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SECTION 11.7 STRATEGY FOR TESTING SERIES |||| 721

11.7 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 1 n p, it is a p-series, which we know to be convergent if p 1 and divergent if p 1.

2. If the series has the form ar n 1 or ar n, it is a geometric series, which converges if r 1 and diverges if r 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 p-series or a geometric series, then one of the comparison tests should be considered. In particular, if an is a rational function or an algebraic function of n (involving roots of polynomials), then the series should be compared with a p-series. 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 an has some negative terms, then we can apply the Comparison Test to an and test for absolute convergence.

4.If you can see at a glance that limn l an 0, then the Test for Divergence should be used.

5.If the series is of the form 1 n 1bn or 1 nbn, 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 thatan 1 an l 1 as n l for all p-series and therefore all rational or algebraic functions of n. Thus the Ratio Test should not be used for such series.

7.If an is of the form bn n, then the Root Test may be useful.

8.If an f n , where x1 f x dx is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisﬁed).

In the following examples we don’t work out all the details but simply indicate which tests should be used.

n 1

EXAMPLE 1

2n 1

n 1

Since an l 21 0 as n l , we should use the Test for Divergence.

n3 1

EXAMPLE 2

n 1

Since an is an algebraic function of n, we compare the given series with a p-series. The

722 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

comparison series for the Limit Comparison Test is bn, where
bn	s		n3 2	1
		n3			M
	3n3		3n3	3n3 2

V EXAMPLE 3 ne n2

n 1

Since the integral x1 xe x2 dx is easily evaluated, we use the Integral Test. The Ratio Test

also works.

EXAMPLE 4

1 n

n 1

Since the series is alternating, we use the Alternating Series Test.

EXAMPLE 5

k 1

Since the series involves k!, we use the Ratio Test.

EXAMPLE 6

n 1

Since the series is closely related to the geometric series 1 3n, we use the Comparison

Test. M

11.7EXERCISES

1–38 Test the series for convergence or divergence.

			1
1.			1

		n			3		n
	n 1	n			3
								n
3.		1 n						n

							n 2
	n 1						n 2
		2	2	n 1
5.		n	2
5.		5				n
	n 1	5

7.n 2 nsln n

9. k 2e k

k 1

1 n 1

11.n ln nn 2

3n n2

	n 1	n!
		n!
	15.	n!

		2 5 8 3n 2
	n 0	2 5 8 3n 2

17. 1 n 21 n

n 1

ln n

19.1 n

n 1 sn

		2n 1 n
2.
2.		n	2n
	n 1	n
						n
4.		1 n				n

				n	2	2
	n 1			n		2
		1
6.		1

		2n 1
	n 1	2n 1
		2 k k!
8.
8.		k 2 !
	k 1	k 2 !

10.	n2e n3
	n 1

12.	sin n

n 1

sin 2n

14.2nn 1 1

n2 1

16.

n 1 n3 1

1 n 1

18. n 2 sn 1

k 5

20.

k 1 5k

2 2n

n 2

21.

22.

n 1

23.

tan 1 n

24.

n sin 1 n

n 1

n2 1

25.

26.

n 1

k ln k

1 n

27.

28.

k 1

n 1

29.

30.

1 j

cosh n

j 5

n 1

j 1

5 k

n! n

31.

32.

k 1

n 1

sin 1 n

33.

34.

n 1

n n cos

35.

36.

n 1

ln n

n 1

n 2

37.

38.

(s2

(s2 1)

n 1

SECTION 11.8 POWER SERIES |||| 723

11.8 POWER SERIES

A power series is a series of the form


1	cn x n c0 c1 x c2 x 2 c3 x 3

n 0

N TRIGONOMETRIC SERIES

A power series is a series in which each term is a power function. A trigonometric series

an cos nx bn sin nx

n 0

is a series whose terms are trigonometric functions. This type of series is discussed on the website

www.stewartcalculus.com

Click on Additional Topics and then on Fourier Series.

N Notice that

n 1 ! n 1 n n 1 . . . 3 2 1

n 1 n!

where x is a variable and the cn’s are constants called the coefﬁcients of the series. For each ﬁxed x, the series (1) is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x. The sum of the series is a function

f x c0 c1 x c2 x 2 cn x n

whose domain is the set of all x for which the series converges. Notice that f resembles a polynomial. The only difference is that f has inﬁnitely many terms.

For instance, if we take cn 1 for all n, the power series becomes the geometric series

x n 1 x x 2 x n

n 0

which converges when 1 x 1 and diverges when x 1 (see Equation 11.2.5). More generally, a series of the form


2	cn x a n c0 c1 x a c2 x a 2

n 0

is called a power series in x a or a power series centered at a or a power series about a. Notice that in writing out the term corresponding to n 0 in Equations 1 and 2 we have adopted the convention that x a 0 1 even when x a. Notice also that when x a, all of the terms are 0 for n 1 and so the power series (2) always converges when x a.

