# POWER SERIES

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مواضيع المحاضرة: POWER SERIES
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### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

1

### POWER SERIES

In previous sections we discussed infinite series of real numbers Here we discuss series of
functions
Definition :
(��) A power series in x is a series of the form
∑ �������� ∞��=� = ��0+ ��1��+ ��2��2+ ⋯ + ��������+ ⋯ (1)
(����) A power series in (��− ��0 )is a series of the form
∑ ����

��=�
(��− ��0 )��= ��0+ ��1 (��− ��0 )+ ��2 (��− ��0 )2+ ⋯ + ���� (��− ��0 )��+ ⋯ . (2)
Where ��0 is a real number
A power series in (��− ��0 ) can be converted to a power series in u by the change of
variables � = ��− ��0 Then ∑ ���� ∞��=� (��− ��0 )��= ∑ �������� ∞��=� for example, consider
∑ (��− 3 )��
��!

��=�
(3)
If � = ��− 3 . then the power series in ( ��− 3 ) given by (3) can be written as ∑ ����
��!
∞��=�
Which is a power series in u
Definition : CONVERGENCE AND DIVERGENCE OF A POWER SERIES
(��) A power series is said to converge at x if the series of real numbers ∑ �������� ∞��=� converges
Otherwise, it is said to diverge at x
(����) A power series is said to converge in a set D of real numbers if it converges, for every
real number x in D
EXAMPLE 1 For what real numbers does the power series
∑ ����
��� ∞��=� = 1+ x
3+ ��2
��+ ��3
��+ ⋯ Converg e?
Solution. The � �ℎ term in this series is ����
3�� Using the ratio test, we find that

### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

2

### lim��→∞

|����+ 1 |
|���� |= lim��→∞
|����+1 |
3��+1
|����
3�� |
= lim��→∞|��
3|=| ��
3 |
We put in the absolute value bars s ince the ratio test only applies to a series of positive terms
However, as the example shows, we can use the ratio test to test for the absolute convergence of
any series of nonzero terms by inserting absolute value bars, thereby making all the terms posit ive
Thus the power series converges absolutely if| ��
3|< 1��| ��|< 3 and diverges if| ��|> 3 The
case| ��|= 3 has to be treated separately for = 3
∑ ����
��� ∞��=� = ∑ ����
��� ∞��=� = ∑ 1�� ∞��=�
Which diverges For �� = −3 , ∑ ����
��� ∞��=� = ∑ (−1)�� ∞��=� .
Which also diverges Thus the series converges in the open interval (−3.3 ) We will show in
Theorem 1 that since the series diverges for �� = 3. it diverges for| ��|> 3.�� that conditional
convergence at any x for which| ��|> 3 is ruled out
EXAMPLE 2 For what of x does the series ∑ ����/(��+ �) ∞��=� converge?
Solution Here ����= ����/(�+ 1) so that

### lim��→∞

�+ 1
�+ 2= 1
lim��→∞
|����+ 1 |
|���� |= lim��→∞ |
����+1
�+ 2
����/(�+ 2|=| ��|lim��→∞
�+ 1
�+ 2=| �� |
Thus the series converges absolutely for| ��|< 1 and diverges for| ��|> 1 �� �� = 1. Then
∑ ����
��+�
∞��=� = ∑ 1
��+�
∞��=�
Which diverges since this series is the harmonic series �� �� = −1. Then
∑ ����
��+ �

��=�
= ∑ (−1)��
�� + � = �− �
�+ �
�− �
�+ ⋯ .

��=�

### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

3

### Which converges conditionally by the alternating series test .In sum , the power series

∑ ����
��+�
∞��=� converges in the half open interval (−1 .1)
The following theorem is of gre at importance in determining the range of values over which a
power series converges
Theorem 1
(��) If ∑ �������� ∞��=� converges a � ��0.��0≠ 0. then it converges absolutely at all x such that
|��|<| ��0 |
(����) If ∑ �������� ∞��=� diverges at ��0 then it diverges at all x such that| ��|>| ��0 |
Proof
(��) Since ∑ ���� ∞��=� ��0�� converges , ������0��→ 0 as �� → ∞ by Theorem 14.4.2. This implies that
for all k sufficiently large | ������0��|< 1 Then if| ��|<| ��0 |an if k is sufficiently large,

|��������|=| ����
��0������
��0��| = | ������0��||��
��0
|
��

Since if| ��|<| ��0|.|��
��0|< 1. and the geometric series ∑| ��
��0|
��
∞��=� converges Thus ∑| �������� ∞��=� |
converges by the comparison test

