Camparison Test - Chapter 11: Infinite Sequences and Series, Engg , Sem Notes | EduRev

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: Camparison Test - Chapter 11: Infinite Sequences and Series, Engg , Sem Notes | EduRev

 Page 1


11.4 Comparison Tests 777
Comparison Tests
We have seen how to determine the convergence of geometric series, p-series, and a few
others. We can test the convergence of many more series by comparing their terms to those
of a series whose convergence is known.
11.4
THEOREM 10 The Comparison Test
Let be a series with no negative terms.
(a) converges if there is a convergent series with for all
for some integer N.
(b) diverges if there is a divergent series of nonnegative terms with
for all for some integer N. n 7 N, a
n
Ú d
n
gd
n
ga
n
n 7 N,
a
n
… c
n
gc
n
ga
n
ga
n
Proof In Part (a), the partial sums of are bounded above by
They therefore form a nondecreasing sequence with a limit 
In Part (b), the partial sums of are not bounded from above. If they were, the par-
tial sums for would be bounded by
and would have to converge instead of diverge.
EXAMPLE 1 Applying the Comparison Test
(a) The series
diverges because its nth term
is greater than the nth term of the divergent harmonic series.
5
5n - 1
=
1
n -
1
5
7
1
n
a
q
n = 1
 
5
5n - 1
gd
n
M
*
= d
1
+ d
2
+
Á
+ d
N
+
a
q
n = N + 1
a
n
gd
n
ga
n
L … M.
M = a
1
+ a
2
+
Á
+ a
N
+
a
q
n = N + 1
c
n
.
ga
n
HISTORICAL BIOGRAPHY
Albert of Saxony
(ca. 1316–1390)
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 777
Page 2


11.4 Comparison Tests 777
Comparison Tests
We have seen how to determine the convergence of geometric series, p-series, and a few
others. We can test the convergence of many more series by comparing their terms to those
of a series whose convergence is known.
11.4
THEOREM 10 The Comparison Test
Let be a series with no negative terms.
(a) converges if there is a convergent series with for all
for some integer N.
(b) diverges if there is a divergent series of nonnegative terms with
for all for some integer N. n 7 N, a
n
Ú d
n
gd
n
ga
n
n 7 N,
a
n
… c
n
gc
n
ga
n
ga
n
Proof In Part (a), the partial sums of are bounded above by
They therefore form a nondecreasing sequence with a limit 
In Part (b), the partial sums of are not bounded from above. If they were, the par-
tial sums for would be bounded by
and would have to converge instead of diverge.
EXAMPLE 1 Applying the Comparison Test
(a) The series
diverges because its nth term
is greater than the nth term of the divergent harmonic series.
5
5n - 1
=
1
n -
1
5
7
1
n
a
q
n = 1
 
5
5n - 1
gd
n
M
*
= d
1
+ d
2
+
Á
+ d
N
+
a
q
n = N + 1
a
n
gd
n
ga
n
L … M.
M = a
1
+ a
2
+
Á
+ a
N
+
a
q
n = N + 1
c
n
.
ga
n
HISTORICAL BIOGRAPHY
Albert of Saxony
(ca. 1316–1390)
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 777
778 Chapter 11: Infinite Sequences and Series
(b) The series
converges because its terms are all positive and less than or equal to the correspon-
ding terms of
The geometric series on the left converges and we have
The fact that 3 is an upper bound for the partial sums of does not
mean that the series converges to 3. As we will see in Section 11.9, the series con-
verges to e.
(c) The series
converges. To see this, we ignore the first three terms and compare the remaining terms
with those of the convergent geometric series The term of
the truncated sequence is less than the corresponding term of the geometric se-
ries. We see that term by term we have the comparison,
So the truncated series and the original series converge by an application of the Com-
parison Test.
The Limit Comparison Test
We now introduce a comparison test that is particularly useful for series in which is a
rational function of n.
a
n
1 +
1
2 +21
+
1
4 +22
+
1
8 +23
+
Á
… 1 +
1
2
+
1
4
+
1
8
+
Á
1>2
n
1>s2
n
+2nd g
q
n=0
s1>2
n
d.
5 +
2
3
+
1
7
+ 1 +
1
2 +21
+
1
4 +22
+
1
8 +23
+
Á
+
1
2
n
+2n
+
Á
g
q
n=0 
s1>n!d
1 +
a
q
n = 0
 