V EXAMPLE 1 For what values of x is the series n!x n convergent?

n 0

SOLUTION We use the Ratio Test. If we let an, as usual, denote the nth term of the series, then an n!x n. If x 0, we have

lim

an 1

lim

n 1 !x n 1

lim n 1

n!x n

n l

By the Ratio Test, the series diverges when x 0. Thus the given series converges only

when x 0.

EXAMPLE 2 For what values of x does the series

converge?

n 1

SOLUTION Let an x 3 n n. Then

an 1

x 3 n 1

n 1

x 3 n

x 3 l x 3

as n l

724 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

National Film Board of Canada

N Notice how closely the computer-generated model (which involves Bessel functions and cosine functions) matches the photograph of a vibrating rubber membrane.

By the Ratio Test, the given series is absolutely convergent, and therefore convergent, when x 3 1 and divergent when x 3 1. Now

x 3 1 &? 1 x 3 1 &? 2 x 4
so the series converges when 2 x 4 and diverges when x 2 or x 4.
The Ratio Test gives no information when x 3 1 so we must consider x 2
and x 4 separately. If we put x 4 in the series, it becomes 1 n, the harmonic
series, which is divergent. If x 2, the series is 1 n n, which converges by the
Alternating Series Test. Thus the given power series converges for 2 x 4.	M

We will see that the main use of a power series is that it provides a way to represent some of the most important functions that arise in mathematics, physics, and chemistry. In particular, the sum of the power series in the next example is called a Bessel function, after the German astronomer Friedrich Bessel (1784–1846), and the function given in Exercise 35 is another example of a Bessel function. In fact, these functions ﬁrst arose when Bessel solved Kepler’s equation for describing planetary motion. Since that time, these functions have been applied in many different physical situations, including the temperature distribution in a circular plate and the shape of a vibrating drumhead.

EXAMPLE 3 Find the domain of the Bessel function of order 0 deﬁned by

1 nx 2n

J0 x

n 0

SOLUTION Let an 1 nx 2n 22n n! 2 . Then

an 1

1 n 1x 2 n 1

22n n! 2

22 n 1 n 1 ! 2

1 nx 2n

x 2n 2

22n n! 2

22n 2 n 1 2 n! 2

x 2n

x 2

l 0 1

for all x

4 n 1 2

Thus, by the Ratio Test, the given series converges for all values of x. In other words, the domain of the Bessel function J0 is , . M

Recall that the sum of a series is equal to the limit of the sequence of partial sums. So when we deﬁne the Bessel function in Example 3 as the sum of a series we mean that, for every real number x,

										n		1 i x 2i
	J0 x lim sn x				where			sn x
	J0 x lim sn x				where			sn x			2		2i		2
			n l							i 0	2			i!
The ﬁrst few partial sums are
s0 x 1				s1 x	1		x 2	s2 x 1						x 2			x 4
s0 x 1				s1 x	1	4		s2 x 1						4			64
						4								4			64
s3 x 1	x 2	x 4		x 6		s4 x 1			x 2			x 4				x 6		x 8

	4	64		2304					4		64				2304			147,456
	4	64		2304					4		64				2304			147,456

								SECTION 11.8 POWER SERIES \|\|\|\| 725
y				s™		Figure 1 shows the graphs of these partial sums, which are polynomials. They are all
y
1						approximations to the function J0, but notice that the approximations become better when
1				s¸		more terms are included. Figure 2 shows a more complete graph of the Bessel function.

						For the power series that we have looked at so far, the set of values of x for which the
				s¢		series is convergent has always turned out to be an interval [a ﬁnite interval for the
				s¢		geometric series and the series in Example 2, the inﬁnite interval , in Example 3,
						and a collapsed interval 0, 0 0 in Example 1]. The following theorem, proved in
0		1			x
0		1			x	Appendix F, says that this is true in general.
				J¸		Appendix F, says that this is true in general.
				J¸
				s¡ s£
				s¡ s£

FIGURE 1						3	THEOREM For a given power series cn x a n, there are only three
Partial sums of the Bessel function J¸						possibilities:		n 0
						possibilities:
						(i)	The series converges only when x a.
						(i)	The series converges only when x a.
y						(ii)	The series converges for all x.
1						(iii)	There is a positive number R such that the series converges if x a R
			y=J¸(x)				and diverges if x a	R.

_10	10					The number R in case (iii) is called the radius of convergence of the power series. By

0x convention, the radius of convergence is R 0 in case (i) and R in case (ii). The

interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In case (i) the interval consists of just a single point a. In case

FIGURE 2 (ii) the interval is , . In case (iii) note that the inequality x a R can be rewritten as a R x a R. When x is an endpoint of the interval, that is, x a R,

anything can happen—the series might converge at one or both endpoints or it might diverge at both endpoints. Thus in case (iii) there are four possibilities for the interval of convergence:

a R, a R a R, a R a R, a R a R, a R

The situation is illustrated in Figure 3.

convergence for |x-a|<R

	a-R	a	a+R
FIGURE 3		divergence for \|x-a\|>R
FIGURE 3		divergence for \|x-a\|>R

We summarize here the radius and interval of convergence for each of the examples already considered in this section.

