n  bn   lim a n  lim bn
n
n
n
lim a n bn  lim a n lim bn
n
n
n
 a  lim a n
lim  n   n
n  b 
lim bn
n
lim bn  0
n
n
lim a n 
lim a n
n
n
lim log a n  log lim a n
n
n
a n  0  lim a n  0
n
Example B4
1
n
Given a n   1  1   , consider lim a n and lim a n .
n 
n

n
Example B5
Given bn 
 1 n
n
, consider lim bn and lim bn .
n 
Adopted from Roy Li’s notes at SPCS
n
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Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
Adopted from Roy Li’s notes at SPCS
Page 4
Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
C. Monotonicity and Boundedness
A sequence is said to be:
 Bounded above iff M , a n  M n  N
 Bounded below iff m, a n  m n  N
 Bounded iff it is bounded above and below.
 M is called a upper bound of the sequence.
 m is called a lower bound of the sequence.
Example C1
1
 
 n
Example C2
Given u1  2 and un1 
n 
2
1  n
n
6(1  un )
, show that un  is bounded below by 2.
7  uu
Example C3
If a1  1 and a n1  2  a n , show that a n  is bounded above.
Adopted from Roy Li’s notes at SPCS
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Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
A sequence is said to be:
an1  an
an1  an
an1  an
an1  an




Monotonic increasing
Monotonic decreasing
Strictly increasing
Strictly decreasing

A strictly increasing (decreasing) sequence is also a monotonic increasing
(decreasing) sequence.

A sequence is said to be monotonic iff it is either monotonic increasing or
monotonic decreasing.
iff
iff
iff
iff
n N
Example C4
Show that the sequence a n  2 n is strictly increasing.
Example C5
Show that the sequence un  in example C2 is strictly decreasing.
Example C6
Show that the sequence a n  in example C3 is decreasing.
Adopted from Roy Li’s notes at SPCS
Page 6
Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
Convergent Sequence
an  is convergent
 lim an exists (and finite)
n
an  is
 AND bounded above  a n  is convergent
an  is
 AND bounded below  a n  is convergent
But the limit may not be equal to the upper / lower bounds.
Example C7
For the sequence un  in example C2, find lim un .
n
Example C8
Evaluate lim a n if 0  a  1 .
n
This method works only if the limit exists, otherwise the result would be incorrect.
e.g. a1  3, an  3an1
Adopted from Roy Li’s notes at SPCS
Page 7
Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
Example C9
1  a n1
1
a1  , a n 
, find lim a n .
n
2
2  a n 1
D. Sandwich Principle
Theorem:
If
bn  a n  c n AND lim bn  lim c n  l ,
n
n
then lim a n  l
n
Note that it is not required to proof the existence of lim a n .
n
Example D1
2  ( 1) n
lim n
n
2n
Adopted from Roy Li’s notes at SPCS
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Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
Example D2
lim n 2 n  3n
n
Example D3
(a)
Using binomial theorem show that
(b)
Hence show that lim n n  1

x
2 1  x   1
n
n n  1
2
n
E. e
The Napierian Constant e
n
n
1
1

e  lim 1    lim   2.71828........
n 
n
n
r0 r!
Adopted from Roy Li’s notes at SPCS
Page 9
Helen Liang Memorial Secondary School (Shatin)
Advanced level Pure Mathematics
Supplementary Lecture Notes
Practice
1.
Let c be a real constant such that 0  c  1
c
c x n2
x1  and x n1  
for n  1,2,
2
2 2
(i) Show that 0  x n  1 .
(ii) Show that  x n  is increasing.
(iii) Find lim x n .
n
2.
Let u1  a and un1 
Show that un 
n( n  1)
for n=1,2,3,...
2 n  un
na  ( n  1)(1  a )
a  n(1  a )
 1 n 1 
a
And hence lim   uk  
n  n!
 1 a
k 1
3.
Evaluate lim n 1n  2 n 10 n .
4.
Given a sequence a1  a2  1 and a n 2  a n1  a n .
n
a n1

n a
n
Show that lim
5.
51
.
2
Two sequences are defined by x1  y1  0 and
x  yn
x n2  y n2
and y n1  n
for n  1,2,
2
xn  yn
(a) Show that x n  x n1  y n1  y n  0 for n  1,2,
(b) Show that lim x n  lim y n .
x n1 
n
6.
n


 
If a  1 , evaluate lim1  a 1  a 2 1  a 4  1  a 2
n
n

Hint:
Adopted from Roy Li’s notes at SPCS
(1  x)(1  x)  1  x 2
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