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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS |||| 697

74. (a) A sequence an is deﬁned recursively by the equation

an 12 an 1 an 2 for n 3, where a1 and a2 can be any real numbers. Experiment with various values of a1 and a2 and use your calculator to guess the limit of the sequence.

(b)Find limn l an in terms of a1 and a2 by expressing an 1 an in terms of a2 a1 and summing a series.

75.Consider the series

	n

	n	1 !
n 1

(a)Find the partial sums s1, s2, s3, and s4. Do you recognize the denominators? Use the pattern to guess a formula for sn.

(b)Use mathematical induction to prove your guess.

(c)Show that the given inﬁnite series is convergent, and ﬁnd its sum.

76.In the ﬁgure there are inﬁnitely many circles approaching the vertices of an equilateral triangle, each circle touching other

circles and sides of the triangle. If the triangle has sides of length 1, ﬁnd the total area occupied by the circles.

11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS

				In general, it is difﬁcult to ﬁnd the exact sum of a series. We were able to accomplish this
				for geometric series and the series										1 n n 1 because in each of those cases we could
				ﬁnd a simple formula for the nth partial sum sn. But usually it isn’t easy to compute
				limn l sn. Therefore, in the next few sections, we develop several tests that enable us to
				determine whether a series is convergent or divergent without explicitly ﬁnding its sum.
				(In some cases, however, our methods will enable us to ﬁnd good estimates of the sum.)
				Our ﬁrst test involves improper integrals.
				We begin by investigating the series whose terms are the reciprocals of the squares of
				the positive integers:
	n	1		the positive integers:			1				1			1			1				1		1
	n
	sn
n
			2
		i	2					n	2			2			2			2			4	2	5	2
	i 1	i							2			2			2			2				2		2
					n 1						1			2			3
5	1.4636			There’s no simple formula for the sum sn of the ﬁrst n terms, but the computer-generated
10	1.5498
10	1.5498			table of values given in the margin suggests that the partial sums are approaching a num-
50	1.6251
50	1.6251			ber near 1.64 as n l and so it looks as if the series is convergent.
100	1.6350
100	1.6350			We can conﬁrm this impression with a geometric argument. Figure 1 shows the curve
500	1.6429
500	1.6429			y 1 x2 and rectangles that lie below the curve. The base of each rectangle is an interval
1000	1.6439
1000	1.6439			of length 1; the height is equal to the value of the function y 1 x2 at the right endpoint
5000	1.6447
5000	1.6447			of the interval. So the sum of the areas of the rectangles is

				1						1			1			1			1							1
				1						1			1			1			1							1

				1		2					2		3	2		4	2		5	2						n	2
				1					2				3			4			5						n 1	n
				y						1
								y=		1
								y=		≈
										≈

area=	1
	1@
0	1		2		3		4	5	x
FIGURE 1	area=	1	area=	1	area=	1	area=	1
FIGURE 1		2@		3@		4@		5@

698 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

If we exclude the ﬁrst rectangle, the total area of the remaining rectangles is smaller than the area under the curve y 1 x2 for x 1, which is the value of the integral x1 1 x2 dx. In Section 7.8 we discovered that this improper integral is convergent and has value 1. So the picture shows that all the partial sums are less than

1	1
	y1		dx 2
12		x2

			Thus the partial sums are bounded. We also know that the partial sums are increasing
			(because all the terms are positive). Therefore the partial sums converge (by the Monotonic
			Sequence Theorem) and so the series is convergent. The sum of the series (the limit of the
			partial sums) is also less than 2:
						12	12	12	12	12 2
						12	12	12	12	12 2
					n 1	n	1	2	3	4
			[The exact sum of this series was found by the Swiss mathematician Leonhard Euler
			(1707–1783) to be	2 6, but the proof of this fact is quite difﬁcult. (See Problem 6 in the
			Problems Plus following Chapter 15.)]
	n		Now let’s look at the series
n	sn 1
	i 1 s	i			1 1			1 1 1 1
	i 1 s				1 1			1 1 1 1
5	3.2317		n 1 sn				s1	s2	s3	s4	s5
10	5.0210		The table of values of sn suggests that the partial sums aren’t approaching a ﬁnite number,
50	12.7524
50	12.7524		so we suspect that the given series may be divergent. Again we use a picture for conﬁrma-
100	18.5896
100	18.5896		tion. Figure 2 shows the curve y 1 x , but this time we use rectangles whose tops lie
500	43.2834
500	43.2834		above the curve.					s
1000	61.8010		above the curve.
5000	139.9681
			y		y=	1
					y=	1
					œ„x
			0		1	area= 1		2		3	4	5	x
						area= 1		area= 1		area= 1	area= 1
	FIGURE 2						œ„1		œ2„	œ3„	œ4„

The base of each rectangle is an interval of length 1. The height is equal to the value of the function y 1 sx at the left endpoint of the interval. So the sum of the areas of all the rectangles is

1	1	1	1	1		1
1	1	1	1	1		1

s1 s2 s3 s4 s5					n 1	sn

This total area is greater than the area under the curve y 1 sx for x 1, which is equal to the integral x1 (1 sx ) dx. But we know from Section 7.8 that this improper integral is divergent. In other words, the area under the curve is inﬁnite. So the sum of the series must be inﬁnite; that is, the series is divergent.

