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Calculus Volume 2

5.1 Sequences

Calculus Volume 25.1 Sequences
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  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

  • 5.1.1. Find the formula for the general term of a sequence.
  • 5.1.2. Calculate the limit of a sequence if it exists.
  • 5.1.3. Determine the convergence or divergence of a given sequence.

In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.

Terminology of Sequences

To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form

a1,a2,a3,…,an,….a1,a2,a3,…,an,….

Each of the numbers in the sequence is called a term. The symbol nn is called the index variable for the sequence. We use the notation

{an}n=1,or simply{an},{an}n=1,or simply{an},

to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number anan exists for each positive integer n,n, we can also define a sequence as a function whose domain is the set of positive integers.

Let’s consider the infinite, ordered list

2,4,8,16,32,….2,4,8,16,32,….

This is a sequence in which the first, second, and third terms are given by a1=2,a1=2, a2=4,a2=4, and a3=8.a3=8. You can probably see that the terms in this sequence have the following pattern:

a1=21,a2=22,a3=23,a4=24,anda5=25.a1=21,a2=22,a3=23,a4=24,anda5=25.

Assuming this pattern continues, we can write the nthnth term in the sequence by the explicit formula an=2n.an=2n. Using this notation, we can write this sequence as

{2n}n=1or{2n}.{2n}n=1or{2n}.

Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the nthnth term anan in terms of the previous term an1.an1. In particular, we can define this sequence as the sequence {an}{an} where a1=2a1=2 and for all n2,n2, each term anan is defined by the recurrence relationan=2an1.an=2an1.

Definition

An infinite sequence{an}{an} is an ordered list of numbers of the form

a1,a2,…,an,….a1,a2,…,an,….

The subscript nn is called the index variable of the sequence. Each number anan is a term of the sequence. Sometimes sequences are defined by explicit formulas, in which case an=f(n)an=f(n) for some function f(n)f(n) defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.

Note that the index does not have to start at n=1n=1 but could start with other integers. For example, a sequence given by the explicit formula an=f(n)an=f(n) could start at n=0,n=0, in which case the sequence would be

a0,a1,a2,….a0,a1,a2,….

Similarly, for a sequence defined by a recurrence relation, the term a0a0 may be given explicitly, and the terms anan for n1n1 may be defined in terms of an1.an1. Since a sequence {an}{an} has exactly one value for each positive integer n,n, it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence {an}{an} consists of all points (n,an)(n,an) for all positive integers n.n. Figure 5.2 shows the graph of {2n}.{2n}.

A graph in quadrant one containing the following points: (1, 2), (2, 4), (3, 8), (4, 16).
Figure 5.2 The plotted points are a graph of the sequence {2n}.{2n}.

Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence

3,7,11,15,19,….3,7,11,15,19,….

You can see that the difference between every consecutive pair of terms is 4.4. Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation

{a1=3an=an1+4forn2.{a1=3an=an1+4forn2.

Note that

a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.

Thus the sequence can also be described using the explicit formula

an=3+4(n1)=4n1.an=3+4(n1)=4n1.

In general, an arithmetic sequence is any sequence of the form an=cn+b.an=cn+b.

In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence

2,23,29,227,281,….2,23,29,227,281,….

We see that the ratio of any term to the preceding term is 13.13. Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as

a1=2an=13·an1forn2.a1=2an=13·an1forn2.

Alternatively, since

a2=13·2a3=(13)(13)(2)=(13)2·2a4=(13)(13)(13)(2)=(13)3·2,a2=13·2a3=(13)(13)(2)=(13)2·2a4=(13)(13)(13)(2)=(13)3·2,

we see that the sequence can be described by using the explicit formula

an=2(13)n1.an=2(13)n1.

The sequence {2n}{2n} that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is 2.2. In general, a geometric sequence is any sequence of the form an=crn.an=crn.

Example 5.1

Finding Explicit Formulas

For each of the following sequences, find an explicit formula for the nthnth term of the sequence.

  1. 12,23,34,45,56,…12,23,34,45,56,…
  2. 34,97,2710,8113,24316,…34,97,2710,8113,24316,…

Solution

  1. First, note that the sequence is alternating from negative to positive. The odd terms in the sequence are negative, and the even terms are positive. Therefore, the nthnth term includes a factor of (−1)n.(−1)n. Next, consider the sequence of numerators {1,2,3,…}{1,2,3,…} and the sequence of denominators {2,3,4,…}.{2,3,4,…}. We can see that both of these sequences are arithmetic sequences. The nthnth term in the sequence of numerators is n,n, and the nthnth term in the sequence of denominators is n+1.n+1. Therefore, the sequence can be described by the explicit formula
    an=(−1)nnn+1.an=(−1)nnn+1.
  2. The sequence of numerators 3,9,27,81,243,…3,9,27,81,243,… is a geometric sequence. The numerator of the nthnth term is 3n3n The sequence of denominators 4,7,10,13,16,…4,7,10,13,16,… is an arithmetic sequence. The denominator of the nthnth term is 4+3(n1)=3n+1.4+3(n1)=3n+1. Therefore, we can describe the sequence by the explicit formula an=3n3n+1.an=3n3n+1.
Checkpoint 5.1

Find an explicit formula for the nthnth term of the sequence {15,17,19,111,…}.{15,17,19,111,…}.

Example 5.2

Defined by Recurrence Relations

For each of the following recursively defined sequences, find an explicit formula for the sequence.

