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Pure Maths/Sequence/p.1
LIMIT OF SEQUENCE
Sequence and Series
A sequence is an arrangement of numbers in a row, e.g. a1 , a2 , a3 , .. . , an , .. .
where an is the
n th term in the sequence and is a function of n which is defined for all positive integral values of n.
an =
e.g.
2 + 3n
5+n
or a n = ( - 2)n+1
The sequence is usually denoted by
 a n
and it is said to be finite (or infinite ) if the number of
terms of the sequence is finite (or infinite).
A series is the sum of all the terms of the sequence :

n
(i) Sn =

r=1
ar
S =
(ii)
- finite series

r=1
ar
- infinite series
Example
1.
Let
 a n
be a sequence of positive numbers such that
 1 + an 
a1 + a 2 + ... + a n = 

 2 
Prove by induction that
2
for n = 1, 2, 3, …
.
a n = 2n - 1 for n = 1, 2, 3, … and hence, find
lim
n 
an
a n 1
.
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2.
Let
u1 = 1, u2 = 3 and
un = un-1 + un-2
for n  3 .
Using Mathematical Induction, or otherwise, show that
un =  n +  n
Hence, find
lim
n 
for n  1
where
 and  are the roots of x2 - x - 1 = 0 .
u n 1
un
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Pure Maths/Sequence/p.2
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Difference Equations
If the sequence  a n  is defined by a recurrence relation relating 3 consecutive terms such as
a  a n+1 + b  a n + c  a n-1 = 0
where a, b, c are real constants
and the first two terms are given, then
if
  ,
a n = C1   n + C 2   n
;
 and  are the roots of ax2 + b x + c = 0
where
if
=,
and


a n = C1 + n C2   n
C1 , C2 are real constants determined
by the first two terms of the sequence.
e.g. Fibonacci sequence :

{ 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , … }
a1 = a 2 = 1 and a n+1 = a n + a n-1
The equation
x2 - x - 1 = 0 has the roots
1 - 5

∴ a n = C1
 2 
n
1 + 5

+ C2 
 2 
for n  2
1- 5
1+ 5
.
and
2
2
n
for some real constants C1 , C2
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Classwork
If k is a non-zero real constant and
 a n
Pure Maths/Sequence/p.3
is a sequence defined by
a n - 2k a n-1 + k 2a n-2 = 0
and
for n  3
a1 = 2k , a 2 = 3k 2 , find a n in terms of n and k.
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Method of Difference
If the r th term of the series can be expressed as the difference of f (r + 1) and f (r ) where f (x) is a
function of x , i.e. ar = f (r + 1) – f (r ) , then
n
Sn =
 f(r + 1) - f(r)
=
f(n + 1) - f(n) + f(n) - f(n - 1) +
=
f(n + 1) - f(1)
r=1
... + f(2) - f(1)
Example
1.
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2. Let
 a n
Pure Maths/Sequence/p.4
be a sequence defined by a1 = 1, a 2 = 2 and a k = 2 a k-1 - a k-2
for k  3 .
a n in terms of n.
Find
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Exercise A
1.
Consider the sequence
un 
in which u1 = 0, un+1 = 2n - un
Using mathematical induction, or otherwise, show that
Hence, find
lim
n
for n = 1, 2, 3, … .
2 un = 2n - 1 + (-1)n
for n = 1, 2, 3, … .
un
.
n
n
1
.
(r
1)
(r
+
3)
r=2
lim
n 
2.
Find
3.
Find the general term a n of the sequence
 a n
which is defined by a1 = 21, a 2 = 1 and
for n  0 .
a n+2 + 3 a n+1 - 4 a n = 0
H.W. W.14.5 Q.3 , 5 , 16
Limit of Sequence
lim a
n n
is the limit of the sequence
 a n .
If the limit exists, then the terms of the sequence will approach a finite value as n tends to infinity
and the sequence is said to be convergent . Otherwise, it is said to be divergent .
e.g. 1.
2.
 n 3  _________________________________________________
 - 3n  ________________________________________________
 a n
A sequence
oscillates , then it is said to oscillate finitely or infinitely according as it is
or is not possible to assign a positive real number k such that
(i) a n =
1
+ (-1)n
n

a n < k for all values of n.
a n < _________
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(ii) a n =

