Skip to Content
OpenStax Logo
Calculus Volume 2

5.3 The Divergence and Integral Tests

Calculus Volume 25.3 The Divergence and Integral Tests
Buy book
  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 5.3.1. Use the divergence test to determine whether a series converges or diverges.
  • 5.3.2. Use the integral test to determine the convergence of a series.
  • 5.3.3. Estimate the value of a series by finding bounds on its remainder term.

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums {Sk}.{Sk}. In practice, explicitly calculating this limit can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or divergence for many types of series. In this section, we discuss two of these tests: the divergence test and the integral test. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.

Divergence Test

For a series n=1ann=1an to converge, the nthnth term anan must satisfy an0an0 as n.n.

Therefore, from the algebraic limit properties of sequences,

limkak=limk(SkSk1)=limkSklimkSk1=SS=0.limkak=limk(SkSk1)=limkSklimkSk1=SS=0.

Therefore, if n=1ann=1an converges, the nthnth term an0an0 as n.n. An important consequence of this fact is the following statement:

Ifan0asn,n=1andiverges.Ifan0asn,n=1andiverges.
5.8

This test is known as the divergence test because it provides a way of proving that a series diverges.

Theorem 5.8

Divergence Test

If limnan=c0limnan=c0 or limnanlimnan does not exist, then the series n=1ann=1an diverges.

It is important to note that the converse of this theorem is not true. That is, if limnan=0,limnan=0, we cannot make any conclusion about the convergence of n=1an.n=1an. For example, limn0(1/n)=0,limn0(1/n)=0, but the harmonic series n=11/nn=11/n diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if an0,an0, the divergence test is inconclusive.

Example 5.13

Using the divergence test

For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.

  1. n=1n3n1n=1n3n1
  2. n=11n3n=11n3
  3. n=1e1/n2n=1e1/n2

Solution

  1. Since n/(3n1)1/30,n/(3n1)1/30, by the divergence test, we can conclude that
    n=1n3n1n=1n3n1

    diverges.
  2. Since 1/n30,1/n30, the divergence test is inconclusive.
  3. Since e1/n210,e1/n210, by the divergence test, the series
    n=1e1/n2n=1e1/n2

    diverges.
Checkpoint 5.12

What does the divergence test tell us about the series n=1cos(1/n2)?n=1cos(1/n2)?

Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums {Sk}{Sk} and showing that S2k>1+k/2S2k>1+k/2 for all positive integers k.k. In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

To illustrate how the integral test works, use the harmonic series as an example. In Figure 5.12, we depict the harmonic series by sketching a sequence of rectangles with areas 1,1/2,1/3,1/4,…1,1/2,1/3,1/4,… along with the function f(x)=1/x.f(x)=1/x. From the graph, we see that

n=1k1n=1+12+13++1k>1k+11xdx.n=1k1n=1+12+13++1k>1k+11xdx.

Therefore, for each k,k, the kthkth partial sum SkSk satisfies

Sk=n=1k1n>1k+11xdx=lnx|1k+1=ln(k+1)ln(1)=ln(k+1).Sk=n=1k1n>1k+11xdx=lnx|1k+1=ln(k+1)ln(1)=ln(k+1).

Since limkln(k+1)=,limkln(k+1)=, we see that the sequence of partial sums {Sk}{Sk} is unbounded. Therefore, {Sk}{Sk} diverges, and, consequently, the series n=11nn=11n also diverges.

This is a graph in quadrant 1 of a decreasing concave up curve approaching the x axis – f(x) = 1/x. Five rectangles are drawn with base 1 over the interval [1, 6]. The height of each rectangle is determined by the value of the function at the left endpoint of the rectangle’s base. The areas for each are marked: 1, 1/2, 1/3, 1/4, and 1/5.
Figure 5.12 The sum of the areas of the rectangles is greater than the area between the curve f(x)=1/xf(x)=1/x and the xx-axis for x1.x1. Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.

Now consider the series n=11/n2.n=11/n2. We show how an integral can be used to prove that this series converges. In Figure 5.13, we sketch a sequence of rectangles with areas 1,1/22,1/32,…1,1/22,1/32,… along with the function f(x)=1/x2.f(x)=1/x2. From the graph we see that

n=1k1n2=1+122+132++1k2<1+1k1x2dx.n=1k1n2=1+122+132++1k2<1+1k1x2dx.

Therefore, for each k,k, the kthkth partial sum SkSk satisfies

Sk=n=1k1n2<1+1k1x2dx=11x|1k=11k+1=21k<2.Sk=n=1k1n2<1+1k1x2dx=11x|1k=11k+1=21k<2.