  1. a1=2,a1=2, an=−3an1an=−3an1 for n2n2
  2. a1=12,a1=12, an=an1+(12)nan=an1+(12)n for n2n2

Solution

  1. Writing out the first few terms, we have
    a1=2a2=−3a1=−3(2)a3=−3a2=(−3)22a4=−3a3=(−3)32.a1=2a2=−3a1=−3(2)a3=−3a2=(−3)22a4=−3a3=(−3)32.

    In general,
    an=2(−3)n1.an=2(−3)n1.
  2. Write out the first few terms:
    a1=12a2=a1+(12)2=12+14=34a3=a2+(12)3=34+18=78a4=a3+(12)4=78+116=1516.a1=12a2=a1+(12)2=12+14=34a3=a2+(12)3=34+18=78a4=a3+(12)4=78+116=1516.

    From this pattern, we derive the explicit formula
    an=2n12n=112n.an=2n12n=112n.
Checkpoint 5.2

Find an explicit formula for the sequence defined recursively such that a1=−4a1=−4 and an=an1+6.an=an1+6.

Limit of a Sequence

A fundamental question that arises regarding infinite sequences is the behavior of the terms as nn gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as n.n. For example, consider the following four sequences and their different behaviors as nn (see Figure 5.3):

  1. {1+3n}={4,7,10,13,…}.{1+3n}={4,7,10,13,…}. The terms 1+3n1+3n become arbitrarily large as n.n. In this case, we say that 1+3n1+3n as n.n.
  2. {1(12)n}={12,34,78,1516,…}.{1(12)n}={12,34,78,1516,…}. The terms 1(12)n11(12)n1 as n.n.
  3. {(−1)n}={1,1,−1,1,…}.{(−1)n}={1,1,−1,1,…}. The terms alternate but do not approach one single value as n.n.
  4. {(−1)nn}={−1,12,13,14,…}.{(−1)nn}={−1,12,13,14,…}. The terms alternate for this sequence as well, but (−1)nn0(−1)nn0 as n.n.
Four graphs in quadrants 1 and 4, labeled a through d. The horizontal axis is for the value of n and the vertical axis is for the value of the term a _n. Graph a has points (1, 4), (2, 7), (3, 10), (4, 13), and (5, 16). Graph b has points (1, 1/2), (2, 3/4), (3, 7/8), and (4, 15/16). Graph c has points (1, -1), (2, 1), (3, -1), (4, 1), and (5, -1). Graph d has points (1, -1), (2, 1/2), (3, -1/3), (4, 1/4), and (5, -1/5).
Figure 5.3 (a) The terms in the sequence become arbitrarily large as n.n. (b) The terms in the sequence approach 11 as n.n. (c) The terms in the sequence alternate between 11 and −1−1 as n.n. (d) The terms in the sequence alternate between positive and negative values but approach 00 as n.n.

From these examples, we see several possibilities for the behavior of the terms of a sequence as n.n. In two of the sequences, the terms approach a finite number as n.n. In the other two sequences, the terms do not. If the terms of a sequence approach a finite number LL as n,n, we say that the sequence is a convergent sequence and the real number LL is the limit of the sequence. We can give an informal definition here.

Definition

Given a sequence {an},{an}, if the terms anan become arbitrarily close to a finite number LL as nn becomes sufficiently large, we say {an}{an} is a convergent sequence and LL is the limit of the sequence. In this case, we write

limnan=L.limnan=L.

If a sequence {an}{an} is not convergent, we say it is a divergent sequence.

From Figure 5.3, we see that the terms in the sequence {1(12)n}{1(12)n} are becoming arbitrarily close to 11 as nn becomes very large. We conclude that {1(12)n}{1(12)n} is a convergent sequence and its limit is 1.1. In contrast, from Figure 5.3, we see that the terms in the sequence 1+3n1+3n are not approaching a finite number as nn becomes larger. We say that {1+3n}{1+3n} is a divergent sequence.

In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.

Definition

A sequence {an}{an} converges to a real number LL if for all ε>0,ε>0, there exists an integer NN such that |anL|<ε|anL|<ε if nN.nN. The number LL is the limit of the sequence and we write

limnan=LoranL.limnan=LoranL.

In this case, we say the sequence {an}{an} is a convergent sequence. If a sequence does not converge, it is a divergent sequence, and we say the limit does not exist.

We remark that the convergence or divergence of a sequence {an}{an} depends only on what happens to the terms anan as n.n. Therefore, if a finite number of terms b1,b2,…,bNb1,b2,…,bN are placed before a1a1 to create a new sequence

b1,b2,…,bN,a1,a2,…,b1,b2,…,bN,a1,a2,…,

this new sequence will converge if {an}{an} converges and diverge if {an}{an} diverges. Further, if the sequence {an}{an} converges to L,L, this new sequence will also converge to L.L.

A graph in quadrant 1 with axes labeled n and a_n instead of x and y, respectively. A positive point N is marked on the n axis. From smallest to largest, points L – epsilon, L, and L + epsilon are marked on the a_n axis, with the same interval epsilon between L and the other two. A blue line y = L is drawn, as are red dotted ones for y = L + epsilon and L – epsilon. Points in quadrant 1 are plotted above and below these lines for x < N. However, past N, the points remain inside the lines y = L + epsilon and L – epsilon, converging on L.
Figure 5.4 As nn increases, the terms anan become closer to L.L. For values of nN,nN, the distance between each point (n,an)(n,an) and the line y=Ly=L is less than ε.ε.