n
1 + (-1)n
2

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1 + (-1)n
3. a n =
n
 a n
Formally, the sequence
Pure Maths/Sequence/p.5
has a limit  if for any positive number ε, however small, there exists
a positive integer N, depending on ε, such that if n > N, then
ε
an   .
< ε i.e. when n   ,
ε
(
 -ε
where
an - 
)
 +ε

lim a =  .
n n
2 + 3n
5+n
e.g. a n =
lim a
n n

lim
n
=
2 + 3n
5+n
=
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= ________________________________
Theorem
1.
The limit of a sequence is unique i.e. if
lim a = 
n n
and
lim a =  ’ , then
n n
 =
’ .
2.
All convergent sequence are bounded, i.e. if the sequence
an  M
positive real number M such that
 a n
is convergent, then there exists a
for all n.
Properties :
If the sequences
1.
lim a n
n
 a n
and
lim a n
 bn  are convergent, i.e. n
 lim a   lim b 
 n n   n n 
lim a b
n n n
3.
an
lim
n b n
=
 lim a 
 n n 
 lim b 
 n n 
lim
=
lim a
n n
n
an
=
an  0
6.
a n  bn

lim
= k
n
k an

lim a
n n
lim
an  0
9.
 0

n
a =  1
lim
n
 

10.
lim b  0
n  n
lim a  n
lim bn
n n
8.

x 

1+
lim


n 
n 
provided that
lim a  0
n n
5.
7.
lim b both exist, then
n n
 bn  = lim a n  lim bn
n
n
2.
4.
and
n
n
for any k  R
an
=
lim a
n  n
if -1 < a < 1
if a = 1
if a > 1
= ex
Note :
if a < -1,
 a n
oscillates infinitely
Pure Maths/Sequence/p.6
Example
1.
3n 2 + 5
n 2n 2 + n - 7
2.
3n 3 + 5n 2 - 1
n 2n 2 + n - 7
lim
=
lim
=
=
=
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3.
lim
n+1 -
n
n
=
=
4.
n2 + n - n
lim
n
=
= ____________________________________________________
5.


1 
1
1
 1 - 2   1 - 2  ...  1 - 2 
n 

2  
3 
n 
lim
= _______________________________________________________________________
= _______________________________________________________________________
2
n
1 
1  1
 1
6. lim
1 + +   + ... +  
n n 
 2
2  2





=
7.
=
=
1
1
 1

+
+ ... +

n  1  2  4 2  3  5
n (n  1)(n + 3) 
lim
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Exercise B
1.
Evaluate the following limits :
(a)

n2 
n 
n 
n + 1
lim
(1)
(b)
 1
2
n
 2 + 2 + ... + 2 
n  n
n
n 
lim
 1
 
 2
Pure Maths/Sequence/p.7
(c )
2.
lim
n
3 n+1 - 3 n
(0)
(d)
lim
n
2  4 2  8 2  ... 
18
24
6 (n + 1)
 12

+
+
+ ... +

n  1  3  4 2  4  5 3  5  6
n (n  2)(n + 3) 
2
(2)
 17 
 
 6
lim
Find
2n
Sandwich Theorem
Let
If
a n , bn , cn
be 3 sequences such that for all positive integers n,
lim a = lim cn =  , then
n n
n
bn  also converges to

i.e.
a n  bn  c n .
lim b =  .
n n
Example
1.

n
n
n 
+
+ ... +


n  n 2 + 1 n 2 + 2
n2 + n 
lim
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2.
n2
.
n 2 n
lim
Find
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3.
 n - 1 

  sin (n !) .
 2
n  n + 1

lim
Find
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Exercise C
1.
Evaluate the following limits :
(a)
 1
1
1 


+
+ ... +
2
2
n  n 2
(n + 1)
(2n) 
lim
(0)
Pure Maths/Sequence/p.8
(b)
2.

lim 
n 
2
+
2
n +1

2


2
n + n
2
+ ... +
2
n +2
Let n be a positive integer.
2n - 1
<
2n
(a) Show that
2n - 1
2n + 1
and deduce that
1 3
2n - 1
  ... 
<
2 4
2n
n
1
(r + 1) (r + 2) (r + 3)
xr =
(a)
If
(b)
Hence, find
1
.
2n + 1
2n - 1
1 3
   ... 
 .
2 4
2n 
lim
(b) Hence find
3.
(2)
, find x r - x r-1 for r > 1.
n
1
r=1 r (r + 1) (r + 2) (r + 3)
lim
n 
Theorem
Suppose a circle is divided into n sectors of equal area.
A
area of △AOB < area of sector AOB < area of △OAT
r
1 2
2
1 2 2
1 2
2
r sin
<
r 
<
r tan
2
n
2
n
2
n