1
2
n
= 1 +
1
1 - s1>2d
= 3.
1 +
a
q
n = 0
 
1
2
n
= 1 + 1 +
1
2
+
1
2
2
+
Á
.
a
q
n = 0
 
1
n!
= 1 +
1
1!
+
1
2!
+
1
3!
+
Á
THEOREM 11 Limit Comparison Test
Suppose that and for all (N an integer).
1. If then and both converge or both diverge.
2.If and converges, then converges.
3.If and diverges, then diverges. ga
n
gb
n
lim
n: q
 
a
n
b
n
= q
ga
n
gb
n
lim
n: q
 
a
n
b
n
= 0
gb
n
ga
n
lim
n: q
 
a
n
b
n
= c 7 0,
n Ú N b
n
7 0 a
n
7 0
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 778
Page 3


11.4 Comparison Tests 777
Comparison Tests
We have seen how to determine the convergence of geometric series, p-series, and a few
others. We can test the convergence of many more series by comparing their terms to those
of a series whose convergence is known.
11.4
THEOREM 10 The Comparison Test
Let be a series with no negative terms.
(a) converges if there is a convergent series with for all
for some integer N.
(b) diverges if there is a divergent series of nonnegative terms with
for all for some integer N. n 7 N, a
n
Ú d
n
gd
n
ga
n
n 7 N,
a
n
… c
n
gc
n
ga
n
ga
n
Proof In Part (a), the partial sums of are bounded above by
They therefore form a nondecreasing sequence with a limit 
In Part (b), the partial sums of are not bounded from above. If they were, the par-
tial sums for would be bounded by
and would have to converge instead of diverge.
EXAMPLE 1 Applying the Comparison Test
(a) The series
diverges because its nth term
is greater than the nth term of the divergent harmonic series.
5
5n - 1
=
1
n -
1
5
7
1
n
a
q
n = 1
 
5
5n - 1
gd
n
M
*
= d
1
+ d
2
+
Á
+ d
N
+
a
q
n = N + 1
a
n
gd
n
ga
n
L … M.
M = a
1
+ a
2
+
Á
+ a
N
+
a
q
n = N + 1
c
n
.
ga
n
HISTORICAL BIOGRAPHY
Albert of Saxony
(ca. 1316–1390)
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 777
778 Chapter 11: Infinite Sequences and Series
(b) The series
converges because its terms are all positive and less than or equal to the correspon-
ding terms of
The geometric series on the left converges and we have
The fact that 3 is an upper bound for the partial sums of does not
mean that the series converges to 3. As we will see in Section 11.9, the series con-
verges to e.
(c) The series
converges. To see this, we ignore the first three terms and compare the remaining terms
with those of the convergent geometric series The term of
the truncated sequence is less than the corresponding term of the geometric se-
ries. We see that term by term we have the comparison,
So the truncated series and the original series converge by an application of the Com-
parison Test.
The Limit Comparison Test
We now introduce a comparison test that is particularly useful for series in which is a
rational function of n.
a
n
1 +
1
2 +21
+
1
4 +22
+
1
8 +23
+
Á
… 1 +
1
2
+
1
4
+
1
8
+
Á
1>2
n
1>s2
n
+2nd g
q
n=0
s1>2
n
d.
5 +
2
3
+
1
7
+ 1 +
1
2 +21
+
1
4 +22
+
1
8 +23
+
Á
+
1
2
n
+2n
+
Á
g
q
n=0 
s1>n!d
1 +
a
q
n = 0
 
1
2
n
= 1 +
1
1 - s1>2d
= 3.
1 +
a
q
n = 0
 
1
2
n
= 1 + 1 +
1
2
+
1
2
2
+
Á
.
a
q
n = 0
 
1
n!
= 1 +
1
1!
+
1
2!
+
1
3!
+
Á
THEOREM 11 Limit Comparison Test
Suppose that and for all (N an integer).
1. If then and both converge or both diverge.
2.If and converges, then converges.
3.If and diverges, then diverges. ga
n
gb
n
lim
n: q
 
a
n
b
n
= q
ga
n
gb
n
lim
n: q
 
a
n
b
n
= 0
gb
n
ga
n
lim
n: q
 
a
n
b
n
= c 7 0,
n Ú N b
n
7 0 a
n
7 0
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 778
11.4 Comparison Tests 779
Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b).
Since there exists an integer N such that for all n
Thus, for 
If converges, then converges and converges by the Direct Compari-
son Test. If diverges, then diverges and diverges by the Direct Com-
parison Test.
EXAMPLE 2 Using the Limit Comparison Test
Which of the following series converge, and which diverge?
(a)
(b)
(c)
Solution
(a)Let For large n, we expect to behave like
since the leading terms dominate for large n, so we let Since
and
diverges by Part 1 of the Limit Comparison Test. We could just as well have
taken but 1 n is simpler. > b
n
= 2>n,
ga
n
lim
n: q
 
a
n
b
n
= lim
n: q
 
2n
2
+ n
n
2
+ 2n + 1
= 2,
a
q
n = 1
b
n
=
a
q
n = 1
 
1
n
 diverges
b
n
= 1>n. 2n>n
2
= 2>n
a
n
a
n
= s2n + 1d>sn
2
+ 2n + 1d.
1 + 2 ln 2
9
+
1 + 3 ln 3
14
+
1 + 4 ln 4
21
+
Á
=
a
q
n = 2
 
