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Calculus Volume 2

5.2 Infinite Series

Calculus Volume 25.2 Infinite Series
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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

  • 5.2.1. Explain the meaning of the sum of an infinite series.
  • 5.2.2. Calculate the sum of a geometric series.
  • 5.2.3. Evaluate a telescoping series.

We have seen that a sequence is an ordered set of terms. If you add these terms together, you get a series. In this section we define an infinite series and show how series are related to sequences. We also define what it means for a series to converge or diverge. We introduce one of the most important types of series: the geometric series. We will use geometric series in the next chapter to write certain functions as polynomials with an infinite number of terms. This process is important because it allows us to evaluate, differentiate, and integrate complicated functions by using polynomials that are easier to handle. We also discuss the harmonic series, arguably the most interesting divergent series because it just fails to converge.

Sums and Series

An infinite series is a sum of infinitely many terms and is written in the form

n=1an=a1+a2+a3+.n=1an=a1+a2+a3+.

But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the limit of partial sums. A partial sum of an infinite series is a finite sum of the form

n=1kan=a1+a2+a3++ak.n=1kan=a1+a2+a3++ak.

To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that 10001000 gallons enters the lake the first week. During the second week, an additional 500500 gallons of oil enters the lake. The third week, 250250 more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after kk weeks. Letting SkSk denote the amount of oil in the lake (measured in thousands of gallons) after kk weeks, we see that

S1=1S2=1+0.5=1+12S3=1+0.5+0.25=1+12+14S4=1+0.5+0.25+0.125=1+12+14+18S5=1+0.5+0.25+0.125+0.0625=1+12+14+18+116.S1=1S2=1+0.5=1+12S3=1+0.5+0.25=1+12+14S4=1+0.5+0.25+0.125=1+12+14+18S5=1+0.5+0.25+0.125+0.0625=1+12+14+18+116.

Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after kk weeks is

Sk=1+12+14+18+116++12k1=n=1k(12)n1.Sk=1+12+14+18+116++12k1=n=1k(12)n1.

We are interested in what happens as k.k. Symbolically, the amount of oil in the lake as kk is given by the infinite series

n=1(12)n1=1+12+14+18+116+.n=1(12)n1=1+12+14+18+116+.

At the same time, as k,k, the amount of oil in the lake can be calculated by evaluating limkSk.limkSk. Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums {Sk}.{Sk}. If the sequence of partial sums {Sk}{Sk} converges, we say that the infinite series converges, and its sum is given by limkSk.limkSk. If the sequence {Sk}{Sk} diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence {Sk}.{Sk}.

First, simplifying some of these partial sums, we see that

S1=1S2=1+12=32S3=1+12+14=74S4=1+12+14+18=158S5=1+12+14+18+116=3116.S1=1S2=1+12=32S3=1+12+14=74S4=1+12+14+18=158S5=1+12+14+18+116=3116.

Plotting some of these values in Figure 5.10, it appears that the sequence {Sk}{Sk} could be approaching 2.

This is a graph in quadrant 1with the x and y axes labeled n and S_n, respectively. From 1 to 5, points are plotted. They increase and seem to converge to 2 and n goes to infinity.
Figure 5.10 The graph shows the sequence of partial sums {Sk}.{Sk}. It appears that the sequence is approaching the value 2.2.

Let’s look for more convincing evidence. In the following table, we list the values of SkSk for several values of k.k.

kk 55 1010 1515 2020
SkSk 1.93751.9375 1.9981.998 1.9999391.999939 1.9999981.999998

These data supply more evidence suggesting that the sequence {Sk}{Sk} converges to 2.2. Later we will provide an analytic argument that can be used to prove that limkSk=2.limkSk=2. For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to 2.2. Since this sequence of partial sums converges to 2,2, we say the infinite series converges to 22 and write

n=1(12)n1=2.n=1(12)n1=2.

Returning to the question about the oil in the lake, since this infinite series converges to 2,2, we conclude that the amount of oil in the lake will get arbitrarily close to 20002000 gallons as the amount of time gets sufficiently large.

This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.

Definition

An infinite series is an expression of the form

n=1an=a1+a2+a3+.n=1an=a1+a2+a3+.

For each positive integer k,k, the sum

Sk=n=1kan=a1+a2+a3++akSk=n=1kan=a1+a2+a3++ak

is called the kthkth partial sum of the infinite series. The partial sums form a sequence {Sk}.{Sk}. If the sequence of partial sums converges to a real number S,S, the infinite series converges. If we can describe the convergence of a series to S,S, we call SS the sum of the series, and we write

n=1an=S.n=1an=S.

If the sequence of partial sums diverges, we have the divergence of a series.

Media

This website shows a more whimsical approach to series.

Note that the index for a series need not begin with n=1n=1 but can begin with any value. For example, the series

n=1(12)n1n=1(12)n1

can also be written as

n=0(12)norn=5(12)n5.n=0(12)norn=5(12)n5.

Often it is convenient for the index to begin at 1,1, so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series

n=21n2.n=21n2.

By introducing the variable m=n1,m=n1, so that n=m+1,n=m+1, we can rewrite the series as

m=11(m+1)2.m=11(m+1)2.