The same sort of geometric reasoning that we used for these two series can be used to prove the following test. (The proof is given at the end of this section.)

case, limn l

SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS |||| 699

N In order to use the Integral Test we need to be able to evaluate x1 f x dx and therefore we have to be able to ﬁnd an antiderivative of f. Frequently this is difﬁcult or impossible, so we need other tests for convergence too.

THE INTEGRAL TEST Suppose f			is a continuous, positive, decreasing function on
1, and let an f n . Then the series n 1 an is convergent if and only if the
improper integral x1 f x dx is convergent. In other words:

(i)	If	f x dx is convergent, then an is convergent.
		y1	n 1

(ii)	If	f x dx is divergent, then an is divergent.
		y1	n 1

NOTE When we use the Integral Test, it is not necessary to start the series or the integral at n 1. For instance, in testing the series

	1			1
n 4	1			1
n 4	n 3			y4 x 3
		2	we use		2	dx

Also, it is not necessary that f be always decreasing. What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N. Then n N an is convergent, so n 1 an is convergent by Note 4 of Section 11.2.

EXAMPLE 1 Test the series

for convergence or divergence.

n 1

SOLUTION The function f x 1 x2 1 is continuous, positive, and decreasing on

1, so we use the Integral Test:

dx lim

t 1

dx lim tan 1x]1t

y1 x2

t l y1 x2 1

t l

lim

tan 1t

t l

Thus x1 1 x2 1 dx is a convergent integral and so, by the Integral Test, the series

1 n2 1 is convergent.

EXAMPLE 2 For what values of p is the series

convergent?

n 1

SOLUTION If p 0, then limn l 1 np . If p 0, then limn l 1 np 1. In either1 np 0, so the given series diverges by the Test for Divergence (11.2.7).

If p 0, then the function f x 1 x p is clearly continuous, positive, and decreasing on 1, . We found in Chapter 7 [see (7.8.2)] that

	1
y1		dx converges if p 1 and diverges if p 1
y1	x p	dx converges if p 1 and diverges if p 1

It follows from the Integral Test that the series 1 np converges if p 1 and diverges if

0 p 1. (For p 1, this series is the harmonic series discussed in Example 7 in
Section 11.2.)	M

The series in Example 2 is called the p-series. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows.

700 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

		1
1 The p-series		1		is convergent if p 1 and divergent if p 1.

		n	p
	n 1	n

EXAMPLE 3
(a) The series
					1		1			1		1		1
					1		1			1		1		1
					3		3			3		3		3
		n 1			n	1			2		3		4
is convergent because it is a p-series with p 3 1.
(b) The series
	1					1					1		1		1
	1					1		1			1		1		1
	1 3					3					3		3		3
	1 3					3					3		3		3
n 1	n				n 1	sn					s2		s3		s4
is divergent because it is a p-series with p 31 1.																M

NOTE We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. In fact,

n 1

2		1

whereas	y1		dx 1
		x2

Therefore, in general,


	an	f x dx
	n 1	y1
			ln n
	EXAMPLE 4 Determine whether the series			converges or diverges.
V
V
		n 1	n

SOLUTION The function f x ln x x is positive and continuous for x 1 because the logarithm function is continuous. But it is not obvious whether or not f is decreasing, so we compute its derivative:

f x	1 x x ln x	1 ln x
	x2	x2

Thus f x 0 when ln x 1, that is, x e. It follows that f is decreasing when x e and so we can apply the Integral Test:

						t
	ln x	dx lim t	ln x	dx lim	ln x 2	1	lim	ln t 2

y1 x		t l y1 x		t l 2			t l 2

Since this improper integral is divergent, the series ln n n is also divergent by the
Integral Test.	M

ESTIMATING THE SUM OF A SERIES

Suppose we have been able to use the Integral Test to show that a series an is convergent and we now want to ﬁnd an approximation to the sum s of the series. Of course, any partial sum sn is an approximation to s because limn l sn s. But how good is such an approximation? To ﬁnd out, we need to estimate the size of the remainder

Rn s sn an 1 an 2 an 3

SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS |||| 701

y=ƒ

	an+1	a	n+2	. . .
			n+2	. . .
0	n			x

The remainder Rn is the error made when sn, the sum of the ﬁrst n terms, is used as an approximation to the total sum.