1 + n ln n
n
2
+ 5
1
1
+
1
3
+
1
7
+
1
15
+
Á
=
a
q
n = 1
 
1
2
n
- 1
3
4
+
5
9
+
7
16
+
9
25
+
Á
=
a
q
n = 1
 
2n + 1
sn + 1d
2
=
a
q
n = 1
 
2n + 1
n
2
+ 2n + 1
ga
n
gsc>2db
n
gb
n
ga
n
gs3c>2db
n
gb
n
 a
c
2
bb
n
6 a
n
6 a
3c
2
bb
n
.
 
c
2
6
a
n
b
n
6
3c
2
,
 -
c
2
6
a
n
b
n
- c 6
c
2
,
n 7 N,
n 7 N Q `
a
n
b
n
- c` 6
c
2
.
c>2 7 0,
Limit definition with
and
replaced by a
n
>b
n
a
n
P= c>2, L = c,
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 779
Page 4


11.4 Comparison Tests 777
Comparison Tests
We have seen how to determine the convergence of geometric series, p-series, and a few
others. We can test the convergence of many more series by comparing their terms to those
of a series whose convergence is known.
11.4
THEOREM 10 The Comparison Test
Let be a series with no negative terms.
(a) converges if there is a convergent series with for all
for some integer N.
(b) diverges if there is a divergent series of nonnegative terms with
for all for some integer N. n 7 N, a
n
Ú d
n
gd
n
ga
n
n 7 N,
a
n
… c
n
gc
n
ga
n
ga
n
Proof In Part (a), the partial sums of are bounded above by
They therefore form a nondecreasing sequence with a limit 
In Part (b), the partial sums of are not bounded from above. If they were, the par-
tial sums for would be bounded by
and would have to converge instead of diverge.
EXAMPLE 1 Applying the Comparison Test
(a) The series
diverges because its nth term
is greater than the nth term of the divergent harmonic series.
5
5n - 1
=
1
n -
1
5
7
1
n
a
q
n = 1
 
5
5n - 1
gd
n
M
*
= d
1
+ d
2
+
Á
+ d
N
+
a
q
n = N + 1
a
n
gd
n
ga
n
L … M.
M = a
1
+ a
2
+
Á
+ a
N
+
a
q
n = N + 1
c
n
.
ga
n
HISTORICAL BIOGRAPHY
Albert of Saxony
(ca. 1316–1390)
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 777
778 Chapter 11: Infinite Sequences and Series
(b) The series
converges because its terms are all positive and less than or equal to the correspon-
ding terms of
The geometric series on the left converges and we have
The fact that 3 is an upper bound for the partial sums of does not
mean that the series converges to 3. As we will see in Section 11.9, the series con-
verges to e.
(c) The series
converges. To see this, we ignore the first three terms and compare the remaining terms
with those of the convergent geometric series The term of
the truncated sequence is less than the corresponding term of the geometric se-
ries. We see that term by term we have the comparison,
So the truncated series and the original series converge by an application of the Com-
parison Test.
The Limit Comparison Test
We now introduce a comparison test that is particularly useful for series in which is a
rational function of n.
a
n
1 +
1
2 +21
+
1
4 +22
+
1
8 +23
+
Á
… 1 +
1
2
+
1
4
+
1
8
+
Á
1>2
n
1>s2
n
+2nd g
q
n=0
s1>2
n
d.
5 +
2
3
+
1
7
+ 1 +
1
2 +21
+
1
4 +22
+
1
8 +23
+
Á
+
1
2
n
+2n
+
Á
g
q
n=0 
s1>n!d
1 +
a
q
n = 0
 