Example 5.7

Evaluating Limits of Sequences of Partial Sums

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

  1. n=1nn+1n=1nn+1
  2. n=1(−1)nn=1(−1)n
  3. n=11n(n+1)n=11n(n+1)

Solution

  1. The sequence of partial sums {Sk}{Sk} satisfies
    S1=12S2=12+23S3=12+23+34S4=12+23+34+45.S1=12S2=12+23S3=12+23+34S4=12+23+34+45.

    Notice that each term added is greater than 1/2.1/2. As a result, we see that
    S1=12S2=12+23>12+12=2(12)S3=12+23+34>12+12+12=3(12)S4=12+23+34+45>12+12+12+12=4(12).S1=12S2=12+23>12+12=2(12)S3=12+23+34>12+12+12=3(12)S4=12+23+34+45>12+12+12+12=4(12).

    From this pattern we can see that Sk>k(12)Sk>k(12) for every integer k.k. Therefore, {Sk}{Sk} is unbounded and consequently, diverges. Therefore, the infinite series n=1n/(n+1)n=1n/(n+1) diverges.
  2. The sequence of partial sums {Sk}{Sk} satisfies
    S1=−1S2=−1+1=0S3=−1+11=−1S4=−1+11+1=0.S1=−1S2=−1+1=0S3=−1+11=−1S4=−1+11+1=0.

    From this pattern we can see the sequence of partial sums is
    {Sk}={−1,0,−1,0,…}.{Sk}={−1,0,−1,0,…}.

    Since this sequence diverges, the infinite series n=1(−1)nn=1(−1)n diverges.
  3. The sequence of partial sums {Sk}{Sk} satisfies
    S1=11·2=12S2=11·2+12·3=12+16=23S3=11·2+12·3+13·4=12+16+112=34S4=11·2+12·3+13·4+14·5=45S5=11·2+12·3+13·4+14·5+15·6=56.S1=11·2=12S2=11·2+12·3=12+16=23S3=11·2+12·3+13·4=12+16+112=34S4=11·2+12·3+13·4+14·5=45S5=11·2+12·3+13·4+14·5+15·6=56.

    From this pattern, we can see that the kthkth partial sum is given by the explicit formula
    Sk=kk+1.Sk=kk+1.

    Since k/(k+1)1,k/(k+1)1, we conclude that the sequence of partial sums converges, and therefore the infinite series converges to 1.1. We have
    n=11n(n+1)=1.n=11n(n+1)=1.
Checkpoint 5.7

Determine whether the series n=1(n+1)/nn=1(n+1)/n converges or diverges.

The Harmonic Series

A useful series to know about is the harmonic series. The harmonic series is defined as

n=11n=1+12+13+14+.n=11n=1+12+13+14+.
5.5

This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums {Sk}{Sk} approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence {Sk}{Sk} in the following table.

kk 1010 100100 10001000 10,00010,000 100,000100,000 1,000,0001,000,000
SkSk 2.928972.92897 5.187385.18738 7.485477.48547 9.787619.78761 12.0901512.09015 14.3927314.39273

Even after 1,000,0001,000,000 terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.

To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:

S1=1S2=1+12S3=1+12+13S4=1+12+13+14.S1=1S2=1+12S3=1+12+13S4=1+12+13+14.

Notice that for the last two terms in S4,S4,

13+14>14+14.13+14>14+14.

Therefore, we conclude that

S4>1+12+(14+14)=1+12+12=1+2(12).S4>1+12+(14+14)=1+12+12=1+2(12).

Using the same idea for S8,S8, we see that

S8=1+12+13+14+15+16+17+18>1+12+(14+14)+(18+18+18+18)=1+12+12+12=1+3(12).S8=1+12+13+14+15+16+17+18>1+12+(14+14)+(18+18+18+18)=1+12+12+12=1+3(12).

From this pattern, we see that S1=1,S1=1, S2=1+1/2,S2=1+1/2, S4>1+2(1/2),S4>1+2(1/2), and S8>1+3(1/2).S8>1+3(1/2). More generally, it can be shown that S2j>1+j(1/2)S2j>1+j(1/2) for all j>1.j>1. Since 1+j(1/2),1+j(1/2), we conclude that the sequence {Sk}{Sk} is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since {Sk}{Sk} is unbounded, it diverges. Thus, the harmonic series diverges.

Algebraic Properties of Convergent Series

Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.

Theorem 5.7

Algebraic Properties of Convergent Series

Let n=1ann=1an and n=1bnn=1bn be convergent series. Then the following algebraic properties hold.

  1. The series n=1(an+bn)n=1(an+bn) converges and n=1(an+bn)=n=1an+n=1bn.n=1(an+bn)=n=1an+n=1bn. (Sum Rule)
  2. The series n=1(anbn)n=1(anbn) converges and n=1(anbn)=n=1ann=1bn.n=1(anbn)=n=1ann=1bn. (Difference Rule)
  3. For any real number c,c, the series n=1cann=1can converges and n=1can=cn=1an.n=1can=cn=1an. (Constant Multiple Rule)

Example 5.8

Using Algebraic Properties of Convergent Series

Evaluate

n=1[3n(n+1)+(12)n2].n=1[3n(n+1)+(12)n2].

