How many nonzero terms of the Maclaurin series for ln1+ x are needed to approx-imate ln 32 with an error of less than 10−4?. Compare your answer with the approximation given by the first
Trang 1f (x)= C0+ C1x−C0
2!x
2−C1 3!x
3+C0 4!x
4+C1 5!x
5−C0 6!x
6−C1 7!x
7+ · · ·
f (x)= C0
1−x
2
2! +x
4
4! − · · · + (−1)n x
2n
(2n)!+ · · ·
cos x
+ C1
x−x
3
3! +x
5
5! −x
7
7! + · · · + (−1)n x
2n +1
(2n+ 1)!+ · · ·
sin x
f (x)= C0cos x+ C1sin x
EXERCISE 30.6 Verify that f (x)= C0cos x+ C1sin x is a solution to the differential equation y = −y
We have shown that if a solution to y = −y has a power series representation, then that solution must be of the form C0cos x+ C1sin x, where C0and C1are constants
In the example just completed, we recognized the Maclaurin series for sin x and cos x
It is entirely possible that we can solve for all the coefficients of a power series and simply have the solution expressed as and defined by the power series expansion There are well-known functions defined by power series that arise in physics, astronomy, and other
applied sciences An example of such functions are the Bessel functions, named after the
astronomer Bessel who came up with them in the early 1800s while working with Kepler’s laws of planetary motion The Bessel function J0(x)is defined by
Jo(x)=
∞
k =0
(−1)k x2k (k!)222k
As is often the case in mathematics, while Bessel functions arose in a particular astronomical problem they are now used in a wide array of situations One such example is in studying the vibrations of a drumhead A graph of the partial sum J0(x)≈13k =0(−1)k x2k
(k!) 2 2 2k is given
in Figure 30.9
.5
–.5 –1
1
2
y
Graph of ∑ (–1)k
k = 0
13 x 2k
(k!)2 22k
Figure 30.9
Trang 2962 CHAPTER 30 Series
Transition to Convergence Tests
Because this chapter began with Taylor polynomials, it was natural to move on to Taylor series directly, without the traditional lead-in of convergence tests for infinite series Taylor’s Theorem enables us to deal with some convergence issues quite efficiently Not only are
we able to show that the series for ex, sin x, and cos x converge, but we can determine that each converges to its generating function Our previous work with geometric series allows us to conclude that the series for 1−x1 converges to its generating function on (−1, 1) When we find a Taylor series by manipulating a known Taylor series, whether
by substitution, differentiation, or integration, we can calculate the radius of convergence But, faced with a generic power series, we have few tools at our disposal with which to determine convergence and divergence More fundamentally, we have no systematic way
of determining the convergence or divergence of an infinite series of the form ak The next section will remedy this situation
P R O B L E M S F O R S E C T I O N 3 0 4
For each series in Problems 1 through 9, determine whether the series converges absolutely, converges conditionally, or diverges.
1.∞k=1(−1)k k!
(k −1)!
2.∞k=1(−1)k +1 k!
(k +1)!
3.∞k=1(−1)k 1
3k
4.∞k=2(−1)n k
ln k
5.∞k=10cos(kn)10k
6.∞k=0 −1112 k
7.∞k=1 1001 sin kπ2
8.∞k=1(−1)k 2k
k
9.∞k=0(−1)n k2−10
2k 2 +5k
10 Is it possible for a geometric series to converge conditionally? If it is possible, produce
an example
11 How many nonzero terms of the Maclaurin series for ln(1+ x) are needed to approx-imate ln 32 with an error of less than 10−4?
12 Approximate 1with error less than 10−5
Trang 313 Arrive at the series for cos x by differentiating the Maclaurin series for sin x.
14 Find the Maclaurin series for arcsin x using the fact that √1
1 −x 2 dx= sin−1x+ C What is the radius of convergence of the series?
In Problems 15 through 17, write the given integral as a power series.