As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences {1+3n}{1+3n} and {(−1)n}{(−1)n} shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The sequence {(−1)n}{(−1)n} diverges because the terms alternate between 11 and −1,−1, but do not approach one value as n.n. On the other hand, the sequence {1+3n}{1+3n} diverges because the terms 1+3n1+3n as n.n. We say the sequence {1+3n}{1+3n} diverges to infinity and write limn(1+3n)=.limn(1+3n)=. It is important to recognize that this notation does not imply the limit of the sequence {1+3n}{1+3n} exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence {5n+2}{5n+2} diverges to negative infinity because −5n+2−5n+2 as n.n. We write this as limn(−5n+2)=.limn(−5n+2)=.

Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence {an}{an} and a related function ff defined on all positive real numbers such that f(n)=anf(n)=an for all integers n1.n1. Since the domain of the sequence is a subset of the domain of f,f, if limxf(x)limxf(x) exists, then the sequence converges and has the same limit. For example, consider the sequence {1n}{1n} and the related function f(x)=1x.f(x)=1x. Since the function ff defined on all real numbers x>0x>0 satisfies f(x)=1x0f(x)=1x0 as x,x, the sequence {1n}{1n} must satisfy 1n01n0 as n.n.

Theorem 5.1

Limit of a Sequence Defined by a Function

Consider a sequence {an}{an} such that an=f(n)an=f(n) for all n1.n1. If there exists a real number LL such that

limxf(x)=L,limxf(x)=L,

then {an}{an} converges and

limnan=L.limnan=L.

We can use this theorem to evaluate limnrnlimnrn for 0r1.0r1. For example, consider the sequence {(1/2)n}{(1/2)n} and the related exponential function f(x)=(1/2)x.f(x)=(1/2)x. Since limx(1/2)x=0,limx(1/2)x=0, we conclude that the sequence {(1/2)n}{(1/2)n} converges and its limit is 0.0. Similarly, for any real number rr such that 0r<1,0r<1, limxrx=0,limxrx=0, and therefore the sequence {rn}{rn} converges. On the other hand, if r=1,r=1, then limxrx=1,limxrx=1, and therefore the limit of the sequence {1n}{1n} is 1.1. If r>1,r>1, limxrx=,limxrx=, and therefore we cannot apply this theorem. However, in this case, just as the function rxrx grows without bound as n,n, the terms rnrn in the sequence become arbitrarily large as n,n, and we conclude that the sequence {rn}{rn} diverges to infinity if r>1.r>1.

We summarize these results regarding the geometric sequence {rn}:{rn}:

rn0if0<r<1rn1ifr=1 rnifr>1.rn0if0<r<1rn1ifr=1 rnifr>1.

Later in this section we consider the case when r<0.r<0.

We now consider slightly more complicated sequences. For example, consider the sequence {(2/3)n+(1/4)n}.{(2/3)n+(1/4)n}. The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences {(2/3)n}{(2/3)n} and {(1/4)n}.{(1/4)n}. As we describe in the following algebraic limit laws, since {(2/3)n}{(2/3)n} and {1/4)n}{1/4)n} both converge to 0,0, the sequence {(2/3)n+(1/4)n}{(2/3)n+(1/4)n} converges to 0+0=0.0+0=0. Just as we were able to evaluate a limit involving an algebraic combination of functions ff and gg by looking at the limits of ff and gg (see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of anan and bnbn by evaluating the limits of {an}{an} and {bn}.{bn}.

Theorem 5.2

Algebraic Limit Laws

Given sequences {an}{an} and {bn}{bn} and any real number c,c, if there exist constants AA and BB such that limnan=Alimnan=A and limnbn=B,limnbn=B, then

  1. limnc=climnc=c
  2. limncan=climnan=cAlimncan=climnan=cA
  3. limn(an±bn)=limnan±limnbn=A±Blimn(an±bn)=limnan±limnbn=A±B
  4. limn(an·bn)=(limnan)·(limnbn)=A·Blimn(an·bn)=(limnan)·(limnbn)=A·B
  5. limn(anbn)=limnanlimnbn=AB,limn(anbn)=limnanlimnbn=AB, provided B0B0 and each bn0.bn0.

Proof

We prove part iii.

Let ϵ>0.ϵ>0. Since limnan=A,limnan=A, there exists a constant positive integer N1N1 such that |an-A|<ε2|an-A|<ε2 for all nN1.nN1. Since limnbn=B,limnbn=B, there exists a constant N2N2 such that |bnB|<ε/2|bnB|<ε/2 for all nN2.nN2. Let NN be the larger of N1N1 and N2.N2. Therefore, for all nN,nN,

|(an+bn)(A+B)||anA|+|bnB|<ε2+ε2=ε.|(an+bn)(A+B)||anA|+|bnB|<ε2+ε2=ε.

The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence {1n2}.{1n2}. As shown earlier, limn1/n=0.limn1/n=0. Similarly, for any positive integer k,k, we can conclude that

limn1nk=0.limn1nk=0.

In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.

Example 5.3

Determining Convergence and Finding Limits

For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.

  1. {53n2}{53n2}
  2. {3n47n2+564n4}{3n47n2+564n4}
  3. {2nn2}{2nn2}
  4. {(1+4n)n}{(1+4n)n}

Solution

  1. We know that 1/n0.1/n0. Using this fact, we conclude that
    limn1n2=limn(1n).limn(1n)=0.limn1n2=limn(1n).limn(1n)=0.

