مواضيع المحاضرة: ABSOLUTE AND CONDITIONAL CONVERGENCE:
قراءة
عرض

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

1

ABSOLUTE AND CONDITIONAL CONVERGENCE:

ALTERNATING SERIES
In Sections 14.5 and 14.6 all the series we dealt with had positive terms in this section we
consider special types of series that have positive and negative terms.
Definition 1 ABSOLUTE CONVERGNCE The series ∑ ��� ∞��=1 is said to converge absolutely if
the series ∑| ��� ∞��=1 | converges
EXAMPLE 1 The series
∑ (−1)��+1
��2

��=1
= 1
12− 1
22+ 1
32− 1
42+ ⋯
Converges absolutely because ∑| (−1)��+1
��2 ∞��=1| = ∑ 1
��2 ∞��=1 converges

EXAMPLE 2 The series

∑ (−1)��+1
��2

��=1
= 1
1− 1
2+ 1
3− 1
4+ 1
5+ ⋯
Does not converge absolutely because ∑ 1
��
∞��=1 diverges
The importance of absolute convergence is given in the theorem below
Theorem 1. If ∑| ��� ∞��=1 | conv erges, then ∑ ��� ∞��=1 also converges That is :
Absolute convergence implies convergence
REMARK. The converse of this theorem is false That is, there are series that are convergent but
not absolutely convergent. We will see examples of this phenomenon sh ortly
Proof Since ��� ≤| ��� |we have
0≤ ���+| ���|≤ 2|��� |
Since ∑| ��� ∞��=1 | converges, we see that ∑ (���+| ��� ∞��=1| ) converges by the comparison test
Then since ���= (���+| ���|)−| ���|.∑ ��� ∞��=1 converges because it is the sum of two convergent
serie s

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

2

EXAMPLE 3 The series ∑ (−1)��+1/��2 ∞��=1 considered in Example 1 converges since it
converges absolutely
Definition 2 ALTERNATING SERIES A series in which successive terms have opposite signs
is called an alternating series
EXAMPLE 4 The series
∑ (−1)��+1
��2

��=1
= 1
1− 1
2+ 1
3− 1
4+ 1
5− 1
6+ ⋯
Is an alternating series
EXAMPLE 5 The series 1+ 1
2− 1
3− 1
4+ 1
5+ 1
6− ⋯ is not an alternating series because two
successive terms have the same sign
Let us consider the series of Example 4:
�= 1− 1
2− 1
3− 1
4+ 1
5− 1
6+ ⋯
Calculating successive partial sums, we find that
�1= 1. �2= 1
2 . �3= 5
6. �4= 7
12 . �5= 47
60 . ⋯
It is clear that this series is not diverging to infinity (indeed, 1
2≤ ��� ≤ 1) and that the partial
sums are getting ''narrowed down'' At this point it is reasonable to suspect that the series
converges. But it does not converge absolutely (since the series of absolute values is the harmonic
series), and we cannot use any of the tests o f the previous section since the terms are not
nonnegative. The result we need is given in the theorem below

Theorem 2 Alternating Series Test Let {���} be a decreasing sequence positive numbers such that
lim��→∞���= 0 Then the alternating series ∑ (−1)��+1 ��� ∞��=1 = �1 − �2+ �3− �4+ ⋯
converges
Proof Looking at the odd numbered partial sums of this series, we find that
�2��+1= (�1 − �2)+ (�3− �4)+ (�5− �6)+ ⋯ + (�2��−1 − �2��)+ �2��+1
Since ��� is decreasing all the terms in parentheses are nonnegative, so that �2��+1 ≥ 0 for every
n Moreover

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

3

�2��+3= �2��+3− �2��+2+ �2��+3= �2��+1− (�2��+2+ �2��+3).
And since �2��+2+ �2��+3 ≥ 0 . we have
�2��+3 ≤ �2��+1
Hence the sequence of odd numbered partial sums is bounded below by 0 and is decreasing a nd is
therefore convergent by Theorem 14.22 Thus �2��+1 converge to L Now let us consider the
sequence of even numbered partial sum We find that �2��+2= �2��+1− �2��+2→ 0 . and since
lim��→∞ �2��+2= lim��→∞�2��+1− lim��→∞�2��+2= ��− 0 = ��
So that the even partial sums also converge to L Since both the odd and even sums converge to L,
we see that the partial sums converge to L, and the proof is complete
EXAMPLE 6 The following alternating series are convergent by the alternating series test
(�)1− 1
2+ 1
3− 1
4+ 1
5− 1
6+ ⋯
(�)1− 1
√2+ 1
√3− 1
√4+ 1
√5
− 1
√6+ 1
√7− ⋯
(�) 1
��� 2− 1
�� � 3+ 1
��� 4− 1
��� 5+ 1
��� 6− ⋯
(�)1− 1
2+ 1
22− 1
23+ 1
24+ 1
25− 1
26+ 1
27 + ⋯
Definition 3 CONDITIONAL CONVERGNCE An alternating series is said to be conditionally
convergent if it is convergent but not absolutely converg ent
In Example 6 all the series are conditionally convergent except the last one, which is absolutely
convergent .It is not difficult to estimate the sum of a convergent alternating series We again
consider the series �= 1− 1
2+ 1
3− 1
4+ 1
5− ⋯
Suppose we wish to approximate S by is nth partial sum ��� Then
�− ���= ± ( 1
�+ 1− 1
�+ 2+ 1
�+ 3− 1
�+ 4+ ⋯ )= ���
But we can esti mate the remainder term ���∶
|���|=| [ 1
�+ 1− ( 1
�+ 2+ 1
�+ 3)− ( 1
�+ 4− 1
�+ 5)− ⋯ ]|≤ 1
�+ 1