15. cos(x2) dx
16. ex3dx
17. 1
1 +x 5 dx
18 Approximate 00.5sin(x2) dx with error less than 10−8 Is your approximation an overestimate, or an underestimate?
19 Approximate 00.1 x
1+x 3 dxwith error less than 10−10
20 Find the Maclaurin series for ln(2+ x) along with its radius of convergence
21 (a) Find the Maclaurin series for ln 11+x−x by subtracting the Maclaurin series for ln(1− x) from that for ln(1 + x)
(b) Show that when x=13, 11+x−x = 2
(c) Use the first four nonzero terms of the series in part (a) to approximate ln 2 Compare your answer with the approximation given by the first four terms of the series for ln(1+ x) evaluated at x = 1, and the value of ln 2 given by a calculator
or computer
22 Show that∞k=0 (2x)k!k is a solution to the differential equation f (x)= 2f (x) What familiar function does this series represent?
23 Show that if f (x)=∞k =0akxk is a power series solution to f (x)= −f (x), then
f (x)=∞k =0(−1)k xk
k! What function does this series represent?
24 Use power series to solve the differential equation f (x)= 9f (x) What familiar function(s) does this series represent?
25 The Bessel function J0(x)is given by J0(x)=∞k =0(−1)k x2k
(k!) 2 2 2k It converges for all x
(a) If the first three nonzero terms of the series are used to approximate J0(0.1), will the approximation be too large, or too small? Give an upper bound for the magnitude
of the error
(b) How many nonzero terms of the series for J0(1) must be used to approximate J0(1) with error less than 10−4?
Trang 4964 CHAPTER 30 Series
30.5 CONVERGENCE TESTS
In this section we focus on ways of determining whether or not a given series converges
We begin by looking at series of constants; in the last subsection we apply our results to the convergence of power series
The Basic Principles
A series∞k=1akconverges to a sum S if the sequence of its partial sums converges to S, where S is a finite number In other words, if limn→∞Sn= S, where Sn=nk =1ak, then the infinite series converges to S Otherwise, the series diverges Note that if limx →∞f (x)= L, then limn →∞f (n)= L The converse is not true
Our first case study was geometric series (Refer to Chapter 18.) For a geometric series
we are able to express Snin closed form and directly compute limn→∞Sn We find that
∞
k =0
arkconverges to a
1− r if|r| < 1 and diverges if|r| ≥ 1
Once we leave the realm of geometric series it can be difficult or impossible to express
Snin closed form, so we generally can’t compute limn→∞Sndirectly Instead, we might determine convergence or divergence by comparing the series in question to a geometric series or an improper integral
We have already established one test for divergence; if the terms of the series don’t tend toward zero then the series diverges
nth Term Test for Divergence.If limn →∞an= 0, then∞n =1andiverges
If limn→∞an= 0, we have no information and must turn our attention back to the sequence of partial sums
D e f i n i t i o n
A sequence{sn} is increasing if sn≤ sn +1for all n≥ 1 It is decreasing if sn≥ sn +1
for all n≥ 1 If a sequence is either increasing or it is decreasing it is said to be
monotonic.
A sequence{sn} is bounded above if there is a constant M such that sn≤ M for all n≥ 1 It is bounded below if there is a constant m such that m ≤ snfor all n≥ 1
A sequence is said to be bounded if it is bounded both above and below.
A bounded sequence may or may not converge It could oscillate between the bounds like {(−1)n} However, if the sequence is bounded and increasing, then its terms must cluster about some number L≤ M A similar statement can be made for a decreasing sequence The following theorem will prove very useful
Trang 5B o u n d e d M o n o t o n i c C o n v e r g e n c e T h e o r e m15
A monotonic sequence converges if it is bounded and diverges otherwise
Suppose the terms of the series∞k=1akare all positive Then the sequence of partial sums
is increasing: sn≤ sn +1= sn+ an +1 Because the terms are positive, the sequence of partial sums is bounded below by zero Therefore, if{Sn} is bounded above, then {Sn} converges and consequently∞k=1akconverges; otherwise they diverge We will use this line of reasoning
repeatedly We’ll refer to it as the Bounded Increasing Partial Sums Theorem.