    Therefore,
    limn(53n2)=limn53limn1n2=53.0=5.limn(53n2)=limn53limn1n2=53.0=5.

    The sequence converges and its limit is 5.5.
  2. By factoring n4n4 out of the numerator and denominator and using the limit laws above, we have
    limn3n47n2+564n4=limn37n2+5n46n44=limn(37n2+5n4)limn(6n44)=(limn(3)limn7n2+limn5n4)(limn6n4limn(4))=(limn(3)7·limn1n2+5·limn1n4)(6·limn1n4limn(4))=37·0+5·06·04=34.limn3n47n2+564n4=limn37n2+5n46n44=limn(37n2+5n4)limn(6n44)=(limn(3)limn7n2+limn5n4)(limn6n4limn(4))=(limn(3)7·limn1n2+5·limn1n4)(6·limn1n4limn(4))=37·0+5·06·04=34.

    The sequence converges and its limit is −3/4.−3/4.
  3. Consider the related function f(x)=2x/x2f(x)=2x/x2 defined on all real numbers x>0.x>0. Since 2x2x and x2x2 as x,x, apply L’Hôpital’s rule and write
    limx2xx2=limx2xln22xTake the derivatives of the numerator and denominator.=limx2x(ln2)22Take the derivatives again.=.limx2xx2=limx2xln22xTake the derivatives of the numerator and denominator.=limx2x(ln2)22Take the derivatives again.=.

    We conclude that the sequence diverges.
  4. Consider the function f(x)=(1+4x)xf(x)=(1+4x)x defined on all real numbers x>0.x>0. This function has the indeterminate form 11 as x.x. Let
    y=limx(1+4x)x.y=limx(1+4x)x.

    Now taking the natural logarithm of both sides of the equation, we obtain
    ln(y)=ln[limx(1+4x)x].ln(y)=ln[limx(1+4x)x].

    Since the function f(x)=lnxf(x)=lnx is continuous on its domain, we can interchange the limit and the natural logarithm. Therefore,
    ln(y)=limx[ln(1+4x)x].ln(y)=limx[ln(1+4x)x].

    Using properties of logarithms, we write
    limx[ln(1+4x)x]=limxxln(1+4x).limx[ln(1+4x)x]=limxxln(1+4x).

    Since the right-hand side of this equation has the indeterminate form ·0,·0, rewrite it as a fraction to apply L’Hôpital’s rule. Write
    limxxln(1+4x)=limxln(1+4/x)1/x.limxxln(1+4x)=limxln(1+4/x)1/x.

    Since the right-hand side is now in the indeterminate form 0/0,0/0, we are able to apply L’Hôpital’s rule. We conclude that
    limxln(1+4/x)1/x=limx41+4/x=4.limxln(1+4/x)1/x=limx41+4/x=4.

    Therefore, ln(y)=4ln(y)=4 and y=e4.y=e4. Therefore, since limx(1+4x)x=e4,limx(1+4x)x=e4, we can conclude that the sequence {(1+4n)n}{(1+4n)n} converges to e4.e4.

Checkpoint 5.3

Consider the sequence {(5n2+1)/en}.{(5n2+1)/en}. Determine whether or not the sequence converges. If it converges, find its limit.

Recall that if ff is a continuous function at a value L,L, then f(x)f(L)f(x)f(L) as xL.xL. This idea applies to sequences as well. Suppose a sequence anL,anL, and a function ff is continuous at L.L. Then f(an)f(L).f(an)f(L). This property often enables us to find limits for complicated sequences. For example, consider the sequence 53n2.53n2. From Example 5.3a. we know the sequence 53n25.53n25. Since xx is a continuous function at x=5,x=5,

limn53n2=limn(53n2)=5.limn53n2=limn(53n2)=5.

Theorem 5.3

Continuous Functions Defined on Convergent Sequences

Consider a sequence {an}{an} and suppose there exists a real number LL such that the sequence {an}{an} converges to L.L. Suppose ff is a continuous function at L.L. Then there exists an integer NN such that ff is defined at all values anan for nN,nN, and the sequence {f(an)}{f(an)} converges to f(L)f(L) (Figure 5.5).

Proof

Let ϵ>0.ϵ>0. Since ff is continuous at L,L, there exists δ>0δ>0 such that |f(x)f(L)|<ε|f(x)f(L)|<ε if |xL|<δ.|xL|<δ. Since the sequence {an}{an} converges to L,L, there exists NN such that |anL|<δ|anL|<δ for all nN.nN. Therefore, for all nN,nN, |anL|<δ,|anL|<δ, which implies |f(an)f(L)|<ε.|f(an)f(L)|<ε. We conclude that the sequence {f(an)}{f(an)} converges to f(L).f(L).

A graph in quadrant 1 with points (a_1, f(a_1)), (a_3, f(a_3)), (L, f(L)), (a_4, f(a_4)), and (a_2, f(a_2)) connected by smooth curves.
Figure 5.5 Because ff is a continuous function as the inputs a1,a2,a3,…a1,a2,a3,… approach L,L, the outputs f(a1),f(a2),f(a3),…f(a1),f(a2),f(a3),… approach f(L).f(L).

Example 5.4

Limits Involving Continuous Functions Defined on Convergent Sequences

Determine whether the sequence {cos(3/n2)}{cos(3/n2)} converges. If it converges, find its limit.

Solution

Since the sequence {3/n2}{3/n2} converges to 00 and cosxcosx is continuous at x=0,x=0, we can conclude that the sequence {cos(3/n2)}{cos(3/n2)} converges and

limncos(3n2)=cos(0)=1.limncos(3n2)=cos(0)=1.

