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97817_11_ch11_p752761.qk_97817_11_ch11_p752761 11/3/10 5:30 PM Page 757
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
SECTION 11.6
757
2 Deﬁnition A series a n is called conditionally convergent if it is convergent
but not absolutely convergent.
Example 2 shows that the alternating harmonic series is conditionally convergent. Thus
it is possible for a series to be convergent but not absolutely convergent. However, the next
theorem shows that absolute convergence implies convergence.
Theorem If a series
3
a n is absolutely convergent, then it is convergent.
PROOF Observe that the inequality
Խ Խ
Խ Խ
0 ഛ an ϩ an ഛ 2 an
is true because a n is either a n or Ϫa n . If a n is absolutely convergent, then a n is
convergent, so 2 a n is convergent. Therefore, by the Comparison Test, (a n ϩ a n ) is
convergent. Then
Խ Խ
Խ Խ
Խ Խ
Խ Խ
͚a
n
Խ Խ) Ϫ ͚ Խ a Խ
͚ (a n ϩ a n
n
is the difference of two convergent series and is therefore convergent.
v
EXAMPLE 3 Determine whether the series
ϱ
͚
n1
cos n
cos 1
cos 2
cos 3
ϩ
ϩ
ϩ иии
n2
12
22
32
is convergent or divergent.
Figure 1 shows the graphs of the terms a n and
partial sums sn of the series in Example 3.
Notice that the series is not alternating but
has positive and negative terms.
SOLUTION This series has both positive and negative terms, but it is not alternating.
(The first term is positive, the next three are negative, and the following three are positive: The signs change irregularly.) We can apply the Comparison Test to the series of
absolute values
ϱ
͚
0.5
n1
͕sn ͖
Խ
Ϳ Ϳ
Խ
ϱ
cos n
cos n
͚
2
n
n2
n1
Խ
Խ
Since cos n ഛ 1 for all n, we have
Խ cos n Խ ഛ
͕a n ͖
0
FIGURE 1
n
n2
1
n2
We know that 1͞n 2 is convergent ( pseries with p 2) and therefore cos n ͞n 2 is
convergent by the Comparison Test. Thus the given series ͑cos n͒͞n 2 is absolutely
convergent and therefore convergent by Theorem 3.
Խ
Խ
The following test is very useful in determining whether a given series is absolutely
convergent.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p752761.qk_97817_11_ch11_p752761 11/3/10 5:30 PM Page 758
758
CHAPTER 11
INFINITE SEQUENCES AND SERIES
The Ratio Test
Ϳ Ϳ
ϱ
a nϩ1
L Ͻ 1, then the series ͚ a n is absolutely convergent
nlϱ
an
n1
(and therefore convergent).
(i) If lim
Ϳ Ϳ
Ϳ Ϳ
ϱ
a nϩ1
a nϩ1
L Ͼ 1 or lim
ϱ, then the series ͚ a n
nlϱ
nlϱ
an
an
n1
is divergent.
(ii) If lim
Ϳ Ϳ
a nϩ1
1, the Ratio Test is inconclusive; that is, no conclusion can
an
be drawn about the convergence or divergence of a n .
(iii) If lim
nlϱ
PROOF
(i) The idea is to compare the given series with a convergent geometric series. Since
L Ͻ 1, we can choose a number r such that L Ͻ r Ͻ 1. Since
lim
nlϱ
Խ
Խ
Ϳ Ϳ
a nϩ1
L
an
LϽr
and
the ratio a nϩ1͞a n will eventually be less than r ; that is, there exists an integer N
such that
a nϩ1
whenever n ജ N
Ͻr
an
Ϳ Ϳ
or, equivalently,
Խa Խ Ͻ Խa Խr
4
nϩ1
whenever n ജ N
n
Putting n successively equal to N , N ϩ 1, N ϩ 2, . . . in 4 , we obtain
Խa Խ Ͻ Խa Խr
Խa Խ Ͻ Խa Խr Ͻ Խa Խr
Խa Խ Ͻ Խa Խr Ͻ Խa Խr
Nϩ1
N
Nϩ2
Nϩ1
N
Nϩ3
Nϩ2
N
2
3
and, in general,
Խa Խ Ͻ Խa Խr
5
Nϩk
N
k
for all k ജ 1
Now the series
ϱ
͚ Խa Խr
N
k
k1
Խ Խ
Խ Խ
Խ Խ
aN r ϩ aN r 2 ϩ aN r 3 ϩ и и и
is convergent because it is a geometric series with 0 Ͻ r Ͻ 1. So the inequality 5
together with the Comparison Test, shows that the series
ϱ
ϱ
͚ Խa Խ ͚ Խa Խ Խa Խ ϩ Խa Խ ϩ Խa Խ ϩ и и и
n
nNϩ1
Nϩk
Nϩ1
Nϩ2
Nϩ3
k1
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p752761.qk_97817_11_ch11_p752761 11/3/10 5:30 PM Page 759
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
SECTION 11.6
759
is also convergent. It follows that the series ϱn1 a n is convergent. (Recall that a finite
number of terms doesn’t affect convergence.) Therefore a n is absolutely convergent.