Since
2
2
sin
<
n
n
lim
cos
n
2
n
O
<
lim

1
n 
cos
2
n

lim
B
r
2
< tan
n
2
sin
n
<
2
n
2
cos
n
2
n
sin
n 
2
n
2
n

T
1
= 1, by Sandwich Theorem,
lim
sin
2
n
= 1
2
n
n 
Note :
lim
1.
n
{ a n + bn } = lim a n
n
provided that both
lim a
n n
e.g. a n = n ; bn = 1 - n
but
2.
lim
n
and

+
lim b
n n
lim b
n n
exist
lim a
n n
and
lim b
n n
do not exist
{ a n + bn } = 1 .
The numbers of the sequences involved should be finite .
lim
n
{ a n + bn + … + k n } =
lim a
n n
+
lim b
n n
+ … +
lim k
n n
Pure Maths/Sequence/p.9
However,
1 + 2 + ... + n
n
n2
lim

1
n n 2
lim
+
2
n n 2
lim
+ … +
n
n n 2
lim
Principle of Monotonic Convergence
a n+1 > a n  for all n  N.
a n+1  a n  for all n  N .
A. A sequence is said to be monotonic (strictly) increasing if a n+1  a n
And it is said to be monotonic (strictly) decreasing if
a n+1  a n
A sequence is said to be monotonic if it is either increasing or decreasing .
e.g. If the sequence
 a n
is defined by
an =
1
1
1
+
+ ... + n
3 + 1 32 + 1
3 +1
for n  1 ,
determine whether it is monotonic or not.
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Classwork
 a n
a n-1
is the sequence defined by a1 = 1 and a n =
a n-1 + 1
for n > 1 .Determine whether it is
increasing or decreasing.
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 a n  is said to be bounded above by a real number M if a n  M for all n  N (or
M is an upper bound of the sequence  a n  ).
A sequence  a n  is said to be bounded below by a real number m if a n  m for all n  N (or
m is an lower bound of the sequence  a n  ).
A sequence  a n  is said to be bounded if and only if there exists a positive constant K such that
B. A sequence
an
 K or - K  a n  K for all n  N .
Note :
If a sequence is bounded above, it has infinitely many upper bounds.
e.g. 1.
an =
2
n2 + 1
 upper bound = __________________________________
lower bound = __________________________________
2.
an = 3 - n
 upper bound = __________________________________
Pure Maths/Sequence/p.10
lower bound = __________________________________
 upper bound = __________________________________
a n = sin n
3.
lower bound = __________________________________
Theorem
1.
If
lim a
n n
exists, then
 a n
forms a bounded sequence.
(Every convergent sequence is bounded.)
2.
A monotonic increasing sequence which is bounded above is convergent, i.e. its limit exists.
Similarly, a monotonic decreasing sequence which is bounded below is also convergent.
Example
1.
If u1 = 1 and un+1 =
un + 1
for n  1 , find
lim u .
n n
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2.
A sequence
 un 
(a) Prove that
(c) Hence, find
2
6 un + 6
is defined by u1 = 4 and un+1 =
2
un + 11
un > 3 for all n  N.
(b) Prove that
for n  1
un+1 - 3
un - 3
<
9
10
for n  1 .
lim u .
n n
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Pure Maths/Sequence/p.11
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(b) _________________________________________________________________________
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(c) _________________________________________________________________________
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Classwork (2001)
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W.12 Q.1, 5
H.W. : W.12 Q.2 , 4 W.13 Q.1(iv)(v)(vii) , 3, 4
Pure Maths/Sequence/p.12
Example
n
1

Define e = lim  1 +  .
n 
n
1

(a) lim  1 +

n 
3n 
n
=
=
2 

(b) lim  1 +

n 
n + 1
-n
=
=
=
W.14 Q.3
W.14.5 Q.4(c), 6, 9, 13, 17
H.W. : W.14 Q.5
W.14.5 Q. 2, 8, 10 , 11, 14, 15
END