Checkpoint 5.4

Determine if the sequence {2n+13n+5}{2n+13n+5} converges. If it converges, find its limit.

Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.

Theorem 5.4

Squeeze Theorem for Sequences

Consider sequences {an},{an}, {bn},{bn}, and {cn}.{cn}. Suppose there exists an integer NN such that

anbncnfor allnN.anbncnfor allnN.

If there exists a real number LL such that

limnan=L=limncn,limnan=L=limncn,

then {bn}{bn} converges and limnbn=Llimnbn=L (Figure 5.6).

Proof

Let ε>0.ε>0. Since the sequence {an}{an} converges to L,L, there exists an integer N1N1 such that |anL|<ε|anL|<ε for all nN1.nN1. Similarly, since {cn}{cn} converges to L,L, there exists an integer N2N2 such that |cnL|<ε|cnL|<ε for all nN2.nN2. By assumption, there exists an integer NN such that anbncnanbncn for all nN.nN. Let MM be the largest of N1,N2,N1,N2, and N.N. We must show that |bnL|<ε|bnL|<ε for all nM.nM. For all nM,nM,

ε<|anL|anLbnLcnL|cnL|<ε.ε<|anL|anLbnLcnL|cnL|<ε.

Therefore, ε<bnL<ε,ε<bnL<ε, and we conclude that |bnL|<ε|bnL|<ε for all nM,nM, and we conclude that the sequence {bn}{bn} converges to L.L.

A graph in quadrant 1 with the line y = L and the x axis labeled as the n axis. Points are plotted above and below the line, converging to L as n goes to infinity. Points a_n, b_n, and c_n are plotted at the same n-value. A_n and b_n are above y = L, and c_n is below it.
Figure 5.6 Each term bnbn satisfies anbncnanbncn and the sequences {an}{an} and {cn}{cn} converge to the same limit, so the sequence {bn}{bn} must converge to the same limit as well.

Example 5.5

Using the Squeeze Theorem

Use the Squeeze Theorem to find the limit of each of the following sequences.

  1. {cosnn2}{cosnn2}
  2. {(12)n}{(12)n}

Solution

  1. Since −1cosn1−1cosn1 for all integers n,n, we have
    1n2cosnn21n2.1n2cosnn21n2.

    Since −1/n20−1/n20 and 1/n20,1/n20, we conclude that cosn/n20cosn/n20 as well.
  2. Since
    12n(12)n12n12n(12)n12n

    for all positive integers n,n, −1/2n0−1/2n0 and 1/2n0,1/2n0, we can conclude that (−1/2)n0.(−1/2)n0.

Checkpoint 5.5

Find limn2nsinnn.limn2nsinnn.

Using the idea from Example 5.5b. we conclude that rn0rn0 for any real number rr such that −1<r<0.−1<r<0. If r<1,r<1, the sequence {rn}{rn} diverges because the terms oscillate and become arbitrarily large in magnitude. If r=−1,r=−1, the sequence {rn}={(−1)n}{rn}={(−1)n} diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.

rn0if|r|<1rn0if|r|<1
5.1
rn1ifr=1rn1ifr=1
5.2
rnifr>1rnifr>1
5.3
{rn}diverges ifr1{rn}diverges ifr1
5.4

Bounded Sequences

We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.

Definition

A sequence {an}{an} is bounded above if there exists a real number MM such that

anManM

for all positive integers n.n.

A sequence {an}{an} is bounded below if there exists a real number MM such that

ManMan

for all positive integers n.n.

A sequence {an}{an} is a bounded sequence if it is bounded above and bounded below.

If a sequence is not bounded, it is an unbounded sequence.

For example, the sequence {1/n}{1/n} is bounded above because 1/n11/n1 for all positive integers n.n. It is also bounded below because 1/n01/n0 for all positive integers n. Therefore, {1/n}{1/n} is a bounded sequence. On the other hand, consider the sequence {2n}.{2n}. Because 2n22n2 for all n1,n1, the sequence is bounded below. However, the sequence is not bounded above. Therefore, {2n}{2n} is an unbounded sequence.

We now discuss the relationship between boundedness and convergence. Suppose a sequence {an}{an} is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms anan that are arbitrarily large in magnitude as nn gets larger. As a result, the sequence {an}{an} cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.

Theorem 5.5

Convergent Sequences Are Bounded

If a sequence {an}{an} converges, then it is bounded.

Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence {(−1)n}{(−1)n} is bounded, but the sequence diverges because the sequence oscillates between 11 and −1−1 and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.

Consider a bounded sequence {an}.{an}. Suppose the sequence {an}{an} is increasing. That is, a1a2a3.a1a2a3. Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that {an}{an} converges. For example, consider the sequence

{12,23,34,45,…}.{12,23,34,45,…}.

Since this sequence is increasing and bounded above, it converges. Next, consider the sequence

{2,0,3,0,4,0,1,12,13,14,…}.{2,0,3,0,4,0,1,12,13,14,…}.

Even though the sequence is not increasing for all values of n,n, we see that −1/2<1/3<1/4<.−1/2<1/3<1/4<. Therefore, starting with the eighth term, a8=−1/2,a8=−1/2, the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.

Definition

A sequence {an}{an} is increasing for all nn0nn0 if

anan+1for allnn0.anan+1for allnn0.