		Series					Radius of convergence	Interval of convergence


Geometric series	x n						R 1	1, 1
	n 0

Example 1	n! x n						R 0	0
	n 0
		x			3	n
Example 2		x			3		R 1	2, 4
Example 2			n				R 1	2, 4
	n 1		n
		1 n x 2n
Example 3							R	,
Example 3		2	2n		2		R	,
	n 0	2		n!

726 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

In general, the Ratio Test (or sometimes the Root Test) should be used to determine the radius of convergence R. The Ratio and Root Tests always fail when x is an endpoint of the interval of convergence, so the endpoints must be checked with some other test.

EXAMPLE 4 Find the radius of convergence and interval of convergence of the series

3 nxn

n 0

n 1

SOLUTION Let an 3 nxn

n 1

. Then

3 n 1xn 1

an 1

n 1

3 nxn

n 2

1 1 n

x l 3 x

as n l

1 2 n

By the Ratio Test, the given series converges if 3 x 1 and diverges if 3 x 1.

Thus it converges if x 31 and diverges if x 31 . This means that the radius of con-

vergence is R 31 .

We know the series converges in the interval ( 13 , 13 ), but we must now test for convergence at the endpoints of this interval. If x 13 , the series becomes

		n	1	n		1	1	1	1	1
	3		( 3 )			1	1	1	1	1

n 0	sn 1				n 0	sn 1	s1 s2 s3 s4

which diverges. (Use the Integral Test or simply observe that it is a p-series with p 12 1.) If x 13 , the series is

		n 1	n		1	n
	3 (3 )				1

	sn 1				sn 1
n 0	sn 1			n 0	sn 1

which converges by the Alternating Series Test. Therefore the given power series converges when 13 x 13 , so the interval of convergence is ( 13 , 13 ]. M

V EXAMPLE 5 Find the radius of convergence and interval of convergence of the series

n x 2 n

n 1

n 0

SOLUTION If an n x 2 n 3n 1, then

an 1

n 1 x 2 n 1

3n 1

3n 2

n x 2 n

x 2

as n l

Using the Ratio Test, we see that the series converges if x 2 3 1 and it diverges ifx 2 3 1. So it converges if x 2 3 and diverges if x 2 3. Thus the radius of convergence is R 3.

SECTION 11.8 POWER SERIES |||| 727

The inequality x 2 3 can be written as 5 x 1, so we test the series at the endpoints 5 and 1. When x 5, the series is

	n 3 n		1
			3	1 nn
	3	n 1
n 0				n 0

which diverges by the Test for Divergence [ 1 nn doesn’t converge to 0]. When x 1, the series is

	n
	n 3	31	n
	n 1
n 0	3		n 0
which also diverges by the Test for Divergence. Thus the series converges only when
5 x 1, so the interval of convergence is 5, 1 .				M

11.8EXERCISES

1.What is a power series?

2.(a) What is the radius of convergence of a power series? How do you ﬁnd it?

(b)What is the interval of convergence of a power series? How do you ﬁnd it?

3–28 Find the radius of convergence and interval of convergence of the series.

		x	n
3.		x

		sn
	n 1	sn
		1			n 1	x	n
5.		1				x
5.				n	3
	n 1			n

x n

7.n!n 0

1 nx n

4.n 1n 0

6. sn x n

n 1

8. n nx n

n 1

									n 2 x n
23.	n! 2x 1 n				24.				n 2 x n
23.	n! 2x 1 n				24.		2 4 6 2n
	n 1					n 1	2 4 6 2n
		4x 1 n					x 2n
25.					26.
25.		n	2		26.		n ln n	2
	n 1	n				n 2	n ln n
				x n
27.
27.		1 3 5		2n 1
	n 1	1 3 5		2n 1
				n! x n
28.
28.		1 3 5		2n 1
	n 1	1 3 5		2n 1

29. If n 0 cn 4n is convergent, does it follow that the following series are convergent?


(a) cn 2 n	(b) cn 4 n
n 0	n 0

n 2 x n

10 nx n

1 n

10.

n 1

2 nx n

x n

11.

12.

n 1

13.

1 n

14.

1 n

n 2

ln n

n 0

2n !

x 2 n

x 3 n

15.

16.

1 n

2n 1

n 0

x 4

17.

18.

x 1 n

n 1

x 2 n

3x 2 n

19.

20.

n 3

n 1

n x 4 n

21.

a n, b 0

22.

n 1

30.Suppose that n 0 cn x n converges when x 4 and diverges when x 6. What can be said about the convergence or diver-

gence of the following series?


(a) cn	(b)	cn 8n
n 0		n 0

(c) cn 3 n	(d)	1 ncn 9n
n 0		n 0

31.If k is a positive integer, ﬁnd the radius of convergence of the series

n! k

kn ! x n

n 0

32.Let p and q be real numbers with p q. Find a power series whose interval of convergence is

(a) p, q	(b)	p, q
(c) p, q	(d)	p, q

33.Is it possible to ﬁnd a power series whose interval of convergence is 0, ? Explain.

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