Solution

We showed earlier that

n=11n(n+1)n=11n(n+1)

and

n=1(12)n1=2.n=1(12)n1=2.

Since both of those series converge, we can apply the properties of Algebraic Properties of Convergent Series to evaluate

n=1[3n(n+1)+(12)n2].n=1[3n(n+1)+(12)n2].

Using the sum rule, write

n=1[3n(n+1)+(12)n2]=n=13n(n+1)+n=1(12)n2.n=1[3n(n+1)+(12)n2]=n=13n(n+1)+n=1(12)n2.

Then, using the constant multiple rule and the sums above, we can conclude that

n=13n(n+1)+n=1(12)n2=3n=11n(n+1)+(12)−1n=1(12)n1=3(1)+(12)−1(2)=3+2(2)=7.n=13n(n+1)+n=1(12)n2=3n=11n(n+1)+(12)−1n=1(12)n1=3(1)+(12)−1(2)=3+2(2)=7.

Checkpoint 5.8

Evaluate n=152n1.n=152n1.

Geometric Series

A geometric series is any series that we can write in the form

a+ar+ar2+ar3+=n=1arn1.a+ar+ar2+ar3+=n=1arn1.
5.6

Because the ratio of each term in this series to the previous term is r, the number r is called the ratio. We refer to a as the initial term because it is the first term in the series. For example, the series

n=1(12)n1=1+12+14+18+n=1(12)n1=1+12+14+18+

is a geometric series with initial term a=1a=1 and ratio r=1/2.r=1/2.

In general, when does a geometric series converge? Consider the geometric series

n=1arn1n=1arn1

when a>0.a>0. Its sequence of partial sums {Sk}{Sk} is given by

Sk=n=1karn1=a+ar+ar2++ark1.Sk=n=1karn1=a+ar+ar2++ark1.

Consider the case when r=1.r=1. In that case,

Sk=a+a(1)+a(1)2++a(1)k1=ak.Sk=a+a(1)+a(1)2++a(1)k1=ak.

Since a>0,a>0, we know akak as k.k. Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for r=1.r=1. For r1,r1, to find the limit of {Sk},{Sk}, multiply Equation 5.6 by 1r.1r. Doing so, we see that

(1r)Sk=a(1r)(1+r+r2+r3++rk1)=a[(1+r+r2+r3++rk1)(r+r2+r3++rk)]=a(1rk).(1r)Sk=a(1r)(1+r+r2+r3++rk1)=a[(1+r+r2+r3++rk1)(r+r2+r3++rk)]=a(1rk).

All the other terms cancel out.

Therefore,

Sk=a(1rk)1rforr1.Sk=a(1rk)1rforr1.

From our discussion in the previous section, we know that the geometric sequence rk0rk0 if |r|<1|r|<1 and that rkrk diverges if |r|>1|r|>1 or r=±1.r=±1. Therefore, for |r|<1,|r|<1, Ska/(1r)Ska/(1r) and we have

n=1arn1=a1rif|r|<1.n=1arn1=a1rif|r|<1.

If |r|1,|r|1, SkSk diverges, and therefore

n=1arn1diverges if|r|1.n=1arn1diverges if|r|1.

Definition

A geometric series is a series of the form

n=1arn1=a+ar+ar2+ar3+.n=1arn1=a+ar+ar2+ar3+.

If |r|<1,|r|<1, the series converges, and

n=1arn1=a1rfor|r|<1.n=1arn1=a1rfor|r|<1.
5.7

If |r|1,|r|1, the series diverges.

Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than n=1n=1 or the exponent involves a linear expression for nn other than n1.n1. As long as we can rewrite the series in the form given by Equation 5.5, it is a geometric series. For example, consider the series

n=0(23)n+2.n=0(23)n+2.

To see that this is a geometric series, we write out the first several terms:

n=0(23)n+2=(23)2+(23)3+(23)4+=49+49·(23)+49·(23)2+.n=0(23)n+2=(23)2+(23)3+(23)4+=49+49·(23)+49·(23)2+.

We see that the initial term is a=4/9a=4/9 and the ratio is r=2/3.r=2/3. Therefore, the series can be written as

n=149·(23)n1.n=149·(23)n1.

Since r=2/3<1,r=2/3<1, this series converges, and its sum is given by

n=149·(23)n1=4/912/3=43.n=149·(23)n1=4/912/3=43.

Example 5.9

Determining Convergence or Divergence of a Geometric Series

Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.

  1. n=1(−3)n+14n1n=1(−3)n+14n1
  2. n=1e2nn=1e2n

Solution

  1. Writing out the first several terms in the series, we have
    n=1(−3)n+14n1=(−3)240+(−3)34+(−3)442+=(−3)2+(−3)2·(−34)+(−3)2·(−34)2+=9+9·(−34)+9·(−34)2+.n=1(−3)n+14n1=(−3)240+(−3)34+(−3)442+=(−3)2+(−3)2·(−34)+(−3)2·(−34)2+=9+9·(−34)+9·(−34)2+.