1
2
n
= 1 +
1
1 - s1>2d
= 3.
1 +
a
q
n = 0
 
1
2
n
= 1 + 1 +
1
2
+
1
2
2
+
Á
.
a
q
n = 0
 
1
n!
= 1 +
1
1!
+
1
2!
+
1
3!
+
Á
THEOREM 11 Limit Comparison Test
Suppose that and for all (N an integer).
1. If then and both converge or both diverge.
2.If and converges, then converges.
3.If and diverges, then diverges. ga
n
gb
n
lim
n: q
 
a
n
b
n
= q
ga
n
gb
n
lim
n: q
 
a
n
b
n
= 0
gb
n
ga
n
lim
n: q
 
a
n
b
n
= c 7 0,
n Ú N b
n
7 0 a
n
7 0
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 778
11.4 Comparison Tests 779
Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b).
Since there exists an integer N such that for all n
Thus, for 
If converges, then converges and converges by the Direct Compari-
son Test. If diverges, then diverges and diverges by the Direct Com-
parison Test.
EXAMPLE 2 Using the Limit Comparison Test
Which of the following series converge, and which diverge?
(a)
(b)
(c)
Solution
(a)Let For large n, we expect to behave like
since the leading terms dominate for large n, so we let Since
and
diverges by Part 1 of the Limit Comparison Test. We could just as well have
taken but 1 n is simpler. > b
n
= 2>n,
ga
n
lim
n: q
 
a
n
b
n
= lim
n: q
 
2n
2
+ n
n
2
+ 2n + 1
= 2,
a
q
n = 1
b
n
=
a
q
n = 1
 
1
n
 diverges
b
n
= 1>n. 2n>n
2
= 2>n
a
n
a
n
= s2n + 1d>sn
2
+ 2n + 1d.
1 + 2 ln 2
9
+
1 + 3 ln 3
14
+
1 + 4 ln 4
21
+
Á
=
a
q
n = 2
 
1 + n ln n
n
2
+ 5
1
1
+
1
3
+
1
7
+
1
15
+
Á
=
a
q
n = 1
 
1
2
n
- 1
3
4
+
5
9
+
7
16
+
9
25
+
Á
=
a
q
n = 1
 
2n + 1
sn + 1d
2
=
a
q
n = 1
 
2n + 1
n
2
+ 2n + 1
ga
n
gsc>2db
n
gb
n
ga
n
gs3c>2db
n
gb
n
 a
c
2
bb
n
6 a
n
6 a
3c
2
bb
n
.
 
c
2
6
a
n
b
n
6
3c
2
,
 -
c
2
6
a
n
b
n
- c 6
c
2
,
n 7 N,
n 7 N Q `
a
n
b
n
- c` 6
c
2
.
c>2 7 0,
Limit definition with
and
replaced by a
n
>b
n
a
n
P= c>2, L = c,
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 779
(b)Let For large n,we expect to behave like so we let
Since
and
converges by Part 1 of the Limit Comparison Test.
(c)Let For large n, we expect to behave like
which is greater than 1 nfor so we take 
Since
and
diverges by Part 3 of the Limit Comparison Test.
EXAMPLE 3Does converge?
Solution Because ln n grows more slowly than for any positive constant c
(Section 11.1, Exercise 91), we would expect to have
for n sufficiently large. Indeed, taking and we have
Since (a p-series with ) converges, converges by Part 2 of
the Limit Comparison Test.
ga
n
p 7 1 gb
n
= gs1>n
5>4
d
 = lim
n: q
 
4
n
1>4
= 0.
 = lim
n: q
 
1>n
s1>4dn
-3>4
 lim
n: q
 
a
n
b
n
= lim
n: q
 
ln n
n
1>4
b
n
= 1>n
5>4
, a
n
= sln nd>n
3>2
ln n
n
3>2
6
n
1>4
n
3>2
=
1
n
5>4
n
c
a
q
n = 1
 
ln n
n
3>2
ga
n
 = q ,
 lim
n: q
 
a
n
b
n
= lim
n: q
 
n + n
2
 ln n
n
2
+ 5
a
q
n = 2
b
n
=
a
q
n = 2
 
1
n
  diverges
b
n
= 1>n. n Ú 3, > sn ln nd>n
2
= sln nd>n,
a
n
a
n
= s1 + n ln nd>sn
2
+ 5d.
ga
n
 = 1,
 = lim
n: q
 
1
1 - s1>2
n
d
 lim
n: q
 
a
n
b
n
= lim
n: q
 
2
n
2
n
- 1
a
q
n = 1
b
n
=
a
q
n = 1
 
1
2
n
  converges
b
n
= 1>2
n
.
1>2
n
, a
n
a
n
= 1>s2
n
- 1d.
780 Chapter 11: Infinite Sequences and Series
l’Hôpital’ s Rule
4100 AWL/Thomas_ch11p746-847  8/25/04  2:41 PM  Page 780
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