A sequence {an}{an} is decreasing for all nn0nn0 if

anan+1for allnn0.anan+1for allnn0.

A sequence {an}{an} is a monotone sequence for all nn0nn0 if it is increasing for all nn0nn0 or decreasing for all nn0.nn0.

We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.

Theorem 5.6

Monotone Convergence Theorem

If {an}{an} is a bounded sequence and there exists a positive integer n0n0 such that {an}{an} is monotone for all nn0,nn0, then {an}{an} converges.

The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 5.7).

A graph in quadrant 1 with the x and y axes labeled n and a_n, respectively. A dotted horizontal is drawn from the a_n axis into quadrant 1. Many points are plotted under the dotted line, increasing in a_n value and converging to the dotted line.
Figure 5.7 Since the sequence {an}{an} is increasing and bounded above, it must converge.

In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.

Example 5.6

Using the Monotone Convergence Theorem

For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.

  1. {4nn!}{4nn!}
  2. {an}{an} defined recursively such that
    a1=2andan+1=an2+12anfor alln2.a1=2andan+1=an2+12anfor alln2.

Solution

  1. Writing out the first few terms, we see that
    {4nn!}={4,8,323,323,12815,…}.{4nn!}={4,8,323,323,12815,…}.

    At first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all n3.n3. We can show this as follows.
    an+1=4n+1(n+1)!=4n+1·4nn!=4n+1·ananifn3.an+1=4n+1(n+1)!=4n+1·4nn!=4n+1·ananifn3.

    Therefore, the sequence is decreasing for all n3.n3. Further, the sequence is bounded below by 00 because 4n/n!04n/n!0 for all positive integers n.n. Therefore, by the Monotone Convergence Theorem, the sequence converges.
    To find the limit, we use the fact that the sequence converges and let L=limnan.L=limnan. Now note this important observation. Consider limnan+1.limnan+1. Since
    {an+1}={a2,a3,a4,…},{an+1}={a2,a3,a4,…},
    the only difference between the sequences {an+1}{an+1} and {an}{an} is that {an+1}{an+1} omits the first term. Since a finite number of terms does not affect the convergence of a sequence,
    limnan+1=limnan=L.limnan+1=limnan=L.

    Combining this fact with the equation
    an+1=4n+1anan+1=4n+1an

    and taking the limit of both sides of the equation
    limnan+1=limn4n+1an,limnan+1=limn4n+1an,

    we can conclude that
    L=0·L=0.L=0·L=0.
  2. Writing out the first several terms,
    {2,54,4140,32813280,…}.{2,54,4140,32813280,…}.

    we can conjecture that the sequence is decreasing and bounded below by 1.1. To show that the sequence is bounded below by 1,1, we can show that
    an2+12an1.an2+12an1.

    To show this, first rewrite
    an2+12an=an2+12an.an2+12an=an2+12an.

    Since a1>0a1>0 and a2a2 is defined as a sum of positive terms, a2>0.a2>0. Similarly, all terms an>0.an>0. Therefore,
    an2+12an1an2+12an1

    if and only if
    an2+12an.an2+12an.

    Rewriting the inequality an2+12anan2+12an as an22an+10,an22an+10, and using the fact that
    an22an+1=(an1)20an22an+1=(an1)20

    because the square of any real number is nonnegative, we can conclude that
    an2+12an1.an2+12an1.

    To show that the sequence is decreasing, we must show that an+1anan+1an for all n1.n1. Since 1an2,1an2, it follows that
    an2+12an2.an2+12an2.

    Dividing both sides by 2an,2an, we obtain
    an2+12anan.an2+12anan.

    Using the definition of an+1,an+1, we conclude that
    an+1=an2+12anan.an+1=an2+12anan.

    Since {an}{an} is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.
    To find the limit, let L=limnan.L=limnan. Then using the recurrence relation and the fact that limnan=limnan+1,limnan=limnan+1, we have
    limnan+1=limn(an2+12an),limnan+1=limn(an2+12an),

    and therefore
    L=L2+12L.L=L2+12L.

    Multiplying both sides of this equation by 2L,2L, we arrive at the equation
    2L2=L2+1.2L2=L2+1.

    Solving this equation for L,L, we conclude that L2=1,L2=1, which implies L=±1.L=±1. Since all the terms are positive, the limit L=1.L=1.
Checkpoint 5.6

Consider the sequence {an}{an} defined recursively such that a1=1,a1=1, an=an1/2.an=an1/2. Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.

Student Project

Fibonacci Numbers

The Fibonacci numbers are defined recursively by the sequence {Fn}{Fn} where F0=0,F0=0, F1=1F1=1 and for n2,n2,

Fn=Fn1+Fn2.Fn=Fn1+Fn2.

Here we look at properties of the Fibonacci numbers.

  1. Write out the first twenty Fibonacci numbers.
  2. Find a closed formula for the Fibonacci sequence by using the following steps.
    1. Consider the recursively defined sequence {xn}{xn} where xo=cxo=c and xn+1=axn.xn+1=axn. Show that this sequence can be described by the closed formula xn=canxn=can for all n0.n0.
    2. Using the result from part a. as motivation, look for a solution of the equation
      Fn=Fn1+Fn2Fn=Fn1+Fn2

      of the form Fn=cλn.Fn=cλn. Determine what two values for λλ will allow FnFn to satisfy this equation.
    3. Consider the two solutions from part b.: λ1λ1 and λ2.λ2. Let Fn=c1λ1n+c2λ2n.Fn=c1λ1n+c2λ2n. Use the initial conditions F0F0 and F1F1 to determine the values for the constants c1c1 and c2c2 and write the closed formula Fn.Fn.
  3. Use the answer in 2 c. to show that
    limnFn+1Fn=1+52.limnFn+1Fn=1+52.