    The initial term a=−3a=−3 and the ratio r=−3/4.r=−3/4. Since |r|=3/4<1,|r|=3/4<1, the series converges to
    91(−3/4)=97/4=367.91(−3/4)=97/4=367.
  2. Writing this series as
    e2n=1(e2)n1e2n=1(e2)n1

    we can see that this is a geometric series where r=e2>1.r=e2>1. Therefore, the series diverges.

Checkpoint 5.9

Determine whether the series n=1(−25)n1n=1(−25)n1 converges or diverges. If it converges, find its sum.

We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.

Example 5.10

Writing Repeating Decimals as Fractions of Integers

Use a geometric series to write 3.263.26 as a fraction of integers.

Solution

Since 3.26=3.262626,3.26=3.262626, first we write

3.262626=3+26100+261000+26100,000+=3+26102+26104+26106+.3.262626=3+26100+261000+26100,000+=3+26102+26104+26106+.

Ignoring the term 3, the rest of this expression is a geometric series with initial term a=26/102a=26/102 and ratio r=1/102.r=1/102. Therefore, the sum of this series is

26/1021(1/102)=26/10299/102=2699.26/1021(1/102)=26/10299/102=2699.

Thus,

3.262626=3+2699=32399.3.262626=3+2699=32399.

Checkpoint 5.10

Write 5.275.27 as a fraction of integers.

Example 5.11

Chapter Opener: Finding the Area of the Koch Snowflake

Define a sequence of figures {Fn}{Fn} recursively as follows (Figure 5.11). Let F0F0 be an equilateral triangle with sides of length 1.1. For n1,n1, let FnFn be the curve created by removing the middle third of each side of Fn1Fn1 and replacing it with an equilateral triangle pointing outward. The limiting figure as nn is known as Koch’s snowflake.

This is a diagram of the Koch snowflake, which it created through iterations. The base case is an equilateral triangle. In each iteration, the middle third of each line segment is replaced with another equilateral triangle pointing outward.
Figure 5.11 The first four figures, F0,F1,F2,andF3,F0,F1,F2,andF3, in the construction of the Koch snowflake.
  1. Find the length LnLn of the perimeter of Fn.Fn. Evaluate limnLnlimnLn to find the length of the perimeter of Koch’s snowflake.
  2. Find the area AnAn of figure Fn.Fn. Evaluate limnAnlimnAn to find the area of Koch’s snowflake.

Solution

  1. Let NnNn denote the number of sides of figure Fn.Fn. Since F0F0 is a triangle, N0=3.N0=3. Let lnln denote the length of each side of Fn.Fn. Since F0F0 is an equilateral triangle with sides of length l0=1,l0=1, we now need to determine N1N1 and l1.l1. Since F1F1 is created by removing the middle third of each side and replacing that line segment with two line segments, for each side of F0,F0, we get four sides in F1.F1. Therefore, the number of sides for F1F1 is
    N1=4·3.N1=4·3.

    Since the length of each of these new line segments is 1/31/3 the length of the line segments in F0,F0, the length of the line segments for F1F1 is given by
    l1=13·1=13.l1=13·1=13.

    Similarly, for F2,F2, since the middle third of each side of F1F1 is removed and replaced with two line segments, the number of sides in F2F2 is given by
    N2=4N1=4(4·3)=42·3.N2=4N1=4(4·3)=42·3.

    Since the length of each of these sides is 1/31/3 the length of the sides of F1,F1, the length of each side of figure F2F2 is given by
    l2=13·l1=13·13=(13)2.l2=13·l1=13·13=(13)2.

    More generally, since FnFn is created by removing the middle third of each side of Fn1Fn1 and replacing that line segment with two line segments of length 13ln113ln1 in the shape of an equilateral triangle, we know that Nn=4Nn1Nn=4Nn1 and ln=ln13.ln=ln13. Therefore, the number of sides of figure FnFn is
    Nn=4n·3Nn=4n·3

    and the length of each side is
    ln=(13)n.ln=(13)n.

    Therefore, to calculate the perimeter of Fn,Fn, we multiply the number of sides NnNn and the length of each side ln.ln. We conclude that the perimeter of FnFn is given by
    Ln=Nn·ln=3·(43)n.Ln=Nn·ln=3·(43)n.

    Therefore, the length of the perimeter of Koch’s snowflake is
    L=limnLn=.L=limnLn=.
  2. Let TnTn denote the area of each new triangle created when forming Fn.Fn. For n=0,n=0, T0T0 is the area of the original equilateral triangle. Therefore, T0=A0=3/4.T0=A0=3/4. For n1,n1, since the lengths of the sides of the new triangle are 1/31/3 the length of the sides of Fn1,Fn1, we have
    Tn=(13)2Tn1=19·Tn1.Tn=(13)2Tn1=19·Tn1.

    Therefore, Tn=(19)n·34.Tn=(19)n·34. Since a new triangle is formed on each side of Fn1,Fn1,
    An=An1+Nn1·Tn=An1+(3·4n1)·(19)n·34=An1+34·(49)n·34.An=An1+Nn1·Tn=An1+(3·4n1)·(19)n·34=An1+34·(49)n·34.