T h e B o u n d e d I n c r e a s i n g P a r t i a l S u m s T h e o r e m
A series∞k=1ak, where ak≥ 0, converges if and only if its sequence of partial sums
is bounded above
Our focus in this section is on the question of convergence versus divergence and not on the sum of a convergent series Therefore, the starting point of the series is not important; the first hundred or thousand terms of the infinite series can be chopped off without impacting convergence issues Keep this in mind when applying the results of this section For example,
if the sequence of partial sums is eventually monotonic, then the Bounded Increasing Partial
Sums Theorem can be applied
In the next few subsections we will discuss convergence tests with the specification that the terms of the series are positive From the observation made above, you can see that what
is really required is that the terms akare positive for all k greater than some fixed number,
or, more generally, have any of the required specifications in the long run
The Integral Test
We revisit the idea of comparing an infinite series and an improper integral in the next example.16
EXAMPLE 30.26 Determine whether the following series converge or diverge
(a)∞k=1 1
k3 =113 +213 +313 +413 + · · · (b)∞k=1 √1
k =√1
1+√1
2+√1
3+ · · ·
SOLUTION In both of these series the terms are positive, decreasing, and going toward zero, but the terms
of the series in part (b) are heading toward zero much more slowly than those in part (a) The values of some partial sums are given in the table on page 966 The information in the table is inconclusive, but it leads us to guess that∞k=1 1
k 3 might converge and∞k=1 √1
k
might diverge
15 A formal proof of this theorem rests on the Completeness Axiom for real numbers, which says that if a nonempty set of real numbers has an upper bound it must have a least upper bound.
16 This was first introduced in Section 29.4.
Trang 6966 CHAPTER 30 Series
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1.0
1.125
1.162037
1.177662
1.185662
1.190291
1.193207
1.195160
1.196531
1.197531
1.198283
1.198862
1.199317
1.199681
1.199977
1.204607
1.204811
1.204982
1.205128
1.205253
1 1.707106 2.284457 2.784457 3.231670 3.639918 4.017883 4.371436 4.704770 5.020997 5.322509 5.611184 5.888534 6.155795 6.413994 6.663994 6.906530 7.142232 7.371648 7.595255
n S n = ∑
n
k =1
1
k3 S n = ∑
n
k =1
1
√k
Partial sums are recorded up to
six decimal planes.
(a) To prove that∞k=1 1
k 3 converges it is enough to show that the increasing sequence of partial sums is bounded We do this by comparing the partial sums to 1∞ 1
x 3 dx, as shown in Figure 30.10
y
x
1
y = (1, 1)
(4, )
The areas of the shaded rectangles correspond to the terms of the series 1
1 3
1
x3
1
4 3
(3, )13 3
(2, )12 3
1
2 3
3 3
+ .This figure is not drawn to scale.
Figure 30.10
Each of the shaded rectangles has a base of length 1 The area = (base) · (height),
so the areas of the rectangles, from left to right, are 1
1 3, 1
2 3, 1
3 3,· · · The sum of the areas of the rectangles is∞k=1 1
k 3 Chop off the first rectangle.nk=2 1
k 3 < 1n 1
x 3 dx; the rectangles lie under the graph of 1
x 3 Consequently, limn→∞nk=2 1
k 3 ≤ 1∞ x13 dx But 1∞ 1
x 3 dxconverges
∞ 1
1
x3dx= lim
b →∞
b 1
x−3dx= lim
b →∞
x−2
−2
b
1
= lim
b →∞
−1 2b2+1
2=1 2
Therefore, the partial sums of∞k=1 1
k3 are bounded by 1+12, and the series converges
by the Bounded Increasing Partial Sums Theorem The sum of the series is greater than
1 and less than 1.5
(b) To prove that∞k=1 √1
k diverges it is enough to show that the sequence of partial sums
is unbounded We do this by comparing the partial sums to 1∞ √1
x dx as shown in Figure 30.11
y
x
y = √x1
(2, √21
(1, 1)
) (3, √31) (4, √41)
The areas of the shaded rectangles correspond to the terms of the series 1
√1 +√21
1
√3 + +
Figure 30.11
Trang 7∞
1
1
√
x dx= lim
b →∞
b 1
x−1 dx= lim
b →∞2x1
b
1
= lim
b →∞2√
b− 2 = ∞ Because we want to show that the partial sums are unbounded, we draw rectangles that lie above the graph of √1x The areas of the shaded rectangles are, from left to right,
1
√
1,√1
2,√1
3,· · · , so the sum of the areas of the shaded rectangles is∞k =1 √1k We see that 1n+1 √1xdx < Sn, so
lim
n →∞
n +1 1
1
√x dx≤ limn→∞Sn
∞ ≤ lim
n →∞Sn The series diverges
REMARKS
To show that a series with positive terms converges we show that the increasing sequence of partial sums is bounded, that is, is less than some constant M To show that it diverges, we show that g(n) < Snwhere limn→∞g(n)= ∞
Suppose we compare the series∞k=1f (k)with the improper integral 1∞f (x) dx If
f (x)is positive, continuous, and decreasing on [1,∞), then by including or omitting the first term of the series, we can depict the area corresponding to the sum as lying above or below the area corresponding to the improper integral (See Figures 30.12 and 30.13.)