We conclude that the sequence of partial sums {Sk}{Sk} is bounded. We also see that {Sk}{Sk} is an increasing sequence:

Sk=Sk1+1k2fork2.Sk=Sk1+1k2fork2.

Since {Sk}{Sk} is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series n=11/n2n=11/n2 converges.

This is a graph in quadrant 1 of the decreasing concave up curve f(x) = 1/(x^2), which approaches the x axis. Rectangles of base 1 are drawn over the interval [0, 5]. The height of each rectangle is determined by the value of the function at the right endpoint of its base. The areas of each are marked: 1, 1/(2^2), 1/(3^2), 1/(4^2) and 1/(5^2).
Figure 5.13 The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve f(x)=1/x2f(x)=1/x2 and the xx-axis for x1.x1. Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.

We can extend this idea to prove convergence or divergence for many different series. Suppose n=1ann=1an is a series with positive terms anan such that there exists a continuous, positive, decreasing function ff where f(n)=anf(n)=an for all positive integers. Then, as in Figure 5.14(a), for any integer k,k, the kthkth partial sum SkSk satisfies

Sk=a1+a2+a3++ak<a1+1kf(x)dx<1+1f(x)dx.Sk=a1+a2+a3++ak<a1+1kf(x)dx<1+1f(x)dx.

Therefore, if 1f(x)dx1f(x)dx converges, then the sequence of partial sums {Sk}{Sk} is bounded. Since {Sk}{Sk} is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if 1f(x)dx1f(x)dx converges, then the series n=1ann=1an also converges. On the other hand, from Figure 5.14(b), for any integer k,k, the kthkth partial sum SkSk satisfies

Sk=a1+a2+a3++ak>1k+1f(x)dx.Sk=a1+a2+a3++ak>1k+1f(x)dx.

If limk1k+1f(x)dx=,limk1k+1f(x)dx=, then {Sk}{Sk} is an unbounded sequence and therefore diverges. As a result, the series n=1ann=1an also diverges. Since ff is a positive function, if 1f(x)dx1f(x)dx diverges, then limk1k+1f(x)dx=.limk1k+1f(x)dx=. We conclude that if 1f(x)dx1f(x)dx diverges, then n=1ann=1an diverges.

This shows two graphs side by side of the same function y = f(x), a decreasing concave up curve approaching the x axis. Rectangles are drawn with base 1 over the intervals [0, 6] and [1, 6]. For the graph on the left, the height of each rectangle is determined by the value of the function at the right endpoint of its base. For the graph on the right, the height of each rectangle is determined by the value of the function at the left endpoint of its base. Areas a_1 through a_6 are marked in the graph on the left, and the same for a_1 to a_5 on the right.
Figure 5.14 (a) If we can inscribe rectangles inside a region bounded by a curve y=f(x)y=f(x) and the xx-axis, and the area bounded by those curves for x1x1 is finite, then the sum of the areas of the rectangles is also finite. (b) If a set of rectangles circumscribes the region bounded by y=f(x)y=f(x) and the xx axis for x1x1 and the region has infinite area, then the sum of the areas of the rectangles is also infinite.
Theorem 5.9

Integral Test

Suppose n=1ann=1an is a series with positive terms an.an. Suppose there exists a function ff and a positive integer NN such that the following three conditions are satisfied:

  1. ff is continuous,
  2. ff is decreasing, and
  3. f(n)=anf(n)=an for all integers nN.nN.
    Then
    n=1anandNf(x)dxn=1anandNf(x)dx

    both converge or both diverge (see Figure 5.14).

Although convergence of Nf(x)dxNf(x)dx implies convergence of the related series n=1an,n=1an, it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,

n=1(1e)n=1e+(1e)2+(1e)3+n=1(1e)n=1e+(1e)2+(1e)3+

is a geometric series with initial term a=1/ea=1/e and ratio r=1/e,r=1/e, which converges to

1/e1(1/e)=1/e(e1)/e=1e1.1/e1(1/e)=1/e(e1)/e=1e1.

However, the related integral 1(1/e)xdx1(1/e)xdx satisfies

1(1e)xdx=1exdx=limb1bexdx=limbex|1b=limb[eb+e−1]=1e.1(1e)xdx=1exdx=limb1bexdx=limbex|1b=limb[eb+e−1]=1e.

Example 5.14

Using the Integral Test

For each of the following series, use the integral test to determine whether the series converges or diverges.

