series module 4 -september 2010
TRANSCRIPT
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29/12/20105. Series I 1
Infinite Series
Module 4
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29/12/20105. Series I 2
Learning Objectives
At the end of the module, students able to
• determine whether a series converges or
diverges using suitable technique for test
of convergence.
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29/12/20105. Series I 3
Techniques to determine whether
series converge or diverge.na
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29/12/20105. Series I 4
Theorem
If the series is convergent, then 1n
na .0lim nn
a
Not always
true
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29/12/20105. Series I 5
If the series is convergent, then 1n
na .0lim nn
a
Always true
Example:
12
3
n n
03
lim2nn
then1
2
3
n nconverges,
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29/12/20105. Series I 6
If the series is convergent, then 1n
na .0lim nn
a
Not always true
Example: The Harmonic Series
1
3
n n
03
limnn
however,1
3
n ndiverges.
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29/12/20105. Series I 7
The Test for Divergence
If does not exist or if then the
series is divergent.
nn
alim ,0lim nn
a
1n
na
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29/12/20105. Series I 8
Example 1
Show that the series diverges. 1
2
2
45n n
n
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29/12/20105. Series I 9
Example 1 - solution
05
1
5
1lim
45lim
2
42
2
nnn n
n
By the Test for Divergence, the series
diverges.
Reminder:What is the difference between sequenceand series?
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29/12/20105. Series I 10
Theorem
If and are convergent series, then so are
the series
na nb
),(),(, nnnnn babaca and and
1 11
1 11
1 1
)()iii(
)()ii(
)i(
n n
nn
n
nn
n n
nn
n
nn
n n
nn
baba
baba
acca
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29/12/20105. Series I 11
From the Test for Divergence
• If the series may converge or diverge.
• If the series must diverge.
0lim nn
a
0lim nn
a
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29/12/20105. Series I 12
Example 2
1
2n
n
12
2
12
3
n n
nn
12
2
45n n
n
Determine whether the following series converge or
diverge?
(i)
(ii)
(iii)
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29/12/20105. Series I 13
Example 2 - solution
11
1
)2(2nn
n n
12
2
12
3
n n
nn
(i)
(ii)
012)2(lim 01n
n, the series diverges
02
1
2lim
12
3lim
2
2
2
2
n
n
n
nn
nn
, the series
diverges.
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29/12/20105. Series I 14
Example 2 - solution
(iii)
05
1
5
1lim
5lim
45lim
22
2
nnn n
n
n
n
Therefore, the series diverges.
12
2
45n n
n
12
2
45n n
n
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29/12/20105. Series I 15
Practice
(i)
1
)1
1(n
n
n
By using the divergence test, show that the
series diverges.
(ii)
1 3k k
k
1kke
k(iii)
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29/12/20105. Series I 16
Remark
If , then the series may converges
or diverges.
Therefore, need to use other test techniques
to confirm.
0lim nn
a
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29/12/20105. Series I 17
Exercise
Exercise 12.2, Page 756No. 21, 25, 29, 32
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29/12/20105. Series I 18
The Comparison Test
A series of positive terms is convergent if its terms
are less than the corresponding terms of a positive
series which is known to be convergent.
The BIG…
convergeThe
SMALL
subset
MUST
converge
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29/12/20105. Series I 19
The Comparison Test
Similarly, the series is divergent if its terms are
greater than the corresponding terms of a series
which is known to be divergent.The SMALL
diverge.The
BIG
???
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29/12/20105. Series I 20
Example 4
Determine whether the series
12 342
5
n nn
converges or diverges.
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29/12/20105. Series I 21
Example 4 - solution
22 2
5
342
5
nnn(based on the
denominator)
Since the series is convergent.1
21
2
1
2
5
2
5
nn nn
By Comparison Test
12 342
5
n nnconvergent.
WHY?
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29/12/20105. Series I 22
Example 5
1
ln
n n
nDetermine whether converges
or diverges.