(ii) If a nϩ1͞a n l L Ͼ 1 or a nϩ1͞a n l ϱ, then the ratio a nϩ1͞a n will eventually be
greater than 1; that is, there exists an integer N such that
Խ Խ
Խ
Խ
Խ
Խ
Խ
Ϳ Ϳ
a nϩ1
Ͼ1
an
Խ
Խ
whenever n ജ N
Խ Խ Խ
This means that a nϩ1 Ͼ a n whenever n ജ N and so
lim a n
0
nlϱ
Therefore a n diverges by the Test for Divergence.
Խ
Խ
NOTE Part (iii) of the Ratio Test says that if lim n l ϱ a nϩ1͞a n 1, the test gives no
information. For instance, for the convergent series 1͞n 2 we have
Ϳ Ϳ
anϩ1
an
1
͑n ϩ 1͒2
n2
1
͑n ϩ 1͒2
n2
1
ͩ ͪ
1
1ϩ
n
2
as n l ϱ
l1
whereas for the divergent series 1͞n we have
1
a nϩ1
nϩ1
n
1
l1
an
1
nϩ1
1
1ϩ
n
n
Ϳ Ϳ
The Ratio Test is usually conclusive if the nth
term of the series contains an exponential or a
factorial, as we will see in Examples 4 and 5.
as n l ϱ
Therefore, if lim n l ϱ a nϩ1͞a n 1, the series a n might converge or it might diverge. In
this case the Ratio Test fails and we must use some other test.
Խ
Խ
ϱ
EXAMPLE 4 Test the series
͚ ͑Ϫ1͒
n
n1
n3
for absolute convergence.
3n
SOLUTION We use the Ratio Test with a n ͑Ϫ1͒nn 3͞3 n:
Estimating Sums
In the last three sections we used various methods for estimating the sum of a series—the
method depended on which test was used to
prove convergence. What about series for
which the Ratio Test works? There are two
possibilities: If the series happens to be an alternating series, as in Example 4, then it is best to
use the methods of Section 11.5. If the terms are
all positive, then use the special methods
explained in Exercise 38.
Ϳ Ϳ
a nϩ1
an
͑Ϫ1͒nϩ1͑n ϩ 1͒3
3 nϩ1
͑n ϩ 1͒3 3 n
ؒ 3
n 3
͑Ϫ1͒ n
3 nϩ1
n
n
3

1
3

ͩ ͪ ͩ ͪ
nϩ1
n
3
1
3
1ϩ
1
n
3
l
1
Ͻ1
3
Thus, by the Ratio Test, the given series is absolutely convergent and therefore
convergent.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p752761.qk_97817_11_ch11_p752761 11/3/10 5:30 PM Page 760
760
CHAPTER 11
INFINITE SEQUENCES AND SERIES
ϱ
v
͚
EXAMPLE 5 Test the convergence of the series
n1
nn
.
n!
SOLUTION Since the terms a n n ͞n! are positive, we don’t need the absolute value
n
signs.
a nϩ1
͑n ϩ 1͒nϩ1 n!
ؒ n
an
͑n ϩ 1͒!
n
͑n ϩ 1͒͑n ϩ 1͒n n!
ؒ n
͑n ϩ 1͒n!
n
ͩ ͪ ͩ ͪ
nϩ1
n
n
n
1
n
1ϩ
le
as n l ϱ
(see Equation 6.4.9 or 6.4*.9). Since e Ͼ 1, the given series is divergent by the Ratio
Test.