y
x
1 2 3 4 5 6 The sum of the area of the shaded
rectangles is a1 + a2 + a3 + a4 + a5
a1
a2
a3
a4 a
5
y = f (x)
(2, f (2))
Figure 30.12
y
x
1 2 3 4 5 6 The sum of the area of the shaded
rectangles is a2 + a3 + a4 + a5 + a6
a2
a3
a4 a
5 a6
y = f (x)
(2, f (2))
Figure 30.13
Trang 8968 CHAPTER 30 Series
Using the reasoning given we can obtain the Integral Test
T h e I n t e g r a l T e s t
Let∞k=1akbe a series such that ak= f (k) for k = 1, 2, 3 , where the function
f is positive, continuous, and decreasing on [1,∞) Then
∞
k =1
ak and
∞ 1
f (x) dx
either both converge or both diverge
The proof of the Integral Test is constructed along the lines of Example 30.26 and makes a nice exercise for the reader Construct the proof for yourself If you have difficulty, consult the proof in Appendix H
EXERCISE 30.7 Use the Integral Test to show that the harmonic series diverges
The harmonic series,∞k=1 1k, and the series from Example 30.26,∞k=1 1
k 3 and∞k=1 √1
k,
are all examples of a class of series referred to as p-series because they can be written in
the form∞k=1k1p, where p is a constant
EXAMPLE 30.27 Show that the p-series∞k=1k1p=11p+21p+31p+41p+ · · · converges if p > 1 and diverges
if p≤ 1
SOLUTION For p≤ 0 the nth term doesn’t tend toward zero:
For p= 0, lim
n →∞
1
np = lim
n →∞1= 1 = 0 For p < 0, lim
n →∞
1
np = ∞ = 0
So for p≤ 0,∞k =1 k1pdiverges by the nth Term Test for Divergence
For p > 0, f (x)=x1p= x−pis positive, continuous, and decreasing on [1,∞) For p= 1, we know∞k =1 1kdiverges Let’s consider p= 1 and apply the Integral Test For p > 0, p= 1,
∞ 1
1
xpdx= lim
b →∞
b 1
x−pdx= lim
b →∞
x−p+1
−p + 1
b
1= lim
b →∞
b−p+1− 1
−p + 1 , so
∞ 1
1
xpdx
converges to p1−1 if p > 1 diverges if p∈ (0, 1)
Therefore∞k=1 k1p converges for p > 1 and diverges for p≤ 1 The conclusion that∞k=1k1pconverges for p > 1 and diverges otherwise will be useful
to keep in mind; we’ll compare other series to p-series Essentially, we’ve shown that for
p >1 the terms 1p tend toward zero rapidly enough to make the series converge If we run
Trang 9into an unfamiliar series whose terms go to zero more rapidly than those of a convergent p-series, we will find that this series converges as well
EXERCISE 30.8 Which of the following series converge?
(a)∞n=1n−1.1 (b) ∞k=1 √51
k 2 (c)∞k=1 3
k √
k (d) ∞n=1 e1n
Answers
(a), (c), and (d) converge (a), (b), and (c) are p-series; (d) is not (d) is a geometric series
Comparison Tests
In Example 30.26(b) we showed∞k=1 √1
kdiverges by comparing it to the divergent integral ∞
1 √1
xdx This approach was taken to illustrate a tool that enabled us to treat all p-series easily However, if we had been concerned only with∞k=1√1
k it would have been simpler
to compare this series with the harmonic series
k≥√kfor k≥ 1, so 1k ≤√1
k for k≥ 1
Comparing∞k=1 √1
k = 1 + √1
2+√1
3+√1
4+ · · · with∞k =1 1k = 1 +12+13+14+ · · · ,
we see that each term of the former series is greater than or equal to the corresponding term
of the harmonic series The harmonic series diverges because the partial sums increase without bound Therefore,∞k=1 √1
k must diverge as well
Having shown that∞k=1√1
k diverges because the partial sums increase without bound,
we can reason that∞k=4 √1
k −1diverges as well For k≥ 4 the latter is, term for term, larger
0 <√1
k<
1
√
k− 1 for k≥ 4.