    Writing out the first few terms A0,A1,A2,A0,A1,A2, we see that
    A0=34A1=A0+34·(49)·34=34+34·(49)·34=34[1+34·(49)]A2=A1+34·(49)2·34=34[1+34·(49)]+34·(49)2·34=34[1+34·(49)+34·(49)2].A0=34A1=A0+34·(49)·34=34+34·(49)·34=34[1+34·(49)]A2=A1+34·(49)2·34=34[1+34·(49)]+34·(49)2·34=34[1+34·(49)+34·(49)2].

    More generally,
    An=34[1+34(49+(49)2++(49)n)].An=34[1+34(49+(49)2++(49)n)].

    Factoring 4/94/9 out of each term inside the inner parentheses, we rewrite our expression as
    An=34[1+13(1+49+(49)2++(49)n1)].An=34[1+13(1+49+(49)2++(49)n1)].

    The expression 1+(49)+(49)2++(49)n11+(49)+(49)2++(49)n1 is a geometric sum. As shown earlier, this sum satisfies
    1+49+(49)2++(49)n1=1(4/9)n1(4/9).1+49+(49)2++(49)n1=1(4/9)n1(4/9).

    Substituting this expression into the expression above and simplifying, we conclude that
    An=34[1+13(1(4/9)n1(4/9))]=34[8535(49)n].An=34[1+13(1(4/9)n1(4/9))]=34[8535(49)n].

    Therefore, the area of Koch’s snowflake is
    A=limnAn=235.A=limnAn=235.

Analysis

The Koch snowflake is interesting because it has finite area, yet infinite perimeter. Although at first this may seem impossible, recall that you have seen similar examples earlier in the text. For example, consider the region bounded by the curve y=1/x2y=1/x2 and the xx-axis on the interval [1,).[1,). Since the improper integral

11x2dx11x2dx

converges, the area of this region is finite, even though the perimeter is infinite.

Telescoping Series

Consider the series n=11n(n+1).n=11n(n+1). We discussed this series in Example 5.7, showing that the series converges by writing out the first several partial sums S1,S2,…,S6S1,S2,…,S6 and noticing that they are all of the form Sk=kk+1.Sk=kk+1. Here we use a different technique to show that this series converges. By using partial fractions, we can write

1n(n+1)=1n1n+1.1n(n+1)=1n1n+1.

Therefore, the series can be written as

n=1[1n1n+1]=(1+12)+(1213)+(1314)+.n=1[1n1n+1]=(1+12)+(1213)+(1314)+.

Writing out the first several terms in the sequence of partial sums {Sk},{Sk}, we see that

S1=112S2=(112)+(1213)=113S3=(112)+(1213)+(1314)=114.S1=112S2=(112)+(1213)=113S3=(112)+(1213)+(1314)=114.

In general,

Sk=(112)+(1213)+(1314)++(1k1k+1)=11k+1.Sk=(112)+(1213)+(1314)++(1k1k+1)=11k+1.

We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since Sk=11/(k+1)Sk=11/(k+1) and 1/(k+1)01/(k+1)0 as k,k, the sequence of partial sums converges to 1,1, and therefore the series converges to 1.1.

Definition

A telescoping series is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.

For example, any series of the form

n=1[bnbn+1]=(b1b2)+(b2b3)+(b3b4)+n=1[bnbn+1]=(b1b2)+(b2b3)+(b3b4)+

is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that

S1=b1b2S2=(b1b2)+(b2b3)=b1b3S3=(b1b2)+(b2b3)+(b3b4)=b1b4.S1=b1b2S2=(b1b2)+(b2b3)=b1b3S3=(b1b2)+(b2b3)+(b3b4)=b1b4.

In general, the kth partial sum of this series is

Sk=b1bk+1.Sk=b1bk+1.

Since the kth partial sum can be simplified to the difference of these two terms, the sequence of partial sums {Sk}{Sk} will converge if and only if the sequence {bk+1}{bk+1} converges. Moreover, if the sequence bk+1bk+1 converges to some finite number B,B, then the sequence of partial sums converges to b1B,b1B, and therefore

n=1[bnbn+1]=b1B.n=1[bnbn+1]=b1B.

In the next example, we show how to use these ideas to analyze a telescoping series of this form.

Example 5.12

Evaluating a Telescoping Series

Determine whether the telescoping series

n=1[cos(1n)cos(1n+1)]n=1[cos(1n)cos(1n+1)]

converges or diverges. If it converges, find its sum.

Solution

By writing out terms in the sequence of partial sums, we can see that

S1=cos(1)cos(12)S2=(cos(1)cos(12))+(cos(12)cos(13))=cos(1)cos(13)S3=(cos(1)cos(12))+(cos(12)cos(13))+(cos(13)cos(14))=cos(1)cos(14).S1=cos(1)cos(12)S2=(cos(1)cos(12))+(cos(12)cos(13))=cos(1)cos(13)S3=(cos(1)cos(12))+(cos(12)cos(13))+(cos(13)cos(14))=cos(1)cos(14).