  1. n=11/n3n=11/n3
  2. n=11/2n1n=11/2n1

Solution

  1. Compare
    n=11n3and11x3dx.n=11n3and11x3dx.

    We have
    11x3dx=limb1b1x3dx=limb[12x2|1b]=limb[12b2+12]=12.11x3dx=limb1b1x3dx=limb[12x2|1b]=limb[12b2+12]=12.

    Thus the integral 11/x3dx11/x3dx converges, and therefore so does the series
    n=11n3.n=11n3.
  2. Compare
    n=112n1and112x1dx.n=112n1and112x1dx.

    Since
    112x1dx=limb1b12x1dx=limb2x1|1b=limb[2b11]=,112x1dx=limb1b12x1dx=limb2x1|1b=limb[2b11]=,

    the integral 11/2x1dx11/2x1dx diverges, and therefore
    n=112n1n=112n1

    diverges.
Checkpoint 5.13

Use the integral test to determine whether the series n=1n3n2+1n=1n3n2+1 converges or diverges.

The p-Series

The harmonic series n=11/nn=11/n and the series n=11/n2n=11/n2 are both examples of a type of series called a p-series.

Definition

For any real number p,p, the series

n=11npn=11np

is called a p-series.

We know the p-series converges if p=2p=2 and diverges if p=1.p=1. What about other values of p?p? In general, it is difficult, if not impossible, to compute the exact value of most pp-series. However, we can use the tests presented thus far to prove whether a pp-series converges or diverges.

If p<0,p<0, then 1/np,1/np, and if p=0,p=0, then 1/np1.1/np1. Therefore, by the divergence test,

n=11/npdiverges ifp0.n=11/npdiverges ifp0.

If p>0,p>0, then f(x)=1/xpf(x)=1/xp is a positive, continuous, decreasing function. Therefore, for p>0,p>0, we use the integral test, comparing

n=11npand11xpdx.n=11npand11xpdx.

We have already considered the case when p=1.p=1. Here we consider the case when p>0,p1.p>0,p1. For this case,

11xpdx=limb1b1xpdx=limb11px1p|1b=limb11p[b1p1].11xpdx=limb1b1xpdx=limb11px1p|1b=limb11p[b1p1].

Because

b1p0ifp>1andb1pifp<1,b1p0ifp>1andb1pifp<1,

we conclude that

11xpdx={1p1ifp>1ifp<1.11xpdx={1p1ifp>1ifp<1.

Therefore, n=11/npn=11/np converges if p>1p>1 and diverges if 0<p<1.0<p<1.

In summary,

n=11np{converges ifp>1diverges ifp1.n=11np{converges ifp>1diverges ifp1.
5.9

Example 5.15

Testing for Convergence of p-series

For each of the following series, determine whether it converges or diverges.

  1. n=11n4n=11n4
  2. n=11n2/3n=11n2/3

Solution

  1. This is a p-series with p=4>1,p=4>1, so the series converges.
  2. Since p=2/3<1,p=2/3<1, the series diverges.
Checkpoint 5.14

Does the series n=11n5/4n=11n5/4 converge or diverge?

Estimating the Value of a Series

Suppose we know that a series n=1ann=1an converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum n=1Nann=1Nan where NN is any positive integer. The question we address here is, for a convergent series n=1an,n=1an, how good is the approximation n=1Nan?n=1Nan? More specifically, if we let

RN=n=1ann=1NanRN=n=1ann=1Nan

be the remainder when the sum of an infinite series is approximated by the NthNth partial sum, how large is RN?RN? For some types of series, we are able to use the ideas from the integral test to estimate RN.RN.

Theorem 5.10

Remainder Estimate from the Integral Test

Suppose n=1ann=1an is a convergent series with positive terms. Suppose there exists a function ff satisfying the following three conditions:

  1. ff is continuous,
  2. ff is decreasing, and
  3. f(n)=anf(n)=an for all integers n1.n1.

Let SNSN be the Nth partial sum of n=1an.n=1an. For all positive integers N,N,

SN+N+1f(x)dx<n=1an<SN+Nf(x)dx.SN+N+1f(x)dx<n=1an<SN+Nf(x)dx.

In other words, the remainder RN=n=1anSN=n=N+1anRN=n=1anSN=n=N+1an satisfies the following estimate:

N+1f(x)dx<RN<Nf(x)dx.N+1f(x)dx<RN<Nf(x)dx.
5.10

This is known as the remainder estimate.