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29/12/20105. Series I 23
Example 5 - solution
3,1ln nn
nn
n 1ln
The series is divergent. (Harmonic series) 1
1
n n
By Comparison Test
1
ln
n n
ndiverges.
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29/12/20105. Series I 24
Practice
By the Comparison Test, determine whether
the series converges or diverges.
1 21
14
4)ii(
4
2)i(
n
n
k
n
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29/12/20105. Series I 25
Remark 1
There are two steps required for the comparisontest to determine whether a series withpositive term converges or diverges:
ka
• Guess whether the series converges or diverges .
• Find a series that proves the guess is correct.
ka
ku
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29/12/20105. Series I 26
Remark 1 - continue
Find a series that proves the guess is correct.
1. If the guess that is divergence, must finda divergence series where .
2. If the guess that is convergence,must finda convergent series whose terms
ku
ka
kukk ua
ka
.kk auku
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29/12/20105. Series I 27
Remark 2
The Comparison Test only applies to series
with nonnegative terms.
Try this test as last resort; other tests are
often easier to apply.
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29/12/20105. Series I 28
The Limit Comparison Test
Given the positive terms series
1n
naand
.1n
nb
If cb
a
n
n
nlim
where c is a finite number and c > 0, then either
both series converge or both diverge.
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29/12/20105. Series I 29
Example 6
By the Limit Comparison Test, determine
whether the series, converges or diverges.
1 12
1
nn
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29/12/20105. Series I 30
Example 6 - solution
1 12
1
nn
Step 1: Find a new series with from the given series.
1 2
1
nn
an
bn
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29/12/20105. Series I 31
Example 6 - solution
1 12
1
nn
Step 2: Compute
1 2
1
nn
an bn
n
n
n b
alim
0112
2lim
2
112
1
limlimn
n
n
n
n
nn
n
n b
a
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29/12/20105. Series I 32
Example 6 - solution
1 12
1
nn
1 2
1
nn
Since the series is a convergent geometric
series, by the Limit Comparison Test, the series
1 2
1
nn
1 12
1
nn
converges.
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29/12/20105. Series I 33
Example 7
15
2
5
32
k k
kk
By the Limit Comparison Test, determine whether
the series converges or diverges.
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29/12/20105. Series I 34
Example 7- solution
15
2
5
32
k k
kk
The highest power of the numerator is 2k2 and
the highest power of the denominator is k5/2 .
Thus,
21
22and
5
32
5
2
5
2
kk
kb
k
kka kk
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29/12/20105. Series I 35
Example 7- solution
21
22and
5
32
5
2
5
2
kk
kb
k
kka kk
Then
)2
.(5
32lim
25
32
lim21
21
5
25
2
k
k
kk
k
k
kk
kk
1
15
2
32
lim52
32lim
5
5
23
25
k
k
k
kk
kk
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29/12/20105. Series I 36
Example 7- solution
21
22and
5
32
5
2
5
2
kk
kb
k
kka kk
Since it is divergent p-series,
therefore by the Limit Comparison Test, 1
5
2
5
32
k k
kk
diverges.
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29/12/20105. Series I 37
Practice
By the Limit Comparison Test, determine whether
the following series converges or diverges.
1
1k7
2
13
5 (ii)
88
624 (i)
kk
kk
kk
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29/12/20105. Series I 38
Remark
The Limit Comparison Test is easier to apply than
the Comparison Test, however it requires skill in
choosing the series for comparison.1n
nb
We must know whetherthis series converges ordiverges.
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29/12/20105. Series I 39
Exercise
Exercise 12.4, Page 770 – 771
No. 1, 3-32 (odd numbers only)
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29/12/20105. Series I 40
The Integral Test
Suppose f (x) is a continuous, positive, decreasing
function on and let ),1[ ).(nfan
Then the series is convergent if and only if
the improper integral is convergent.
That is
1n
na
1
)( dxxf
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29/12/20105. Series I 41
The Integral Test
1n
na
1
)( dxxf(i) If is convergent, then is
convergent.