(����)Suppose| ��|>| ��0 |and ∑ ������0�� ∞��=� diverges If ∑ ������0�� ∞��=� did converge, then by
part (��).∑ ������0�� ∞��=� would also converge This contradiction completes the proof of the theorem
Theorem 1 is very useful for it enables us to place all power series in one of three categories:
Definition 3 RADIUS OF CONVERGNCE
Category 1: ∑ �������� ∞��=� converges only at 0
Category 2: ∑ �������� ∞��=� converges for all real numbers
Category 3: There exists appositive real number R, called the radius of convergence of the power
series, such that ∑ �������� ∞��=� converg es if| ��|< �� and diverges if| ��|> �� ����� = �� ���� ��� = −��
the series may converge or diverge
We can extend the notion of radius of convergence to Categories 1 and 2:

### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

4

### 1∙ In Category 1 we say that the radius of convergence is 0

2∙ In Category 2 we say that the radius of convergence is ∞
NOTE. The series in Examples 1 and 2 both fall into Category 3 Example 1, �� = 3 ; and in
Example 2, , �� = 1
EXAMPLE 3 For what values of x does the series ∑ ��!���� ∞��=� converge?
Solution Here
lim��→∞|����+1
����
|= lim��→∞|(�+ 1)!����+1
�!����| =| ��|lim��→∞(�+ 1)= ∞
So that if �� ≠ 0. the series diverges Thus �� = 0 and the series falls into Category 1
EXAMPLE 4 For what values of x does the series ∑ ����/��! ∞��=� converg es?
Sol .: Here lim��→∞|����+1
����| = lim��→∞|����+1(��+1)!
����
��!
|=| ��|lim��→∞ 1
��+1= 0.
So that the series converges for every real number x Here �� = ∞ and the series falls into
Category 2
In going thr ough these examples, we find that there is an easy way to calculate the radius of
convergence .
Theorem 2. Consider the power series ∑ �������� ∞��=� and suppose th at lim��→∞ =| ����+1
���� | exists and is
equal to L or that lim��→∞|����|1/�� exists and is equal to L
(��) If ��= ∞ . then �� = 0 and the series falls into Category 1
(����) If ��= ∞ . the �� = ∞ and the series fal ls into Category 2
(������ ) If 0< ��< ∞ . the �� = 1/�� and the series falls into Category 3
Definition 4 INTERVAL OF CONVERGENCE The interval of convergence of a power series
is the interval over which the power series converges
Using Theo rem 2, we can calculate the interval of convergence of power series in one or two
steps
(��) Calculate �� If �� = 0 . the series converges only at 0; and if �� = ∞ . the interval of
convergence is (−∞ .∞ )

### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

5

(����) If 0< �� < ∞ . check the values �� = −�� and �� = �� T hen the interval of convergence
is(−��.��).[−��.��).[−��.��].�� [ −��.��]. depending on the convergence or divergence of the
series at �� = �� and �� = −��
NOTE In example 1, the interval of convergence is (−3.3) and i n Example 2 the interval
of convergence is [−1. 1)
EXAMPLE 5 : Find the radius of convergence and interval of convergence of the power
series ∑ 2������/�� (��+ 2) ∞��=0
So L. : Here ����= 2��/�� (�+ 2) and
��= lim��→∞ |����+1
����
|= lim��→∞ |
2��+1
�� (�+ 3)
2��
�� (�+ 2)
|= 2lim��→∞
�� (�+ 2)
�� (�+ 3)= 2
Thus R = 1
L= 1
2 For x= 1
2 .
∑ 2������
�� (��+ 2)

��=0
= ∑ 1
�� (��+ 2)

