1. A sequence can be thought of as a list of numbers
written in a definite order:
The number a1 is called the first term, a2 is the
second term, and in general an is the
nth term.
A function whose domain is the set of positive
integers
a1, a2, a3, a4, ,a‧‧‧ n,‧‧‧

2. NOTATION The sequence {a1 ,a2 ,a3 , . . .}
is also denoted by
{an} or ∞
=1}{ nna

3. In the following examples, we give three
descriptions of the sequence:
1. Using the preceding notation
2. Using the defining formula
3. Writing out the terms of the sequence
Example 1

4. Preceding
Notation
Defining
Formula
Terms of
Sequence
( 1) ( 1)
3
n
n
n − +
 
 
( 1) ( 1)
3
n
n n
n
a
− +
=
2 3 4 5
, , , ,...,
3 9 27 81
( 1) ( 1)
,...
3
n
n
n
 
− −  
 
− + 
  
( 1) ( 1)
3
n
n n
n
a
− +
=( 1) ( 1)
3
n
n
n − +
 
 
2 3 4 5
, , , ,...,
3 9 27 81
( 1) ( 1)
,...
3
n
n
n
 
− −  
 
− + 
  
{ } 3
3
n
n
∞
=
−
3
( 3)
na n
n
= −
≥
{ }0,1, 2, 3,..., 3,...n −

5. CONVERGENT AND DIVERGENT SEQUENCE
CONVERGENT----
The sequence (An) in R is said to converge if there exists a
number L for all ε > 0, there exists a natural number N such
that L−ε < an < L+ε for all n ≥ N.The sequence has a limit , we
can say that the sequence is convergent
DIVERGENT----
If a sequence has no limit , we can say that the sequence is
called divergent sequence

6. Example
●
Ex. Is the series convergent or divergent?
●
Sol. Since
it is a series with thus is divergent.
●
Ex. Find the sum of the series
●
Sol.
2 1
1
2 3n n
n
∞
−
=
∑
4/3 1r = >
2 1 1 1
1 1 1
4
2 3 4 3 4( )
3
n n n n n
n n n
∞ ∞ ∞
− − −
= = =
= =∑ ∑ ∑
1 1 1
.
1 3 2 4 ( 2)n n
+ + + +
× × +
L L
1
1 1 1 1 1 1 1
( ) (1 )
2 2 2 2 1 2
n
n
k
s
k k n n=
= − = + − −
+ + +
∑
3
lim
4
n
n
s s
→∞
⇒ = =

7. Example
●
Ex. Show that the series
is divergent?
●
Sol.
1
1 1 1 1
1
2 3 4n n
∞
=
= + + + +∑ L
1 2
1 1 1 2 1
1, 1 ,
2 3 4 4 2
s s= = + + > =
1 1 1 1 4 1 1 1 8 1
,
5 6 7 8 8 2 9 16 16 2
+ + + > = + + > =L
2
1
2
n
n
s > + → ∞

8. Example
●
Ex. Determine whether the series converges
or diverges.
●
Sol. The improper integral
So the series diverges.
2
1
1
ln ln
2
x x
dx
x
∞
∞  
= = ∞ 
 
∫
1
ln
n
n
n
∞
=
∑
1
ln
n
n
n
∞
=
∑

9. Test for convergence or divergence of:
2
3
n
na n
 
=  ÷
 
1
2
3
n
n
n
∞
=
 
 ÷
 
∑
1
1
2
( 1)
3
n
n
na
+
+
 
+  ÷
 
=
1
1
2
( 1)
2
3
3
n
n
n
n
n
a
a
n
+
+
 
+ 
=
 
 ÷

÷
 

1
1 2
3
n n
n
n
+ −
+  
=  ÷
 
1 2
3
n
n
+  
=  ÷
 
1
lim n
n
n
a
a
+
→∞
2 1
lim
3 n
n
n→∞
+
=
2
3
=
Since this ratio is less than 1, the
series converges.

10. Test for convergence or divergence of:
2
2
n n
n
a =
2
11
( 1)
2nn
n
a ++
+
=
2
2
1
1
( 1)
2
2
n
n
n
n
na
n
a +
+
+
=
( )
2
1 2
1 2
2
n
n
n
n+
+
= ×
2
2 1
( 1) 2
2
n
n
n
n +
+
=
1
lim n
n
n
a
a
+
→∞
Since this ratio is less than 1, the
series converges.
2
1 2n
n
n∞
=
∑
2
2
1 2 1
2
n n
n
+ +
= ×
2
2
1 2 1
lim
2 n
n
n
n
→∞
+ +
=
The ratio of the leading
coefficients is 1
1
2
=

11. LIMITS OF SEQUENCES
A sequence {an} is called:
– Increasing, if an < an+1 for all n ≥ 1,
that is, a1 < a2 < a3 < · · ·
– Decreasing, if an > an+1 for all n ≥ 1
– Monotonic, if it is either increasing or decreasing

12. Example
Ex. Find the limit
Sol.
.
1
2
1
1
1
lim
222






+
++
+
+
+∞→
nnnnn
L
nn
n
nnnnnnnn +
=
+
++
+
≥
+
++
+
+
+ 222222
111
2
1
1
1
LL
11
1
1
11
2
1
1
1
222222
+
=
+
++
+
≤
+
++
+
+
+ n
n
nnnnnn
LL
1
1
1
1
lim
1
lim,1
1
1
1
limlim
2
22
=
+
=
+
=
+
=
+ ∞→∞→∞→∞→
n
n
n
n
nn
n
nnnn

15. SEQUENCES

16. SEQUENCES

17. DECREASING SEQUENCES
Q-Show that the sequence is decreasing
Sequence is decresing beacause
and so an > an+1 for all n ≥ 1.
Example 11
3
5n
 
 
+ 
3 3 3
5 ( 1) 5 6n n n
> =
+ + + +

18. Q-Show that the sequence is decreasing.
2
1
n
n
a
n
=
+
2 2
1
( 1) 1 1
n n
n n
+
<
+ + +
We must show that an+1 < an,
that is,
The inequality is equivalent to the one we get by
cross-multiplication:
2 2
2 2
3 2 3 2
2
1
( 1)( 1) [( 1) 1]
( 1) 1 1
1 2 2
1
n n
n n n n
n n
n n n n n n
n n
+
< ⇔ + + < + +
+ + +
⇔ + + + < + +
⇔ < +

19. BOUNDED SEQUENCES
A sequence {an} is bounded:
– Above, if there is a number M such that an ≤ M
for all n ≥ 1
– Below, if there is a number m such that m ≤ an
for all n ≥ 1
– If it is bounded above and below

20. 1.The sequence an = n is bounded below (an > 0)
but not above.
2.The sequence an = n/(n+1) is bounded
because 0 < an < 1 for all n.

21. Theorem
A convergent sequence of real numbers is
bounded
Proof:
Suppose that lim (xn) = x and let ε := 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.