(ii) If is divergent, then is
divergent.
1
)( dxxf1n
na
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29/12/20105. Series I 42
Example
Test the series
12 1
1
n n
for convergence or divergence.
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29/12/20105. Series I 43
Example - solution
Let
12 1
1
n n
,1
1)(
2xxf
We know that this function is continuous,
positive, and decreasing on
So, can use the Integral Test.
).,1[
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29/12/20105. Series I 44
Example - solution
t
tdx
xdx
x1
2
1
2 1
1lim
1
1
t
tx
1
1tanlim
4421tantanlim 11 t
t
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29/12/20105. Series I 45
Example - solution
1
2 1
1dx
xThus, is convergent integral,
so by the Integral Test, the series
is convergent.
12 1
1
n n
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29/12/20105. Series I 46
Exercises
Exercise 12.3, Page 765
No. 1-24 (odd only)
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29/12/20105. Series I 47
An Alternating Series
An alternating series is a series whose terms
are alternately positive and negative.
...6
1
5
1
4
1
3
1
2
11
)1(
1
1
n
n
n
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29/12/20105. Series I 48
The Alternating Series Test
(AST)
If the alternating series
.convergent is series the then
(ii)
all fori.e., (i)
satisfies
n0lim
)...(,
)0(...)1(
1231
4321
1
1
n
nn
n
n
n
n
b
nbbbbb
bbbbbb
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29/12/20105. Series I 49
Example 1
...6
1
5
1
4
1
3
1
2
11
)1(
1
1
n
n
n
Determine whether the series converge or
diverge by AST.
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29/12/20105. Series I 50
Example 1 - solution
...6
1
5
1
4
1
3
1
2
11
)1(
1
1
n
n
n
01
limlim (ii)
1
1
1because (i)
n
1
nb
nnbb
nn
nn
Therefore, by the Alternating Series Test
(AST) the series is convergent.
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29/12/20105. Series I 51
Example 2
Determine whether the series converges
or diverges.
1
1
)1(
3)1(
k
k
kk
k
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29/12/20105. Series I 52
Example 2 - solution
1
1
)1(
3)1(
k
k
kk
k
(i) To show bk+1< bk
)1(
3
)2)(1(
4
kk
k
kk
k
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29/12/20105. Series I 53
Example 2 - solution
(i) To show bk+1< bk
)1(
3
)2)(1(
4
kk
k
kk
k
)2)(1(
)3)(2(
)2)(1(
)4(
kkk
kk
kkk
kk
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29/12/20105. Series I 54
Example 2 - solution
(i) To show bk+1< bk
)2)(1(
)65(4 22
kkk
kkkk
0)2)(1(
)6(
)2)(1(
6
kkk
k
kkk
k
therefore, bk+1< bk
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29/12/20105. Series I 55
Example 2 - solution
(ii) To show 0)1(
3lim
kk
k
k
01
lim3
lim)1(
3lim
1
31
2
2
k
kk
kkk kk
k
kk
k
since condition (i) and (ii) are satisfied, by
AST the series converges.
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29/12/20105. Series I 56
Practice
By the Alternating Series Test, determine
whether the following series converges
or diverges.
1
1
12
1
)1( (ii)
)1( (i)
k
kk
n
n
e
n
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29/12/20105. Series I 57
Remark
The Alternating Series Test only applies to
the alternating series.
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29/12/20105. Series I 58
Exercises
Exercise 12.5, Page 775 – 776
No. 1, 2-20 (odd numbers only), 33
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29/12/20105. Series I 59
Type of Convergence
For an alternating series, there are two types
of convergence:
(i) Absolutely Convergence;
(ii) Conditionally Convergence.
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29/12/20105. Series I 60
Absolutely Convergence
A series is called absolutely
convergent
if the series of absolute value
is convergent.