We use the same notation and ideas as in the Integral Test, assuming that f is decreasing on n, . Comparing the areas of the rectangles with the area under y f x for x n in Figure 3, we see that

Rn an 1 an 2 y f x dx

FIGURE 3
y	y=ƒ
	y=ƒ
	an+1	an+2	. . .
0	n+1		x
	n+1

FIGURE 4

Similarly, we see from Figure 4 that

Rn an 1 an 2 y f x dx

n 1

So we have proved the following error estimate.

2 REMAINDER ESTIMATE FOR THE INTEGRAL TEST Suppose		f k ak , where f
is a continuous, positive, decreasing function for x n and an is convergent. If
Rn s sn, then

yn 1 f x dx Rn yn f x dx

V EXAMPLE 5

(a)Approximate the sum of the series 1 n3 by using the sum of the ﬁrst 10 terms. Estimate the error involved in this approximation.

(b)How many terms are required to ensure that the sum is accurate to within 0.0005?

SOLUTION In both parts (a) and (b) we need to know xn f x dx. With f x 1 x3, which satisﬁes the conditions of the Integral Test, we have

												t
	1		dx lim				1			n			lim				1				1			1
			dx lim										lim
yn x3				t l			2x2							t l				2t2			2n2			2n2
(a)
		1				1			1					1						1
		1		s10		1			1					1						1			1.1975
		3		s10		3			3					3								3	1.1975
n 1		n			1			2					3						10
According to the remainder estimate in (2), we have
								1								1			1
				R10 y10							dx
				R10 y10				x3			dx				2 10 2				200

So the size of the error is at most 0.005.

(b) Accuracy to within 0.0005 means that we have to ﬁnd a value of n such that Rn 0.0005. Since

			1		1
	Rn yn			dx
	Rn yn		x3	dx	2n2
we want	1		0.0005

		2n2

702 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

Solving this inequality, we get

n2 1 1000 or n s1000 31.6 0.001

We need 32 terms to ensure accuracy to within 0.0005.

If we add sn to each side of the inequalities in (2), we get


3	sn yn 1	f x dx s sn yn f x dx

because sn Rn s. The inequalities in (3) give a lower bound and an upper bound for s. They provide a more accurate approximation to the sum of the series than the partial sum sn does.

EXAMPLE 6 Use (3) with n 10 to estimate the sum of the series .

n 1 n3

SOLUTION The inequalities in (3) become

s10 y11

dx s s10

y10

From Example 5 we know that

2n2

s10

s s10

2 11 2

2 10 2

Using s10 1.197532, we get

1.201664 s 1.202532

If we approximate s by the midpoint of this interval, then the error is at most half the length of the interval. So

	1
		1.2021 with error 0.0005	M
	3
n 1	n

If we compare Example 6 with Example 5, we see that the improved estimate in (3) can be much better than the estimate s sn. To make the error smaller than 0.0005 we had to use 32 terms in Example 5 but only 10 terms in Example 6.

PROOF OF THE INTEGRAL TEST

We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and 2 for the series 1 n2 and 1 sn . For the general series an, look at Figures 5 and 6. The area of the ﬁrst shaded rectangle in Figure 5 is the value of f at the right endpoint of 1, 2 ,

y y=ƒ

	a™	a£	a¢	a∞	an
0	1 2 3 4 5 . . .				n x

FIGURE 5

	SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS \|\|\|\| 703
that is,	f 2 a2. So, comparing the areas of the shaded rectangles with the area under
y f x from 1 to n, we see that
	n
4	a2 a3 an y1 f x dx

(Notice that this inequality depends on the fact that f is decreasing.) Likewise, Figure 6 shows that

				n
y	y=ƒ	5		y1	f x dx a1 a2 an 1

		(i)	If y1	f x dx is convergent, then (4) gives
		an-1			n
					n	n f x dx
	a¡ a™ a£ a¢				ai	n f x dx	f x dx
0	1 2 3 4 5 . . .	n x			i 2	y1	y1
0	1 2 3 4 5 . . .	n x

since f x 0. Therefore

FIGURE 6

sn a1 ai a1 y f x dx M, say

i 2	1

Since sn M for all n, the sequence sn is bounded above. Also sn 1 sn an 1 sn

since an 1 f n 1 0. Thus sn is an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem (11.1.12). This means that an is convergent.

(ii) If x1 f x dx is divergent, then x1n f x dx l as n l because f x 0. But (5) gives

	n 1
n	f x dx ai sn 1
y1	i 1
and so sn 1 l . This implies that sn l and so an diverges.		M

11.3EXERCISES

1. Draw a picture to show that

	1			1

			y1			dx
	n	1.3		x	1.3
n 2

3– 8 Use the Integral Test to determine whether the series is convergent or divergent.