In general,

Sk=cos(1)cos(1k+1).Sk=cos(1)cos(1k+1).

Since 1/(k+1)01/(k+1)0 as kk and cosxcosx is a continuous function, cos(1/(k+1))cos(0)=1.cos(1/(k+1))cos(0)=1. Therefore, we conclude that Skcos(1)1.Skcos(1)1. The telescoping series converges and the sum is given by

n=1[cos(1n)cos(1n+1)]=cos(1)1.n=1[cos(1n)cos(1n+1)]=cos(1)1.
Checkpoint 5.11

Determine whether n=1[e1/ne1/(n+1)]n=1[e1/ne1/(n+1)] converges or diverges. If it converges, find its sum.

Student Project

Euler’s Constant

We have shown that the harmonic series n=11nn=11n diverges. Here we investigate the behavior of the partial sums SkSk as k.k. In particular, we show that they behave like the natural logarithm function by showing that there exists a constant γγ such that

n=1k1nlnkγask.n=1k1nlnkγask.

This constant γγ is known as Euler’s constant.

  1. Let Tk=n=1k1nlnk.Tk=n=1k1nlnk. Evaluate TkTk for various values of k.k.
  2. For TkTk as defined in part 1. show that the sequence {Tk}{Tk} converges by using the following steps.
    1. Show that the sequence {Tk}{Tk} is monotone decreasing. (Hint: Show that ln(1+1/k>1/(k+1))ln(1+1/k>1/(k+1))
    2. Show that the sequence {Tk}{Tk} is bounded below by zero. (Hint: Express lnklnk as a definite integral.)
    3. Use the Monotone Convergence Theorem to conclude that the sequence {Tk}{Tk} converges. The limit γγ is Euler’s constant.
  3. Now estimate how far TkTk is from γγ for a given integer k.k. Prove that for k1,k1, 0<Tkγ1/k0<Tkγ1/k by using the following steps.
    1. Show that ln(k+1)lnk<1/k.ln(k+1)lnk<1/k.
    2. Use the result from part a. to show that for any integer k,k,
      TkTk+1<1k1k+1.TkTk+1<1k1k+1.
    3. For any integers kk and jj such that j>k,j>k, express TkTjTkTj as a telescoping sum by writing
      TkTj=(TkTk+1)+(Tk+1Tk+2)+(Tk+2Tk+3)++(Tj1Tj).TkTj=(TkTk+1)+(Tk+1Tk+2)+(Tk+2Tk+3)++(Tj1Tj).

      Use the result from part b. combined with this telescoping sum to conclude that
      TkTj<1k1j.TkTj<1k1j.
    4. Apply the limit to both sides of the inequality in part c. to conclude that
      Tkγ1k.Tkγ1k.
    5. Estimate γγ to an accuracy of within 0.001.0.001.

Section 5.2 Exercises

Using sigma notation, write the following expressions as infinite series.

67.

1+12+13+14+1+12+13+14+

68.

11+11+11+11+

69.

112+1314+...112+1314+...

70.

sin1+sin1/2+sin1/3+sin1/4+sin1+sin1/2+sin1/3+sin1/4+

Compute the first four partial sums S1,…,S4S1,…,S4 for the series having nthnth term anan starting with n=1n=1 as follows.

71.

an=nan=n

72.

an=1/nan=1/n

73.

an=sin(nπ/2)an=sin(nπ/2)

74.

an=(−1)nan=(−1)n

In the following exercises, compute the general term anan of the series with the given partial sum Sn.Sn. If the sequence of partial sums converges, find its limit S.S.

75.

Sn=11n,Sn=11n, n2n2

76.

Sn=n(n+1)2,Sn=n(n+1)2, n1n1

77.

Sn=n,n2Sn=n,n2

78.

Sn=2(n+2)/2n,n1Sn=2(n+2)/2n,n1

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

79.

n=1nn+2n=1nn+2

80.

n=1(1(−1)n))n=1(1(−1)n))

81.

n=11(n+1)(n+2)n=11(n+1)(n+2) (Hint: Use a partial fraction decomposition like that for n=11n(n+1).)n=11n(n+1).)

82.

n=112n+1n=112n+1 (Hint: Follow the reasoning for n=11n.)n=11n.)

Suppose that n=1an=1,n=1an=1, that n=1bn=−1,n=1bn=−1, that a1=2,a1=2, and b1=−3.b1=−3. Find the sum of the indicated series.

83.

n=1(an+bn)n=1(an+bn)

84.

n=1(an2bn)n=1(an2bn)

85.

n=2(anbn)n=2(anbn)

86.

n=1(3an+14bn+1)n=1(3an+14bn+1)

State whether the given series converges and explain why.

87.

n=11n+1000n=11n+1000 (Hint: Rewrite using a change of index.)

88.

n=11n+1080n=11n+1080 (Hint: Rewrite using a change of index.)

89.

1+110+1100+11000+1+110+1100+11000+

90.

1+eπ+e2π2+e3π3+1+eπ+e2π2+e3π3+

91.