1
)1(n
n
n a
1
)1(n
n
n a
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29/12/20105. Series I 61
Example 1
The series ...4
1
3
1
2
11
)1(222
12
1
n
n
n
is absolutely convergent because
...4
1
3
1
2
11
1)1(222
12
12
1
nn
n
nnis convergent.
Why?
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29/12/20105. Series I 62
Conditionally Convergence
A series is called conditionally
convergent if the series is convergent
however the series diverges
(i.e., not absolutely convergent).
n
n a)1(
1
)1(n
n
n a
![Page 63: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/63.jpg)
29/12/20105. Series I 63
Example 2
The series ...4
1
3
1
2
11
)1(
1
1
n
n
n
is convergent , however the series
...4
1
3
1
2
11
1)1(
1 1
1
n n
n
nnis divergent.
...4
1
3
1
2
11
)1(
1
1
n
n
n
conditionally
convergent.
![Page 64: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/64.jpg)
29/12/20105. Series I 64
Theorem
If a series is called absolutely convergent,
then the series is convergent.
1
)1(n
n
n a
thenconvergent absolutely is )1( If1n
n
n a1
)1(n
n
n a
is convergent.
![Page 65: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/65.jpg)
29/12/20105. Series I 65
The Ratio Test
,1lim 1 La
a
n
n
n1
)1(n
n
n a(i) If then the series
is absolutely converges (and therefore
converges).
1
)1(n
n
n aGiven an alternating series
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29/12/20105. Series I 66
The Ratio Test
,1lim 1 La
a
n
n
n1
)1(n
n
n a(ii) If then the series
diverges.
Note: The series maybe
convergent or divergent.
1
)1(n
n
n aGiven an alternating series
1
)1(n
n
n a
![Page 67: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/67.jpg)
29/12/20105. Series I 67
The Ratio Test
,1lim 1
n
n
n a
a(iii) If then the Ratio Test
is inconclusive.
1
)1(n
n
n aGiven an alternating series
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29/12/20105. Series I 68
Example 1
Using the ratio test, determine whether the
given series converge or diverge.
1
1
1
12
1 (iii)
4
)!2( (ii)
!
1 (i)
k
kk
n
k
k
n
![Page 69: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/69.jpg)
29/12/20105. Series I 69
Example 1- solution
101
1lim
)!1(
!limlimlim
!
1 (i)
!1
)!1(1
1
1
n
n
n
a
a
n
n
nn
n
nn
n
n
n
By the Ratio Test, the series converges.
![Page 70: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/70.jpg)
29/12/20105. Series I 70
Example 1- solution
By the Ratio Test, the series diverges.
)12)(22(lim4
1
4
1.
)!2(
)!22(lim
4
]!2[4
)]!1(2[
limlim
4
)!2( (ii)
11
1
kkk
k
k
k
a
a
k
kk
k
k
kk
k
k
kk
![Page 71: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/71.jpg)
29/12/20105. Series I 71
Example 1- solution
By the Ratio Test, the series may be
converge or diverges.
112
12lim
12
1
1)1(2
1
limlim
12
1 (iii)
1
1k
k
k
k
k
a
a
k
kkk
k
k
![Page 72: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/72.jpg)
29/12/20105. Series I 72
Example 2
Determine whether the series below, is
absolutely convergent, conditionally
convergent or divergent.
1
3
3)1(
nn
n n
![Page 73: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/73.jpg)
29/12/20105. Series I 73
Example 2 - solution
1
3
3)1(
nn
n nn
n
n
na
3)1(
3
31
3
3
1
31
1 3.
3
)1(lim
3)1(
3
)1()1(
limlimn
n
n
n
a
a n
nn
n
n
n
n
nn
n
n
![Page 74: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/74.jpg)
29/12/20105. Series I 74
Example 2 - solution
3
1lim
3
1
)1(
3
1lim
3.
3
)1(lim
3
3
3
3
31
3
n
n
n
n
n
n
n
n
n
nn
By the Ratio Test, the series converge
absolutely ( the series is also convergent).
![Page 75: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/75.jpg)
29/12/20105. Series I 75
Remark
May like to use the Ratio Test,when an
involves factorials or nth power.