NOTE Although the Ratio Test works in Example 5, an easier method is to use the Test
for Divergence. Since
nn
n ؒ n ؒ n ؒ иии ؒ n
an
ജn
n!
1 ؒ 2 ؒ 3 ؒ иии ؒ n
it follows that a n does not approach 0 as n l ϱ. Therefore the given series is divergent by
the Test for Divergence.
The following test is convenient to apply when n th powers occur. Its proof is similar to
the proof of the Ratio Test and is left as Exercise 41.
The Root Test
Խ Խ
n
(i) If lim s
a n L Ͻ 1, then the series
nlϱ
ϱ
͚a
n
is absolutely convergent
n1
(and therefore convergent).
Խ Խ
Խ Խ
n
n
(ii) If lim s
a n L Ͼ 1 or lim s
a n ϱ, then the series
nlϱ
nlϱ
ϱ
͚a
n
is divergent.
n1
Խ Խ
n
(iii) If lim s
a n 1, the Root Test is inconclusive.
nlϱ
Խ Խ
n
If lim n l ϱ s
a n 1, then part (iii) of the Root Test says that the test gives no information. The series a n could converge or diverge. (If L 1 in the Ratio Test, don’t try the
Root Test because L will again be 1. And if L 1 in the Root Test, don’t try the Ratio Test
because it will fail too.)
ϱ
v
EXAMPLE 6 Test the convergence of the series
͚
n1
SOLUTION
an
Խ Խ
n
an
s
ͩ
2n ϩ 3
3n ϩ 2
ͪ
ͩ
2n ϩ 3
3n ϩ 2
ͪ
n
.
n
3
2n ϩ 3
n
2
l Ͻ1
3n ϩ 2
2
3
3ϩ
n
2ϩ
Thus the given series converges by the Root Test.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p752761.qk_97817_11_ch11_p752761 11/3/10 5:30 PM Page 761
SECTION 11.6
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
761
Rearrangements
The question of whether a given convergent series is absolutely convergent or conditionally
convergent has a bearing on the question of whether infinite sums behave like finite sums.
If we rearrange the order of the terms in a finite sum, then of course the value of the sum
remains unchanged. But this is not always the case for an infinite series. By a rearrangement of an infinite series a n we mean a series obtained by simply changing the order of
the terms. For instance, a rearrangement of a n could start as follows:
a 1 ϩ a 2 ϩ a 5 ϩ a 3 ϩ a 4 ϩ a 15 ϩ a 6 ϩ a 7 ϩ a 20 ϩ и и и
It turns out that
if a n is an absolutely convergent series with sum s,
then any rearrangement of a n has the same sum s.
However, any conditionally convergent series can be rearranged to give a different sum. To
illustrate this fact let’s consider the alternating harmonic series
1 Ϫ 12 ϩ 13 Ϫ 14 ϩ 15 Ϫ 16 ϩ 17 Ϫ 18 ϩ и и и ln 2
6
(See Exercise 36 in Section 11.5.) If we multiply this series by 12 , we get
1
2
Ϫ 14 ϩ 16 Ϫ 18 ϩ и и и 12 ln 2
Inserting zeros between the terms of this series, we have
Adding these zeros does not affect the sum of
the series; each term in the sequence of partial
sums is repeated, but the limit is the same.
0 ϩ 12 ϩ 0 Ϫ 14 ϩ 0 ϩ 16 ϩ 0 Ϫ 18 ϩ и и и 12 ln 2
7
Now we add the series in Equations 6 and 7 using Theorem 11.2.8:
1 ϩ 13 Ϫ 12 ϩ 15 ϩ 17 Ϫ 14 ϩ и и и 32 ln 2
8
Notice that the series in 8 contains the same terms as in 6 , but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are
different. In fact, Riemann proved that
if a n is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of a n that has a sum equal to r.
A proof of this fact is outlined in Exercise 44.
11.6
Exercises
1. What can you say about the series a n in each of the following
cases?
Ϳ Ϳ
Ϳ Ϳ
a nϩ1
8
(a) lim
nlϱ
an
(c) lim
nlϱ
Ϳ Ϳ
ϱ
5.
n0
a nϩ1
0.8
(b) lim
nlϱ
an
ϱ
7.