1+πe+π2e4+π3e6+π4e8+1+πe+π2e4+π3e6+π4e8+

92.

1π3+π29π327+1π3+π29π327+

For anan as follows, write the sum as a geometric series of the form n=1arn.n=1arn. State whether the series converges and if it does, find the value of an.an.

93.

a1=−1a1=−1 and an/an+1=−5an/an+1=−5 for n1.n1.

94.

a1=2a1=2 and an/an+1=1/2an/an+1=1/2 for n1.n1.

95.

a1=10a1=10 and an/an+1=10an/an+1=10 for n1.n1.

96.

a1=1/10a1=1/10 and an/an+1=−10an/an+1=−10 for n1.n1.

Use the identity 11y=n=0yn11y=n=0yn to express the function as a geometric series in the indicated term.

97.

x1+xx1+x in xx

98.

x1x3/2x1x3/2 in xx

99.

11+sin2x11+sin2x in sinxsinx

100.

sec2xsec2x in sinxsinx

Evaluate the following telescoping series or state whether the series diverges.

101.

n=121/n21/(n+1)n=121/n21/(n+1)

102.

n=11n131(n+1)13n=11n131(n+1)13

103.

n=1(nn+1)n=1(nn+1)

104.

n=1(sinnsin(n+1))n=1(sinnsin(n+1))

Express the following series as a telescoping sum and evaluate its nth partial sum.

105.

n=1ln(nn+1)n=1ln(nn+1)

106.

n=12n+1(n2+n)2n=12n+1(n2+n)2 (Hint: Factor denominator and use partial fractions.)

107.

n=2ln(1+n1)lnnln(n+1)n=2ln(1+n1)lnnln(n+1)

108.

n=1(n+2)n(n+1)2n+1n=1(n+2)n(n+1)2n+1 (Hint: Look at 1/(n2n).)1/(n2n).)

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

109.

Let an=f(n)2f(n+1)+f(n+2),an=f(n)2f(n+1)+f(n+2), in which f(n)0f(n)0 as n.n. Find n=1an.n=1an.

110.

an=f(n)f(n+1)f(n+2)+f(n+3),an=f(n)f(n+1)f(n+2)+f(n+3), in which f(n)0f(n)0 as n.n. Find n=1an.n=1an.

111.

Suppose that an=c0f(n)+c1f(n+1)+c2f(n+2)+c3f(n+3)+c4f(n+4),an=c0f(n)+c1f(n+1)+c2f(n+2)+c3f(n+3)+c4f(n+4), where f(n)0f(n)0 as n.n. Find a condition on the coefficients c0,…,c4c0,…,c4 that make this a general telescoping series.

112.

Evaluate n=11n(n+1)(n+2)n=11n(n+1)(n+2) (Hint: 1n(n+1)(n+2)=12n1n+1+12(n+2))1n(n+1)(n+2)=12n1n+1+12(n+2))

113.

Evaluate n=22n3n.n=22n3n.

114.

Find a formula for n=11n(n+N)n=11n(n+N) where NN is a positive integer.

115.

[T] Define a sequence tk=n=1k1(1/k)lnk.tk=n=1k1(1/k)lnk. Use the graph of 1/x1/x to verify that tktk is increasing. Plot tktk for k=1100k=1100 and state whether it appears that the sequence converges.

116.

[T] Suppose that NN equal uniform rectangular blocks are stacked one on top of the other, allowing for some overhang. Archimedes’ law of the lever implies that the stack of NN blocks is stable as long as the center of mass of the top (N1)(N1) blocks lies at the edge of the bottom block. Let xx denote the position of the edge of the bottom block, and think of its position as relative to the center of the next-to-bottom block. This implies that (N1)x=(12x)(N1)x=(12x) or x=1/(2N).x=1/(2N). Use this expression to compute the maximum overhang (the position of the edge of the top block over the edge of the bottom block.) See the following figure.

This is a diagram of the edge of a table with several blocks stacked on the edge. Each block is pushed slightly to the right, and over the edge of the table. An arrow is drawn with arrowheads at both ends from the edge of the table to a line drawn down from the edge of the highest block.

Each of the following infinite series converges to the given multiple of ππ or 1/π.1/π.

In each case, find the minimum value of NN such that the NthNth partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to 1515 decimals place, π=3.141592653589793....π=3.141592653589793....

117.

[T] π=−3+n=1n2nn!2(2n)!,π=−3+n=1n2nn!2(2n)!, error <0.0001<0.0001

118.

[T] π2=k=0k!(2k+1)!!=k=02kk!2(2k+1)!,π2=k=0k!(2k+1)!!=k=02kk!2(2k+1)!, error <10−4<10−4

119.

[T] 98012π=49801k=0(4k)!(1103+26390k)(k!)43964k,98012π=49801k=0(4k)!(1103+26390k)(k!)43964k, error <10−12<10−12

120.

[T] 112π=k=0(−1)k(6k)!(13591409+545140134k)(3k)!(k!)36403203k+3/2,112π=k=0(−1)k(6k)!(13591409+545140134k)(3k)!(k!)36403203k+3/2, error <10−15<10−15

121.