We illustrate Remainder Estimate from the Integral Test in Figure 5.15. In particular, by representing the remainder RN=aN+1+aN+2+aN+3+RN=aN+1+aN+2+aN+3+ as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by Nf(x)dxNf(x)dx and bounded below by N+1f(x)dx.N+1f(x)dx. In other words,

RN=aN+1+aN+2+aN+3+>N+1f(x)dxRN=aN+1+aN+2+aN+3+>N+1f(x)dx

and

RN=aN+1+aN+2+aN+3+<Nf(x)dx.RN=aN+1+aN+2+aN+3+<Nf(x)dx.

We conclude that

N+1f(x)dx<RN<Nf(x)dx.N+1f(x)dx<RN<Nf(x)dx.

Since

n=1an=SN+RN,n=1an=SN+RN,

where SNSN is the NthNth partial sum, we conclude that

SN+N+1f(x)dx<n=1an<SN+Nf(x)dx.SN+N+1f(x)dx<n=1an<SN+Nf(x)dx.
This shows two graphs side by side of the same decreasing concave up function y = f(x) that approaches the x axis in quadrant 1. Rectangles are drawn with a base of 1 over the intervals N through N + 4. The heights of the rectangles in the first graph are determined by the value of the function at the right endpoints of the bases, and those in the second graph are determined by the value at the left endpoints of the bases. The areas of the rectangles are marked: a_(N + 1), a_(N + 2), through a_(N + 4).
Figure 5.15 Given a continuous, positive, decreasing function ff and a sequence of positive terms anan such that an=f(n)an=f(n) for all positive integers n,n, (a) the areas aN+1+aN+2+aN+3+<Nf(x)dx,aN+1+aN+2+aN+3+<Nf(x)dx, or (b) the areas aN+1+aN+2+aN+3+>N+1f(x)dx.aN+1+aN+2+aN+3+>N+1f(x)dx. Therefore, the integral is either an overestimate or an underestimate of the error.

Example 5.16

Estimating the Value of a Series

Consider the series n=11/n3.n=11/n3.

  1. Calculate S10=n=1101/n3S10=n=1101/n3 and estimate the error.
  2. Determine the least value of NN necessary such that SNSN will estimate n=11/n3n=11/n3 to within 0.001.0.001.

Solution

  1. Using a calculating utility, we have
    S10=1+123+133+143++11031.19753.S10=1+123+133+143++11031.19753.

    By the remainder estimate, we know
    RN<N1x3dx.RN<N1x3dx.

    We have
    101x3dx=limb10b1x3dx=limb[12x2]Nb=limb[12b2+12N2]=12N2.101x3dx=limb10b1x3dx=limb[12x2]Nb=limb[12b2+12N2]=12N2.

    Therefore, the error is R10<1/2(10)2=0.005.R10<1/2(10)2=0.005.
  2. Find NN such that RN<0.001.RN<0.001. In part a. we showed that RN<1/2N2.RN<1/2N2. Therefore, the remainder RN<0.001RN<0.001 as long as 1/2N2<0.001.1/2N2<0.001. That is, we need 2N2>1000.2N2>1000. Solving this inequality for N,N, we see that we need N>22.36.N>22.36. To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is N=23.N=23.
Checkpoint 5.15

For n=11n4,n=11n4, calculate S5S5 and estimate the error R5.R5.

Section 5.3 Exercises

For each of the following sequences, if the divergence test applies, either state that limnanlimnan does not exist or find limnan.limnan. If the divergence test does not apply, state why.

138.

an=nn+2an=nn+2

139.

an=n5n23an=n5n23

140.

an=n3n2+2n+1an=n3n2+2n+1

141.

an=(2n+1)(n1)(n+1)2an=(2n+1)(n1)(n+1)2

142.

an=(2n+1)2n(3n2+1)nan=(2n+1)2n(3n2+1)n

143.

an=2n3n/2an=2n3n/2

144.

an=2n+3n10n/2an=2n+3n10n/2

145.

an=e−2/nan=e−2/n

146.

an=cosnan=cosn

147.

an=tannan=tann

148.

an=1cos2(1/n)sin2(2/n)an=1cos2(1/n)sin2(2/n)

149.

an=(11n)2nan=(11n)2n

150.

an=lnnnan=lnnn

151.

an=(lnn)2nan=(lnn)2n

State whether the given pp-series converges.

152.

n=11nn=11n

153.

n=11nnn=11nn

154.

n=11n23n=11n23

155.

n=11n43n=11n43

156.

n=1nenπn=1nenπ

157.

n=1nπn2en=1nπn2e

Use the integral test to determine whether the following sums converge.