	1		1
3.		4.
	5		5

n 1	sn	n 1	n

What can you conclude about the series?

2.Suppose f is a continuous positive decreasing function for x 1 and an f n . By drawing a picture, rank the following three quantities in increasing order:

6 f x dx	5	6
6 f x dx	ai	ai
y1	i 1	i 2

			1				1
5.			1		6.		1

				3
		2n	1	3			sn 4
	n 1	2n	1			n 1	sn 4
							n 2
7.	ne n				8.		n 2
7.	ne n				8.		n 1
	n 1					n 1	n 1

704 |||| CHAPTER 11 INFINITE SEQUENCES AND SERIES

9–26 Determine whether the series is convergent or divergent.

			2
9.			2																		10.		n 1.4 3n 1.2
9.		n	0.85																		10.		n 1.4 3n 1.2
	n 1	n																					n 1
11.	1		1						1						1				1
11.	1																		125
			8							27			64						125
12.	1					1							1						1			1
12.	1
			2 s2										3 s3						4 s4			5 s5
13.	1		1						1					1				1
13.	1
			3							5			7				9
14.	1			1							1					1				1
14.	5
	5				8					11			14						17
		5 2 s																								n2
		5 2 s										n														n2
15.																					16.
15.							n	3													16.			n	3	1
	n 1						n																n 1	n		1
						1																		3n 2
17.						1															18.			3n 2
17.		n	2			4															18.			n n 1
	n 1	n				4																	n 1	n n 1
		ln n																								1
19.																					20.
19.		n			3																20.			n	2	4n 5
	n 1	n																					n 1	n		4n 5
					1																					1
21.					1																22.					1

		n ln n																						n ln n			2
	n 2	n ln n																					n 2	n ln n
		e	1 n																						2
23.		e																			24.			n
23.			2																		24.			n
	n 1	n																					n 3	e
						1																				n
25.						1															26.					n
25.		n	3			n															26.			n	4	1
	n 1	n				n																	n 1	n		1

27–30 Find the values of

	1
27.	1

	n ln n	p
n 2	n ln n

29. n 1 n2 p

n 1

p for which the series is convergent.

	1
28.	1

	n ln n ln ln n	p
n 3	n ln n ln ln n

ln n

30.p

n 1 n

31. The Riemann zeta-function is deﬁned by

		1
	x	1
	x	x
	n 1	n

and is used in number theory to study the distribution of prime numbers. What is the domain of ?

32.(a) Find the partial sum s10 of the series n 1 1 n4. Estimate the error in using s10 as an approximation to the sum of the series.

(b)Use (3) with n 10 to give an improved estimate of the sum.

(c)Find a value of n so that sn is within 0.00001 of the sum.

33.(a) Use the sum of the ﬁrst 10 terms to estimate the sum of the series n 1 1 n2. How good is this estimate?

(b)Improve this estimate using (3) with n 10.

(c)Find a value of n that will ensure that the error in the approximation s sn is less than 0.001.

34.Find the sum of the series n 1 1 n5 correct to three decimal places.

35.Estimate n 1 2n 1 6 correct to ﬁve decimal places.

36.How many terms of the series n 2 1 n ln n 2 would you need to add to ﬁnd its sum to within 0.01?

37.Show that if we want to approximate the sum of the seriesn 1 n 1.001 so that the error is less than 5 in the ninth decimal place, then we need to add more than 1011,301 terms!

CAS 38. (a) Show that the series n 1 ln n 2 n 2 is convergent.

(b)Find an upper bound for the error in the approximation s sn.

(c)What is the smallest value of n such that this upper bound is less than 0.05?

(d)Find sn for this value of n.

39.(a) Use (4) to show that if sn is the nth partial sum of the harmonic series, then

sn 1 ln n

(b)The harmonic series diverges, but very slowly. Use part (a) to show that the sum of the ﬁrst million terms is less than 15 and the sum of the ﬁrst billion terms is less than 22.

40.Use the following steps to show that the sequence

tn 1 1 1 1 ln n
2 3	n

has a limit. (The value of the limit is denoted by and is called Euler’s constant.)

(a) Draw a picture like Figure 6 with f x 1 x and interpret tn as an area [or use (5)] to show that tn 0 for all n.

(b) Interpret

1 tn tn 1 ln n 1 ln n n 1

as a difference of areas to show that tn tn 1 0. Therefore, tn is a decreasing sequence.

(c)Use the Monotonic Sequence Theorem to show that tn is convergent.

41.Find all positive values of b for which the series n 1 b ln n converges.

42.Find all values of c for which the following series converges.

	c	1

		n 1
n 1 n

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