[T] A fair coin is one that has probability 1/21/2 of coming up heads when flipped.

  1. What is the probability that a fair coin will come up tails nn times in a row?
  2. Find the probability that a coin comes up heads for the first time after an even number of coin flips.
122.

[T] Find the probability that a fair coin is flipped a multiple of three times before coming up heads.

123.

[T] Find the probability that a fair coin will come up heads for the second time after an even number of flips.

124.

[T] Find a series that expresses the probability that a fair coin will come up heads for the second time on a multiple of three flips.

125.

[T] The expected number of times that a fair coin will come up heads is defined as the sum over n=1,2,…n=1,2,… of nn times the probability that the coin will come up heads exactly nn times in a row, or n/2n+1.n/2n+1. Compute the expected number of consecutive times that a fair coin will come up heads.

126.

[T] A person deposits $10$10 at the beginning of each quarter into a bank account that earns 4%4% annual interest compounded quarterly (four times a year).

  1. Show that the interest accumulated after nn quarters is $10(1.01n+110.01n).$10(1.01n+110.01n).
  2. Find the first eight terms of the sequence.
  3. How much interest has accumulated after 22 years?
127.

[T] Suppose that the amount of a drug in a patient’s system diminishes by a multiplicative factor r<1r<1 each hour. Suppose that a new dose is administered every NN hours. Find an expression that gives the amount A(n)A(n) in the patient’s system after nn hours for each nn in terms of the dosage dd and the ratio r.r. (Hint: Write n=mN+k,n=mN+k, where 0k<N,0k<N, and sum over values from the different doses administered.)

128.

[T] A certain drug is effective for an average patient only if there is at least 11 mg per kg in the patient’s system, while it is safe only if there is at most 22 mg per kg in an average patient’s system. Suppose that the amount in a patient’s system diminishes by a multiplicative factor of 0.90.9 each hour after a dose is administered. Find the maximum interval NN of hours between doses, and corresponding dose range dd (in mg/kg) for this NN that will enable use of the drug to be both safe and effective in the long term.

129.

Suppose that an0an0 is a sequence of numbers. Explain why the sequence of partial sums of anan is increasing.

130.

[T] Suppose that anan is a sequence of positive numbers and the sequence SnSn of partial sums of anan is bounded above. Explain why n=1ann=1an converges. Does the conclusion remain true if we remove the hypothesis an0?an0?

131.

[T] Suppose that a1=S1=1a1=S1=1 and that, for given numbers S>1S>1 and 0<k<1,0<k<1, one defines an+1=k(SSn)an+1=k(SSn) and Sn+1=an+1+Sn.Sn+1=an+1+Sn. Does SnSn converge? If so, to what? (Hint: First argue that Sn<SSn<S for all nn and SnSn is increasing.)

132.

[T] A version of von Bertalanffy growth can be used to estimate the age of an individual in a homogeneous species from its length if the annual increase in year n+1n+1 satisfies an+1=k(SSn),an+1=k(SSn), with SnSn as the length at year n,n, SS as a limiting length, and kk as a relative growth constant. If S1=3,S1=3, S=9,S=9, and k=1/2,k=1/2, numerically estimate the smallest value of nn such that Sn8.Sn8. Note that Sn+1=Sn+an+1.Sn+1=Sn+an+1. Find the corresponding nn when k=1/4.k=1/4.

133.

[T] Suppose that n=1ann=1an is a convergent series of positive terms. Explain why limNn=N+1an=0.limNn=N+1an=0.

134.

[T] Find the length of the dashed zig-zag path in the following figure.

No-Alt-Text
135.

[T] Find the total length of the dashed path in the following figure.

This is a triangle drawn in quadrant 1 with vertices at (1, 1), (0, 0), and (1, 0). The zigzag line is drawn starting at (0.5, 0) and goes to the middle of the hypotenuse, the midpoint between that point and the vertical leg, the midpoint of the upper half of the hypotenuse, the midpoint between that point and the vertical leg, and so on until it converges on the top vertex.
136.

[T] The Sierpinski triangle is obtained from a triangle by deleting the middle fourth as indicated in the first step, by deleting the middle fourths of the remaining three congruent triangles in the second step, and in general deleting the middle fourths of the remaining triangles in each successive step. Assuming that the original triangle is shown in the figure, find the areas of the remaining parts of the original triangle after NN steps and find the total length of all of the boundary triangles after NN steps.

No-Alt-Text
137.

[T] The Sierpinski gasket is obtained by dividing the unit square into nine equal sub-squares, removing the middle square, then doing the same at each stage to the remaining sub-squares. The figure shows the remaining set after four iterations. Compute the total area removed after NN stages, and compute the length the total perimeter of the remaining set after NN stages.

This is a black square with many smaller squares removed from it, leaving behind blank spaces in a pattern of squares. There are four iterations of the removal process. At the first, the central 1/9 square area is removed. Each side is 1/3 of that of the next larger square. Next, eight smaller squares are removed around this one. Eight smaller squares are removed from around each of those – 64 in total. Eight even smaller ones are removed from around each of those 64.
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