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

4

That is, the error is less than the first term that we left out! For example,

|�− �20 | ≤ 1
21 ≈ 0∙0476
Theorem 3 :If is convergent alternating series with monotone dec reasing terms, then for any n,
| �− ��� |≤| ���+1 |(1)
EXAMPLE 7: The series ∑ (−1)��+1
(��+1) ∞��=1 = 1
���� 2− 1
�� �� 3+ 1
���� 4− 1
���� 5+ ⋯
Can be approximated by with an error of less than 1/��� (�+ 2) for example, with
�= 10 .1/��� (�+ 2) = 1/��� 12 ≈ 0∙4 Hence the sum
∑ (−1)��+1
(��+ 1)

��=1
= 1
��� 2− 1
�� � 3+ ⋯
Can be approximated by
�10 = 1
��� 2− 1
�� � 3+ 1
��� 4− 1
��� 5+ 1
��� 6 − 1
��� 7 + 1
��� 8 − 1
��� 9 + 1
��� 10 − 1
��� 11 ≈ 0∙7197 .
With an error of less than 0.4
By modifying Theorem 3 we can significantly improve on the last result
Theorem 4 :Suppose that the hypotheses of Theorem 3 hold and that, in addition, the sequence {|���−
���+1|} is monotone decreasing Let ���= ���−1− (−1)′′ 1
2��� Then
|�− ��� |≤ 1
2| ���− ���+1 |(2)
EXAMPLE 8: We can improve the estimate in Example 7. We nay approximate
∑ (−1)��+1 /��� (��+ 1)�� ∞��=1
1
���� 2− 1
�� �� 3+ 1
���� 4− 1
���� 5+ 1
����6 − 1
����7 + 1
����8 − 1
����9 + 1
���� 10 − 1
2����11 ≈ 0∙9282
With �= 10 (so that �+ 1= 11 ) �10 = �9− 1
2( 1
����11) .
Which is precisely the sum given above Thus
|�− �10|< 1
2| �10 − �11| = 1
2( 1
��� 11 − 1
��� 12 )≈ 0∙0073
This result is a considerable improvement

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

5

Note that in order to justify this result , we must verify that| ���− ���+1 |is monotone decreasing But
here| ���− ���+1|= 1
���� (��+1)− 1
���� (��+2)
Let �(�)= 1
���� (��+1)− 1
���� (��+2)
Then �′(�)= − 1
(��+1)����2(��+1)− 1
(��+2)����2(��+2)< 0
Thus f is a decreasing function , which shows that � (�+ 1)< �(�)
There is one fascinating fact about an alternating series that is conditionally but not absolutely convergent:
By reordering the terms of a conditionally convergent alte rnating series, the new series of real number
Let us illustrate this fact with the series
�= 1− 1
2+ 1
3− 1
4+ 1
5− 1
6+ ⋯
The odd - numbered terms sum to a divergent series:
1+ 1
3+ 1
5+ 1
7+ ⋯ (3)
The even - numbered terms are likewise a divergent series
− 1
2− 1
4− 1
6− ⋯ (4)
(��) Choose enough terms from the series (3) so that the sum exceeds 1.5. We can do so since the series
diverges 1+ 1
3+ 1
5= 1∙53333 ⋯
(����) Add enough negative terms form (4) so tha t the sum is now just under 1.5
1+ 1
3+ 1
5− 1
2= 1∙0333 ⋯
(������ ) Add more terms from (3) until 1.5 is exceeded
1+ 1
3+ 1
5− 1
2+ 1
7= 1
9+ 1
11 + 1
13 + 1
15 = 1∙5218
(����) Again subtract terms from (4) until the sum is under 1.5.
1+ 1
3+ 1
5− 1
2+ 1
7+ 1
9+ 1
11 + 1
13 + 1
15 − 1
4= 1∙2718
We continue the process ''converge '' to 1.5. Since the terms in each series are decreasing to 0, the amount
above or below 1.5 will approach 0 and the partial sums converge