![Page 76: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/76.jpg)
29/12/20105. Series I 76
Exercises
Exercise 12.6, Page 781-782
No. 1, 3,7,9
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29/12/20105. Series I 77
Exercises
Use the Ratio Test to determine whether the
series converges. If the test is inconclusive,
then say so.
12
1
13
1
15 4. )
2
1( 3.
! 2.
!
4 1.
nk
k
nk
k
n
nk
n
n
k
![Page 78: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/78.jpg)
29/12/20105. Series I 78
The Root Test
.convergent is
series then the,1lim If (i)1
nn
nn
n aLa
divergent. is
series then the,limor ,1lim If (ii)1
nnn
nn
nn
n aaLa
ve.inconclusi,1lim If (iii)n
nna
![Page 79: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/79.jpg)
29/12/20105. Series I 79
Power Series
It is a series of the form
1n
n
n xa(i)
1
)1(n
n
n
n xa(ii)
(iii)
1
)()1(n
n
n
n cbxa
![Page 80: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/80.jpg)
29/12/20105. Series I 80
Power Series
You need to know how to find
radius of convergence and interval of
convergence.
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29/12/20105. Series I 81
Example 1
Use the Root Test to determine whether the
following series converge or diverge.
1
2
))1(ln(
1 (ii)
12
54 (i)
nn
k
k
n
k
k
![Page 82: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/82.jpg)
29/12/20105. Series I 82
Example 1 - solution
1212
54lim
12
54lim
12
54 (i)
2
k
k
k
k
k
k
k
k
k
k
k
k
By the Root Test the series diverges.
![Page 83: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/83.jpg)
29/12/20105. Series I 83
Example 1 - solution
10)1ln(
1lim
))1(ln(
1lim
))1(ln(
1 (ii)
1
nn
n
nn
nn
nn
By the Root Test the series converges
(Absolutely convergence).
![Page 84: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/84.jpg)
29/12/20105. Series I 84
Example 2
Test the convergence of the series
.23
32)1(
1
1
k
k
k
k
k
![Page 85: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/85.jpg)
29/12/20105. Series I 85
Example 2 - solution
.23
32)1(
1
1
k
k
k
k
k
3
2
23
32lim
23
32lim
k
k
k
k
k
k
k
k
By the Root Test, the series absolutely converges
(and therefore converges).
![Page 86: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/86.jpg)
29/12/20105. Series I 86
Remark
May like to use the Root Test, when an ,
involves nth powers.
![Page 87: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/87.jpg)
29/12/20105. Series I 87
Exercises
Exercise 12.6, Page 781 – 782
No. 1, 2 – 28 ( odd numbers only),
29,31.
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29/12/20105. Series I 88
Exercises
Use the Root Test to determine whether the
series converges. If the test is inconclusive,
then say so.
1 1
11
)1( 4. 6
3.
200 2.
12
23 1.
k n
kk
k
n
n
k
k
ek
n
k
k
![Page 89: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/89.jpg)
29/12/20105. Series I 89
Practice
Using the Ratio Test, determine whether
the given series converge or diverge.
1
1
! (ii)
!
3 (i)
k
k
k
k
k
k
k
![Page 90: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/90.jpg)
29/12/20105. Series I 90
Practice
Use the Root Test to determine whether the
series converges or diverges.
1
1
1 (ii)
12
13 (i)
k
k-k
n
n
)-e(
n-
n
![Page 91: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/91.jpg)
29/12/20105. Series I 91
Practice
Classify whether each series is conditionally
convergent, absolutely convergent or
divergent.
14 3
1
1
1)1( (ii)
159
13)1( (i)
k
k
n
n
k
n
n
![Page 92: Series Module 4 -September 2010](https://reader034.vdocuments.mx/reader034/viewer/2022051709/577ccf151a28ab9e788ed666/html5/thumbnails/92.jpg)
29/12/20105. Series I 92
END of MODULE