    The number ϕ=(1+5)/2ϕ=(1+5)/2 is known as the golden ratio (Figure 5.8 and Figure 5.9).
    This is a photo of a sunflower, particularly the curves of the seeds at its middle. The number of spirals in each direction is always a Fibonacci number.
    Figure 5.8 The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)

    This is a photo of the Parthenon, an ancient Greek temple that was designed with the proportions of the Golden Rule. The entire temple’s front side fits perfectly into a rectangle with those proportions, as do the columns, the level between the columns and the roof, and a portion of the trim below the roof.
    Figure 5.9 The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)

Section 5.1 Exercises

Find the first six terms of each of the following sequences, starting with n=1.n=1.

1.

an=1+(−1)nan=1+(−1)n for n1n1

2.

an=n21an=n21 for n1n1

3.

a1=1a1=1 and an=an1+nan=an1+n for n2n2

4.

a1=1,a1=1, a2=1a2=1 and an+2=an+an+1an+2=an+an+1 for n1n1

5.

Find an explicit formula for anan where a1=1a1=1 and an=an1+nan=an1+n for n2.n2.

6.

Find a formula anan for the nthnth term of the arithmetic sequence whose first term is a1=1a1=1 such that an1an=17an1an=17 for n1.n1.

7.

Find a formula anan for the nthnth term of the arithmetic sequence whose first term is a1=−3a1=−3 such that an1an=4an1an=4 for n1.n1.

8.

Find a formula anan for the nthnth term of the geometric sequence whose first term is a1=1a1=1 such that an+1an=10an+1an=10 for n1.n1.

9.

Find a formula anan for the nthnth term of the geometric sequence whose first term is a1=3a1=3 such that an+1an=1/10an+1an=1/10 for n1.n1.

10.

Find an explicit formula for the nthnth term of the sequence whose first several terms are {0,3,8,15,24,35,48,63,80,99,…}.{0,3,8,15,24,35,48,63,80,99,…}. (Hint: First add one to each term.)

11.

Find an explicit formula for the nthnth term of the sequence satisfying a1=0a1=0 and an=2an1+1an=2an1+1 for n2.n2.

Find a formula for the general term anan of each of the following sequences.

12.

{1,0,−1,0,1,0,−1,0,…}{1,0,−1,0,1,0,−1,0,…} (Hint: Find where sinxsinx takes these values)

13.

{1,1/3,1/5,1/7,…}{1,1/3,1/5,1/7,…}

Find a function f(n)f(n) that identifies the nthnth term anan of the following recursively defined sequences, as an=f(n).an=f(n).

14.

a1=1a1=1 and an+1=anan+1=an for n1n1

15.

a1=2a1=2 and an+1=2anan+1=2an for n1n1

16.

a1=1a1=1 and an+1=(n+1)anan+1=(n+1)an for n1n1

17.

a1=2a1=2 and an+1=(n+1)an/2an+1=(n+1)an/2 for n1n1

18.

a1=1a1=1 and an+1=an/2nan+1=an/2n for n1n1

Plot the first NN terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.

19.

[T] a1=1,a1=1, a2=2,a2=2, and for n2,n2, an=12(an1+an2);an=12(an1+an2); N=30N=30

20.

[T] a1=1,a1=1, a2=2,a2=2, a3=3a3=3 and for n4,n4, an=13(an1+an2+an3),an=13(an1+an2+an3), N=30N=30

21.

[T] a1=1,a1=1, a2=2,a2=2, and for n3,n3, an=an1an2;an=an1an2; N=30N=30

22.

[T] a1=1,a1=1, a2=2,a2=2, a3=3,a3=3, and for n4,n4, an=an1an2an3;an=an1an2an3; N=30N=30

Suppose that limnan=1,limnan=1, limnbn=−1,limnbn=−1, and 0<bn<an0<bn<an for all n.n. Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.

23.

limn(3an4bn)limn(3an4bn)

24.

limn(12bn12an)limn(12bn12an)

25.

limnan+bnanbnlimnan+bnanbn

26.

limnanbnan+bnlimnanbnan+bn

Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.

27.

n22nn22n

28.

(n1)2(n+1)2(n1)2(n+1)2

29.

nn+1nn+1

30.

n1/nn1/n (Hint: n1/n=e1nlnn)n1/n=e1nlnn)

For each of the following sequences, whose nthnth terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

31.

n/2n,n/2n, n2n2

32.

ln(1+1n)ln(1+1n)

33.

sinnsinn

34.

cos(n2)cos(n2)

35.

n1/n,n1/n, n3n3

36.

n−1/n,n−1/n, n3n3

37.

tanntann

38.

Determine whether the sequence defined as follows has a limit. If it does, find the limit.

a1=2,a1=2, a2=22,a2=22, a3=222a3=222 etc.

39.

Determine whether the sequence defined as follows has a limit. If it does, find the limit.

a1=3,a1=3, an=2an1,an=2an1, n=2,3,….n=2,3,….

Use the Squeeze Theorem to find the limit of each of the following sequences.

40.

nsin(1/n)nsin(1/n)

41.

cos(1/n)11/ncos(1/n)11/n

42.

an=n!nnan=n!nn

43.

an=sinnsin(1/n)an=sinnsin(1/n)

For the following sequences, plot the first 2525 terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.