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

6

We will indicate in Section 14.10 that without rearranging, we have

∑ (−1)��+1
��

��=1
= 1− 1
2+ 1
3− 1
4+ 1
5− 1
6+ ⋯ = ��� 2≈ 0∙693147 (5)
We again illustrate what can happen when the terms in the series (5) are rearranged † Consider the series
�∗ = 1− 1
2− 1
4+ 1
3− 1
6− 1
8+ 1
5− 1
10 − 1
12 + ⋯ (6)
This series is the rearrangement of the series ( 5) in which each odd term is followed by two even terms
Inserting paren theses, we have
�∗ = (1− 1
2)− 1
4+ (1
3− 1
6)− 1
8+ (1
5− 1
10)− 1
12 + ⋯
= 1
2− 1
4+ 1
6− 1
8+ 1
10 − 1
12 + ⋯
= 1
2(1− 1
2+ 1
3− 1
4+ 1
5− 1
6+ ⋯ = 1
2 ��� 2

PROBLEMS

In problems 1 - 30, determine whether the given series is absolutely convergent, conditionall y convergent
or divergent
1∙∑ (−1)��

��=�
2∙ ∑ (−1)��+1
���

��=�
3∙ ∑ (−1)��
�������

��=�

4∙∑ (−1)��

��
��

��=�
5∙∑ (−1)�� ��
�����

��=�
6∙ ∑ (−1)����� ��
��

��=�

7∙ ∑ (−1)��+1

��� − �

��=�
8∙ ∑ ����� ����


��=�
9∙ ∑ ����� ����


��=�

10 ∙∑ (−�)��

��!

��=�
11 ∙∑ ��!
(−�)��

��=�
12 ∙ ∑ (−�)��
���

��=�

13 ∙ ∑ ���

(−�)��

��=�
14 ∙ ∑ (−�)��+�
√��(��− �)

��=�
15 ∙∑ (−�)�����
���+ �

��=�

Lecture no .2-5 by Hussein J. A bdulHussein

Advanced Calculus I I Al Muthanna University, Coll ege of Science

7

16 ∙∑

����� (����
� )
���

��=�
17 ∙ ∑
����� (����
� )
���

��=�
18 ∙ ∑ (−�)�� (��+ �)
��(��+ �)

��=�

19 ∙ ∑ (−�)�� ��(��+ �)
(��+ �)�

��=�
20 ∙ ∑ (−�)�� ��(��+ �)
(��+ �)�

��=�
21 ∙ ∑ (−�)�����
��

��=�

22 ∙ ∑ (−�)��+�

��!

��=�
23 ∙ ∑ (−�)������
��!

��=�
24 ∙∑ (−�)��√��
��+ �

��=�

25 ∙∑ (−�)��(���+ �)
���+ �

��=�
26 ∙∑ (−�)��
√��� �� �

��=�
27 ∙∑ (−�)�����
�+ ���

��=�

28 ∙∑ (−�)��


��=�
(1+ 1
��)
��
29 ∙ ∑ (−�)��
��√��� ��

��=�
30 ∙ ∑ (−�)�� ���
���+ ����+ ��− �

��=�

In Problems 32 - 37, use the result of Theorem 3 or Theorem 4 to estimate the given sum to within the
indicated accuracy
32 ∙∑ (−�)��+�
��!

��=�
;����� < 0∙001 33 ∙∑ (−�)��+�
���

��=�
;����� < 0∙01
34 ∙∑ (−�)��+�
���

��=�
;����� < 0∙0001 35 ∙ ∑ (−�)��+�
����� ��

��=�
;����� < 0∙05
36 ∙ ∑ (−�)��+�
���

��=�
;����� < 0∙0001 37 ∙ ∑ (−�)��+�
√��

��=�
;����� < 0∙1
38. Find the first ten terms of a rearrangement of the series ∑ (−�)��+�
�� ∞��=� that converges to 0
39. Find the first ten terms of a rearrangement of the series ∑ (−�)��+�
�� ∞��=� that converges to 0.3
40.Explain why there is no rearrangement of the series ∑ (−�)��
��� ∞��=� that converges to −�
41.Prove that if ∑ ���� ∞��=� is a convergent series of nonzero terms , then ∑ �/|���� |∞��=� diverges
42. Show that if ∑ ���� ∞��=� is absolutely convergent, then ∑ ����� ∞��=� ��� convergent for any integer �≥ 1
43. Give an example of a sequence {����} such that ∑ ����� ∞��=� converges but ∑ ���� ∞��=� diverges
44. Give an example of a sequence {����} such that ∑ ���� ∞��=� converges but ∑ ����� ∞��=� diverges


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