2.
͚
n1
ϱ
3.
͚
a nϩ1
1
an
n1
ϱ
6.
2 k
3
͚ ͑Ϫ1͒
n1
ϱ
11.
͚
n1
ϱ
4.
͚ ͑Ϫ1͒
nϪ1
n1
n
n ϩ4
2
ϱ
13.
͚
n1
͚
n0
ϱ
͚ k( )
8.
͚
n1
ϱ
9.
͑Ϫ2͒ n
n2
n
5n
͑Ϫ1͒ n
5n ϩ 1
k1
2–30 Determine whether the series is absolutely convergent,
conditionally convergent, or divergent.
ϱ
͚
n
͑1.1͒ n
n4
͑Ϫ3͒ n
͑2n ϩ 1͒!
n!
100 n
ϱ
10.
͚ ͑Ϫ1͒
n
n1
ϱ
͑Ϫ1͒n e 1͞n
n3
12.
10 n
͑n ϩ 1͒4 2nϩ1
14.
͚
n1
ϱ
͚
n1
n
sn 3 ϩ 2
sin 4n
4n
n 10
͑Ϫ10͒ nϩ1
1. Homework Hints available at stewartcalculus.com
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p762771.qk_97817_11_ch11_p762771 11/3/10 5:31 PM Page 762
762
CHAPTER 11
ϱ
15.
͚
n1
ϱ
17.
͚
n2
ϱ
19.
͚
n1
16.
͑Ϫ1͒n
ln n
18.
n ϩ1
2n 2 ϩ 1
1ϩ
n1
ϱ
25.
͚
n1
20.
ϱ
22.
͚
n2
n2
1
n
͚
n1
n
ϱ
24.
͚
n1
ϱ
n 100 100 n
n!
27. 1 Ϫ
͚
ϱ
ͩ ͪ
͚ͩ
ͪ
ϱ
n1
n1
2
͚
͚
ϱ
cos͑n͞3͒
n!
n1
23.
ϱ
͑Ϫ1͒ n arctan n
n2
ϱ
21.
INFINITE SEQUENCES AND SERIES
26.
͚
n1
3 Ϫ cos n
n 2͞3 Ϫ 2
n!
nn
͑Ϫ2͒
nn
29.
͚
n1
(b) Deduce that lim n l ϱ x n͞n! 0 for all x.
ͩ ͪ
Ϫ2n
nϩ1
5n
͑2n͒!
͑n!͒ 2
n
n1
38. Let a n be a series with positive terms and let rn a nϩ1 ͞a n.
Suppose that lim n l ϱ rn L Ͻ 1, so a n converges by the
Ratio Test. As usual, we let Rn be the remainder after n terms,
that is,
Rn a nϩ1 ϩ a nϩ2 ϩ a nϩ3 ϩ и и и
(a) If ͕rn ͖ is a decreasing sequence and rnϩ1 Ͻ 1, show, by
summing a geometric series, that
2
2n
n!
2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͒
n!
͚ ͑Ϫ1͒
͑n!͒2
͑kn͒!
37. (a) Show that ϱn0 x n͞n! converges for all x.
n
1ؒ3ؒ5
1ؒ3ؒ5ؒ7
1ؒ3
ϩ
Ϫ
ϩ иии
3!
5!
7!
1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒
ϩ ͑Ϫ1͒ nϪ1
ϩ иии
͑2n Ϫ 1͒!
ϱ
30.
ϱ
͚
n1
2ؒ6
2 ؒ 6 ؒ 10
2 ؒ 6 ؒ 10 ؒ 14
2
28.
ϩ
ϩ
ϩ
ϩ иии
5
5ؒ8
5 ؒ 8 ؒ 11
5 ؒ 8 ؒ 11 ؒ 14
ϱ
36. For which positive integers k is the following series convergent?
Rn ഛ
a nϩ1
1 Ϫ rnϩ1
(b) If ͕rn ͖ is an increasing sequence, show that
Rn ഛ
a nϩ1
1ϪL
39. (a) Find the partial sum s5 of the series ϱn1 1͑͞n2 n͒. Use Exer
cise 38 to estimate the error in using s5 as an approximation
to the sum of the series.