Similarly, knowing that the geometric series∞k=1 21k converges, we can conclude that
∞
k =1 2 k1+1converges as well 0 < 1
2 k +1<
1
2 k, so the partial sums of∞k=1 1
2 k +1 are smaller than the corresponding partial sums of the geometric series The partial sums of∞2k =1 2 k1+1
are increasing and bounded Therefore∞k=1 2k1
+1 converges We can generalize this line
of reasoning to compare pairs of series with positive terms
T h e C o m p a r i s o n T e s t
Also known as the Direct Comparison Test
i Let 0≤ ak≤ bkfor all k (or all k≥ N for some constant N)
If
∞
k =1
bkconverges, then
∞
k =1
akconverges as well
ii Let 0≤ ck≤ akfor all k (or all k≥ N for some constant N)
If
∞
k =1
ckdiverges, then
∞
k =1
akdiverges as well
Trang 10970 CHAPTER 30 Series
y
x
y = x
y = lnx
Figure 30.14
Less formally, this says that we can compare series whose terms are positive as follows.
We can show that a series is convergent by showing that its terms are (eventually) smaller than the terms of a series known to be convergent We can show that a series is divergent
by showing that its terms are (eventually) larger than the terms of a series known to be divergent.
Proof
i Let Sn=nk =1akand ˆSn=nk =1bkand suppose 0≤ ak≤ bkfor all k
Then{Sn} and { ˆSn} and both increasing sequences, bounded below by 0 Sn≤ ˆSnfor all
n.∞k=1bkconverges, so{ ˆSn} is bounded by some M Then {Sn} is likewise bounded
by M, and hence∞k=1akconverges by the Bounded Increasing Partial Sums Theorem Because the first N terms of a series don’t affect whether or not it converges, we can draw the same conclusion if 0≤ ak≤ bkfor all k≥ N
ii Let ˜Sn=nk =1ckand suppose 0≤ ck≤ akfor all k
Then{Sn} and { ˜Sn} are both increasing sequences and ˜Sn≤ Snfor all n But∞k=1ck diverges By the Bounded Increasing Partial Sums Theorem{ ˜Sn} is unbounded, and therefore{Sn} is unbounded and∞k =1akdiverges
EXAMPLE 30.28 Does∞k=2 ln k1 converge or diverge?
SOLUTION Compare with the harmonic series
k >ln k for all k≥ 2 (See Figure 30.14.) Therefore 0≤ 1k <ln k1 for all k≥ 2 The series∞k =1 1k diverges; therefore∞k=2 ln k1 diverges by the Comparison Test
To use the Comparison Test effectively we need to have some familiar series on call The series most commonly used for comparison are the following:
The geometric series
∞
k =1
ark converges for |r| < 1 diverges for|r| ≥ 1
The p-series
∞
k =1
1
kp
converges for p > 1 diverges for p≤ 1
Suppose we’re working with the series∞k=110k1
−2 Basically this “looks like”∞k=1101k =
∞
k =1 101
k
, which converges Our series should behave similarly But 10k>10k− 2, so
1
10 k <10k1
−2 We can’t apply the Direct Comparison Test This is frustrating, because we know that the tail of the series is what really matters, and for large k, 10k≈ 10k− 2 The Limit Comparison Test gets us out of this bind
... sin x and cos xIt is entirely possible that we can solve for all the coefficients of a power series and simply have the solution expressed as and defined by the power series expansion... In both of these series the terms are positive, decreasing, and going toward zero, but the terms
of the series in part (b) are heading toward zero much more slowly than those in part (a)... converges if and only if its sequence of partial sums
is bounded above
Our focus in this section is on the question of convergence versus divergence and not on the sum of a convergent