44.

[T] an=sinnan=sinn

45.

[T] an=cosnan=cosn

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

46.

an=tan−1(n2)an=tan−1(n2)

47.

an=(2n)1/nn1/nan=(2n)1/nn1/n

48.

an=ln(n2)ln(2n)an=ln(n2)ln(2n)

49.

an=(12n)nan=(12n)n

50.

an=ln(n+2n23)an=ln(n+2n23)

51.

an=2n+3n4nan=2n+3n4n

52.

an=(1000)nn!an=(1000)nn!

53.

an=(n!)2(2n)!an=(n!)2(2n)!

Newton’s method seeks to approximate a solution f(x)=0f(x)=0 that starts with an initial approximation x0x0 and successively defines a sequence xn+1=xnf(xn)f(xn).xn+1=xnf(xn)f(xn). For the given choice of ff and x0,x0, write out the formula for xn+1.xn+1. If the sequence appears to converge, give an exact formula for the solution x,x, then identify the limit xx accurate to four decimal places and the smallest nn such that xnxn agrees with xx up to four decimal places.

54.

[T] f(x)=x22,f(x)=x22, x0=1x0=1

55.

[T] f(x)=(x1)22,f(x)=(x1)22, x0=2x0=2

56.

[T] f(x)=ex2,f(x)=ex2, x0=1x0=1

57.

[T] f(x)=lnx1,f(x)=lnx1, x0=2x0=2

58.

[T] Suppose you start with one liter of vinegar and repeatedly remove 0.1L,0.1L, replace with water, mix, and repeat.

  1. Find a formula for the concentration after nn steps.
  2. After how many steps does the mixture contain less than 10%10% vinegar?
59.

[T] A lake initially contains 20002000 fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by 6%6% each month. However, factoring in all causes, 150150 fish are lost each month.

  1. Explain why the fish population after nn months is modeled by Pn=1.06Pn1150Pn=1.06Pn1150 with P0=2000.P0=2000.
  2. How many fish will be in the pond after one year?
60.

[T] A bank account earns 5%5% interest compounded monthly. Suppose that $1000$1000 is initially deposited into the account, but that $10$10 is withdrawn each month.

  1. Show that the amount in the account after nn months is An=(1+.05/12)An110;An=(1+.05/12)An110; A0=1000.A0=1000.
  2. How much money will be in the account after 11 year?
  3. Is the amount increasing or decreasing?
  4. Suppose that instead of $10,$10, a fixed amount dd dollars is withdrawn each month. Find a value of dd such that the amount in the account after each month remains $1000.$1000.
  5. What happens if dd is greater than this amount?
61.

[T] A student takes out a college loan of $10,000$10,000 at an annual percentage rate of 6%,6%, compounded monthly.

  1. If the student makes payments of $100$100 per month, how much does the student owe after 1212 months?
  2. After how many months will the loan be paid off?
62.

[T] Consider a series combining geometric growth and arithmetic decrease. Let a1=1.a1=1. Fix a>1a>1 and 0<b<a.0<b<a. Set an+1=a.anb.an+1=a.anb. Find a formula for an+1an+1 in terms of an,an, a,a, and bb and a relationship between aa and bb such that anan converges.

63.

[T] The binary representation x=0.b1b2b3...x=0.b1b2b3... of a number xx between 00 and 11 can be defined as follows. Let b1=0b1=0 if x<1/2x<1/2 and b1=1b1=1 if 1/2x<1.1/2x<1. Let x1=2xb1.x1=2xb1. Let b2=0b2=0 if x1<1/2x1<1/2 and b2=1b2=1 if 1/2x<1.1/2x<1. Let x2=2x1b2x2=2x1b2 and in general, xn=2xn1bnxn=2xn1bn and bn1=0bn1=0 if xn<1/2xn<1/2 and bn1=1bn1=1 if 1/2xn<1.1/2xn<1. Find the binary expansion of 1/3.1/3.

64.

[T] To find an approximation for π,π, set a0=2+1,a0=2+1, a1=2+a0,a1=2+a0, and, in general, an+1=2+an.an+1=2+an. Finally, set pn=3.2n2an.pn=3.2n2an. Find the first ten terms of pnpn and compare the values to π.π.

For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of NN integers a1,a2,…,aNa1,a2,…,aN by fixing two special integers KK and MM and letting an+1an+1 be the remainder after dividing K.anK.an into M,M, then creates a bit sequence of zeros and ones whose nthnth term bnbn is equal to one if anan is odd and equal to zero if anan is even. If the bits bnbn are pseudorandom, then the behavior of their average (b1+b2++bN)/N(b1+b2++bN)/N should be similar to behavior of averages of truly randomly generated bits.

65.

[T] Starting with K=16,807K=16,807 and M=2,147,483,647,M=2,147,483,647, using ten different starting values of a1,a1, compute sequences of bits bnbn up to n=1000,n=1000, and compare their averages to ten such sequences generated by a random bit generator.

66.

[T] Find the first 10001000 digits of ππ using either a computer program or Internet resource. Create a bit sequence bnbn by letting bn=1bn=1 if the nthnth digit of ππ is odd and bn=0bn=0 if the nthnth digit of ππ is even. Compute the average value of bnbn and the average value of dn=|bn+1bn|,dn=|bn+1bn|, n=1,...,999.n=1,...,999. Does the sequence bnbn appear random? Do the differences between successive elements of bnbn appear random?

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