(b) Find a value of n so that sn is within 0.00005 of the sum.
Use this value of n to approximate the sum of the series.
2 n n!
5 ؒ 8 ؒ 11 ؒ и и и ؒ ͑3n ϩ 2͒
40. Use the sum of the first 10 terms to approximate the sum of
the series
ϱ
31. The terms of a series are defined recursively by the equations
a1 2
a nϩ1
5n ϩ 1
an
4n ϩ 3
Determine whether a n converges or diverges.
a1 1
a nϩ1
sn
an
33–34 Let ͕bn͖ be a sequence of positive numbers that converges
to 12. Determine whether the given series is absolutely convergent.
ϱ
͚
n1
bnn cos n
n
ϱ
34.
͚
n1
͑Ϫ1͒ n n!
n b1 b 2 b 3 и и и bn
n
35. For which of the following series is the Ratio Test inconclusive
(that is, it fails to give a definite answer)?
ϱ
(a)
͚
n1
ϱ
(c)
͚
n1
ϱ
1
n3
(b)
͚
n1
͑Ϫ3͒
ϱ
nϪ1
sn
(d)
͚
n1
n
2n
sn
1 ϩ n2
n
2n
Use Exercise 38 to estimate the error.
41. Prove the Root Test. [Hint for part (i): Take any number r such
Խ Խ
Determine whether a n converges or diverges.
33.
n1
that L Ͻ r Ͻ 1 and use the fact that there is an integer N such
n
that s
a n Ͻ r whenever n ജ N .]
32. A series a n is defined by the equations
2 ϩ cos n
͚
42. Around 1910, the Indian mathematician Srinivasa Ramanujan
discovered the formula
1
2 s2
9801
ϱ
͚
n0
͑4n͒!͑1103 ϩ 26390n͒
͑n!͒ 4 396 4n
William Gosper used this series in 1985 to compute the first
17 million digits of .
(a) Verify that the series is convergent.
(b) How many correct decimal places of do you get if you
use just the first term of the series? What if you use two
terms?
43. Given any series a n , we define a series aϩn whose terms are
all the positive terms of a n and a series aϪn whose terms
are all the negative terms of a n. To be specific, we let
aϩn
Խ Խ
an ϩ an
2
aϪn
Խ Խ
a n Ϫ an
2
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p762771.qk_97817_11_ch11_p762771 11/3/10 5:31 PM Page 763
SECTION 11.7
Notice that if a n Ͼ 0, then aϩn a n and aϪn 0, whereas if
a n Ͻ 0, then aϪn a n and aϩn 0.
(a) If a n is absolutely convergent, show that both of the
series aϩn and aϪn are convergent.
(b) If a n is conditionally convergent, show that both of the
series aϩn and aϪn are divergent.
763
Take just enough positive terms aϩn so that their sum is greater
than r. Then add just enough negative terms aϪn so that the
cumulative sum is less than r. Continue in this manner and use
Theorem 11.2.6.]
45. Suppose the series a n is conditionally convergent.
(a) Prove that the series n 2 a n is divergent.
(b) Conditional convergence of a n is not enough to determine whether na n is convergent. Show this by giving an
example of a conditionally convergent series such that
na n converges and an example where na n diverges.
44. Prove that if a n is a conditionally convergent series and
r is any real number, then there is a rearrangement of a n
whose sum is r. [Hints: Use the notation of Exercise 43.
11.7
STRATEGY FOR TESTING SERIES
Strategy for Testing Series
We now have several ways of testing a series for convergence or divergence; the problem
is to decide which test to use on which series. In this respect, testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a given
series, but you may find the following advice of some use.
It is not wise to apply a list of the tests in a specific order until one finally works. That
would be a waste of time and effort. Instead, as with integration, the main strategy is to
classify the series according to its form.
1. If the series is of the form
1͞n p, it is a pseries, which we know to be convergent
if p Ͼ 1 and divergent if p ഛ 1.
2. If the series has the form
Խ Խ
ar nϪ1 or ar n, it is a geometric series, which converges
Խ Խ
if r Ͻ 1 and diverges if r ജ 1. Some preliminary algebraic manipulation may
be required to bring the series into this form.
