This document discusses sequences and their properties. It defines a sequence as a list of numbers written in a definite order. The nth term of a sequence is denoted as an. It provides examples of describing sequences using notation, defining formulas, and listing terms. It defines convergent and divergent sequences and gives examples testing for convergence or divergence. It also discusses bounded sequences and decreasing sequences, giving examples and proofs.

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.