158.

n=11n+5n=11n+5

159.

n=11n+53n=11n+53

160.

n=21nlnnn=21nlnn

161.

n=1n1+n2n=1n1+n2

162.

n=1en1+e2nn=1en1+e2n

163.

n=12n1+n4n=12n1+n4

164.

n=21nln2nn=21nln2n

Express the following sums as pp-series and determine whether each converges.

165.

n=12lnnn=12lnn (Hint: 2lnn=1/nln22lnn=1/nln2.)

166.

n=13lnnn=13lnn (Hint: 3lnn=1/nln33lnn=1/nln3.)

167.

n=1n2−2lnnn=1n2−2lnn

168.

n=1n3−2lnnn=1n3−2lnn

Use the estimate RNNf(t)dtRNNf(t)dt to find a bound for the remainder RN=n=1ann=1NanRN=n=1ann=1Nan where an=f(n).an=f(n).

169.

n=110001n2n=110001n2

170.

n=110001n3n=110001n3

171.

n=1100011+n2n=1100011+n2

172.

n=1100n/2nn=1100n/2n

[T] Find the minimum value of NN such that the remainder estimate N+1f<RN<NfN+1f<RN<Nf guarantees that n=1Nann=1Nan estimates n=1an,n=1an, accurate to within the given error.

173.

an=1n2,an=1n2, error <10−4<10−4

174.

an=1n1.1,an=1n1.1, error <10−4<10−4

175.

an=1n1.01,an=1n1.01, error <10−4<10−4

176.

an=1nln2n,an=1nln2n, error <10−3<10−3

177.

an=11+n2,an=11+n2, error <10−3<10−3

In the following exercises, find a value of NN such that RNRN is smaller than the desired error. Compute the corresponding sum n=1Nann=1Nan and compare it to the given estimate of the infinite series.

178.

an=1n11,an=1n11, error <10−4,<10−4, n=11n11=1.000494n=11n11=1.000494

179.

an=1en,an=1en, error <10−5,<10−5, n=11en=1e1=0.581976n=11en=1e1=0.581976

180.

an=1en2,an=1en2, error <10−5,<10−5, n=1n/en2=0.40488139857n=1n/en2=0.40488139857

181.

an=1/n4,an=1/n4, error <10−4,<10−4, n=11/n4=π4/90=1.08232...n=11/n4=π4/90=1.08232...

182.

an=1/n6,an=1/n6, error <10−6,<10−6, n=11/n4=π6/945=1.01734306...,n=11/n4=π6/945=1.01734306...,

183.

Find the limit as nn of 1n+1n+1++12n.1n+1n+1++12n. (Hint: Compare to n2n1tdt.)n2n1tdt.)

184.

Find the limit as nn of 1n+1n+1++13n1n+1n+1++13n

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

185.

In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number Hk=(1+12+13++1k).Hk=(1+12+13++1k). Recall that Tk=HklnkTk=Hklnk is decreasing. Compute T=limkTkT=limkTk to four decimal places. (Hint: 1k+1<kk+11xdx1k+1<kk+11xdx.)

186.

[T] Complete sampling with replacement, sometimes called the coupon collector’s problem, is phrased as follows: Suppose you have NN unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps E(N)E(N) that it takes to draw each unique item at least once. It turns out that E(N)=N.HN=N(1+12+13++1N).E(N)=N.HN=N(1+12+13++1N). Find E(N)E(N) for N=10,20,and50.N=10,20,and50.

187.

[T] The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has nn cards, then the probability that the insertion will be below the card initially at the bottom (call this card B)B) is 1/n.1/n. Thus the expected number of top random insertions before BB is no longer at the bottom is n. Once one card is below B,B, there are two places below BB and the probability that a randomly inserted card will fall below BB is 2/n.2/n. The expected number of top random insertions before this happens is n/2.n/2. The two cards below BB are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.

188.

Suppose a scooter can travel 100100 km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel 100HN100HN km, where HN=1+1/2++1/N.HN=1+1/2++1/N.

189.

Show that for the remainder estimate to apply on [N,)[N,) it is sufficient that f(x)f(x) be decreasing on [N,),[N,), but ff need not be decreasing on [1,).[1,).

190.

[T] Use the remainder estimate and integration by parts to approximate n=1n/enn=1n/en within an error smaller than 0.0001.0.0001.

191.

Does n=21n(lnn)pn=21n(lnn)p converge if pp is large enough? If so, for which p?p?

192.

[T] Suppose a computer can sum one million terms per second of the divergent series n=1N1n.n=1N1n. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed 100.100.

193.

[T] A fast computer can sum one million terms per second of the divergent series n=2N1nlnn.n=2N1nlnn. Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed 100.100.

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction
Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.