3. If the series has a form that is similar to a pseries or a geometric series, then
one of the comparison tests should be considered. In particular, if a n is a rational
function or an algebraic function of n (involving roots of polynomials), then the
series should be compared with a pseries. Notice that most of the series in Exercises 11.4 have this form. (The value of p should be chosen as in Section 11.4 by
keeping only the highest powers of n in the numerator and denominator.) The comparison tests apply only to series with positive terms, but if a n has some negative
terms, then we can apply the Comparison Test to a n and test for absolute
convergence.
Խ Խ
4. If you can see at a glance that lim n l ϱ a n
0, then the Test for Divergence should
be used.
5. If the series is of the form
͑Ϫ1͒nϪ1bn or ͑Ϫ1͒nbn , then the Alternating Series
Test is an obvious possibility.
6. Series that involve factorials or other products (including a constant raised to the
nth power) are often conveniently tested using the Ratio Test. Bear in mind that
nϩ1͞a n l 1 as n l ϱ for all pseries and therefore all rational or algebraic
functions of n. Thus the Ratio Test should not be used for such series.
Խa
Խ
7. If a n is of the form ͑bn ͒n, then the Root Test may be useful.
8. If a n f ͑n͒, where
x1ϱ f ͑x͒ dx is easily evaluated, then the Integral Test is effective
(assuming the hypotheses of this test are satisfied).
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_11_ch11_p762771.qk_97817_11_ch11_p762771 11/3/10 5:31 PM Page 764
764
CHAPTER 11
INFINITE SEQUENCES AND SERIES
In the following examples we don’t work out all the details but simply indicate which
tests should be used.
ϱ
v
EXAMPLE 1
nϪ1
2n ϩ 1
͚
n1
Since a n l
0 as n l ϱ, we should use the Test for Divergence.
1
2
ϱ
͚
EXAMPLE 2
n1
sn 3 ϩ 1
3n ϩ 4n 2 ϩ 2
3
Since a n is an algebraic function of n, we compare the given series with a pseries. The
comparison series for the Limit Comparison Test is bn , where
bn
ϱ
v
EXAMPLE 3
n 3͞2
1
sn 3
3͞2
3n 3
3n 3
3n
͚ ne
Ϫn 2
n1
Since the integral x1ϱ xeϪx dx is easily evaluated, we use the Integral Test. The Ratio Test
also works.
2
ϱ
͚ ͑Ϫ1͒
n
EXAMPLE 4
n1
n3
n ϩ1
4
Since the series is alternating, we use the Alternating Series Test.
ϱ
v
EXAMPLE 5
͚
k1
2k
k!
Since the series involves k!, we use the Ratio Test.
ϱ
EXAMPLE 6
͚
n1
1
2 ϩ 3n
Since the series is closely related to the geometric series 1͞3 n, we use the Comparison
Test.
Exercises
11.7
1–38 Test the series for convergence or divergence.
ϱ
1.
͚
n1
1
n ϩ 3n
ϱ
n
3. ͚ ͑Ϫ1͒
nϩ2
n1
n
ϱ
5.
ϱ
8.
k1
2 Ϫk
e
n
n
4. ͚ ͑Ϫ1͒ 2
n ϩ2
n1
1
nsln n
͚k
͑2n ϩ 1͒
n 2n
n
ϱ
ϱ
ϱ
ϱ
6.
͚
͚
n1
͚
n2
9.
2.
n 2 2 nϪ1
͑Ϫ5͒ n
n1
7.
ϱ
͚
n1
ϱ
͚
k1
ϱ
10.
1
2n ϩ 1
2 k k!
͑k ϩ 2͒!
͚ne
n1
2 Ϫn 3
11.
͚
n1
ϱ
13.
͚
n1
ϱ
15.
͚
k1
ϱ
17.
͚
n1
ϱ
18.
͚
n2
ͩ
1
1
ϩ n
n3
3
ͪ
ϱ
12.
͚
k1
ϱ
3n n2
n!
14.
2 kϪ1 3 kϩ1
kk
16.
1
ksk 2 ϩ 1
͚
sin 2n
1 ϩ 2n
ϱ
n2 ϩ 1
n3 ϩ 1
n1
͚
n1
1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒
2 ؒ 5 ؒ 8 ؒ и и и ؒ ͑3n Ϫ 1͒
͑Ϫ1͒ nϪ1
sn Ϫ 1
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.