��=0
.
Which diverges by comparison with the harmonic series since 1
����(��+2)> 1
��+2 ��� �� ≥ 1
�� �� = − 1
2 . then ∑ 2������
����(��+2)
∞��=0 = ∑ (−1)��
����(��+2)
∞��=0
Which converges by the alternating series test Thus the interval of convergence is [ − 1
2.1
2)
EXAMPLE 6 Find the radius of convergence and interval of convergence of the power series
∑ (−1)��(��− 3)��/(��+ 1)2 ∞��=0
Sol .: We make the substitution � = ��− 3 The series then becomes ∑ (−1)�����/(��+ 1)2 ∞��=0 and
��= lim��→∞ |����+1
����| = lim��→∞
(��+1)2
(��+2)2= 1 , So that �� = 1 �� � = 1 .
∑ (−1)������
(��+1)2 ∞��=0 = ∑ 1
(��+1)2 ∞��=0 , Which converges �� � = 1 .

(��+ 1)2

��=0
= ∑ (−1)��
(��+ 1)2

��=0
.

### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

6

### Which also converges Thus the interval of convergence of the

series ∑ (−1)������
(��+1)2 ∞��=0 ��� [−1 .1] Since � = ��− 3 . the original series converges for −1≤ ��−
3≤ 1.�� 2 ≤ �� ≤ 4 Hence the inte rval of convergence is [2 .4]
EXAMPLE 7 Find the radius of convergence and interval of convergence of the power
series ∑ ��2��= 1+ ��2+ ��4+ ⋯ ∞��=0 .
So l.: ∑ ��2��= 1+ 0∙��+ 1∙��2+ 0∙��3+ 1∙��4+ 0��5+ 1∙��6+∙⋯ ∞��=0
This example illustrates the pitfalls of blindly applying formulas We have ��0= 1.��1= 0.��2=
1.��3= 0.⋯ Thus the ratio ����+1
���� is even and is undefined if n is odd The simplest thing to do here
is to apply the ratio test directly T he ratio of convergence terms is ��2��+2
��2�� = ��2 Thus the series
converges if| ��|< 1. Then ��2= 1 and the series diverges Finally. the interval of convergence
is (−1.1)
PROBLEMS
Find the radius of convergen ce and interval of convergence of the given power series
1∙∑ ����
6��

��=0
2∙ ∑ (−1)������
8��

��=0
3∙∑ (��+ 1)��
3��

��=0

4��

��=0
5∙∑ (3��)��

��=0
6∙∑ ����
��2+ 1

��=0

��3+ 3

��=0
8∙∑ ����
(���� )2

��=2
9∙ ∑ (��− 17 )��
��!

��=0

### 10 ∙ ∑ ����

�� �� ��

��=2
11 ∙ ∑

��=0
��2�� 12 ∙ ∑

��=1
��2��
��
13 ∙ ∑ (−1)����2��
����

��=1
14 ∙ ∑ ���� ��
�� (��+ 1)

��=1
15 ∙ ∑ (−1)������ ��
√��+ 1

��=0

### 16 ∙ ∑ ����

����

��=1
∗17 ∙∑ ����
(�� ��)��

��=1
[ ����� ���� �ℎ� ���� ���� ]

### Lecture no .2-6 by Hussein J. Ab dulHussein

Advanced Calculus I I Al Muthanna University, College of Science

7

### 18 ∙∑ 3������

��5

��=1
19 ∙∑ (−2��)��
��4

��=1
20 ∙ ∑ (2��+ 3)��
��!

��=0

### 21 ∙ ∑ (2��+ 3)��

5��

��=0
22 ∙∑ (2��− 5)��
32��

��=0
23 ∙ ∑ (��
15 )
�� ∞
��=0
����
24 ∙ ∑ (−1)����2��

��=0
25 ∙∑ (−1)����2��+1

��=0
26∙ ∑ (�� ��)(��+ 3)��
��+ 1

��=1

∗27 ∙ ∑ ����(��+ 1)�� ∞��=1 28 ∙∑ ����
��! ���� ∞��=1
29 ∙∑ (��+ 10 )��
(��+ 1)3��

��=0
30 ∙∑ ��!
����

��=1
���� 31 ∙ ∑ [1+ (−1)��]
��

��=0
����
32 ∙ ∑ [1+ (−1)��]
��!

��=0
���� 33 ∙ ∑ [1+ (−1)��]
��

��=1
����

رفعت المحاضرة من قبل: Cruz Maldanado
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