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Calculus Volume 2

5.5 Alternating Series

Calculus Volume 25.5 Alternating 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

  • 5.5.1. Use the alternating series test to test an alternating series for convergence.
  • 5.5.2. Estimate the sum of an alternating series.
  • 5.5.3. Explain the meaning of absolute convergence and conditional convergence.

So far in this chapter, we have primarily discussed series with positive terms. In this section we introduce alternating series—those series whose terms alternate in sign. We will show in a later chapter that these series often arise when studying power series. After defining alternating series, we introduce the alternating series test to determine whether such a series converges.

The Alternating Series Test

A series whose terms alternate between positive and negative values is an alternating series. For example, the series

n=1(12)n=12+1418+116n=1(12)n=12+1418+116
5.11

and

n=1(−1)n+1n=112+1314+n=1(−1)n+1n=112+1314+
5.12

are both alternating series.

Definition

Any series whose terms alternate between positive and negative values is called an alternating series. An alternating series can be written in the form

n=1(−1)n+1bn=b1b2+b3b4+n=1(−1)n+1bn=b1b2+b3b4+
5.13

or

n1(−1)nbn=b1+b2b3+b4n1(−1)nbn=b1+b2b3+b4
5.14

Where bn0bn0 for all positive integers n.

Series (1), shown in Equation 5.11, is a geometric series. Since |r|=|1/2|<1,|r|=|1/2|<1, the series converges. Series (2), shown in Equation 5.12, is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.

To prove this, we look at the sequence of partial sums {Sk}{Sk} (Figure 5.17).

Proof

Consider the odd terms S2k+1S2k+1 for k0.k0. Since 1/(2k+1)<1/2k,1/(2k+1)<1/2k,

S2k+1=S2k112k+12k+1<S2k1.S2k+1=S2k112k+12k+1<S2k1.

Therefore, {S2k+1}{S2k+1} is a decreasing sequence. Also,

S2k+1=(112)+(1314)++(12k112k)+12k+1>0.S2k+1=(112)+(1314)++(12k112k)+12k+1>0.

Therefore, {S2k+1}{S2k+1} is bounded below. Since {S2k+1}{S2k+1} is a decreasing sequence that is bounded below, by the Monotone Convergence Theorem, {S2k+1}{S2k+1} converges. Similarly, the even terms {S2k}{S2k} form an increasing sequence that is bounded above because

S2k=S2k2+12k112k>S2k2S2k=S2k2+12k112k>S2k2

and

S2k=1+(12+13)++(12k2+12k1)12k<1.S2k=1+(12+13)++(12k2+12k1)12k<1.

Therefore, by the Monotone Convergence Theorem, the sequence {S2k}{S2k} also converges. Since

S2k+1=S2k+12k+1,S2k+1=S2k+12k+1,

we know that

limkS2k+1=limkS2k+limk12k+1.limkS2k+1=limkS2k+limk12k+1.

Letting S=limkS2k+1S=limkS2k+1 and using the fact that 1/(2k+1)0,1/(2k+1)0, we conclude that limkS2k=S.limkS2k=S. Since the odd terms and the even terms in the sequence of partial sums converge to the same limit S,S, it can be shown that the sequence of partial sums converges to S,S, and therefore the alternating harmonic series converges to S.S.

It can also be shown that S=ln2,S=ln2, and we can write

n=1(−1)n+1n=112+1314+=ln(2).n=1(−1)n+1n=112+1314+=ln(2).
This graph demonstrates the alternating hamanic series in the first quadrant. The highest line 1 is drawn to S1, the next line -1/2 is drawn to S2, the next line +1/3 is drawn to S3, the line -1/4 is drawn to S4, and the last line +1/5 is drawn to S5. The odd terms are decreasing and bounded below, and the even terms are increasing and bounded above. It seems to be converging to S, which is in the middle of S2, S4 and S5, S3, S1.
Figure 5.17 For the alternating harmonic series, the odd terms S2k+1S2k+1 in the sequence of partial sums are decreasing and bounded below. The even terms S2kS2k are increasing and bounded above.

More generally, any alternating series of form (3) (Equation 5.13) or (4) (Equation 5.14) converges as long as b1b2b3b1b2b3 and bn0bn0 (Figure 5.18). The proof is similar to the proof for the alternating harmonic series.

This diagram illustrates an alternating series in quadrant 1. The highest line b1 is drawn out to S1, the next line –b2 is drawn back to S2, the next line b3 is drawn out to S3, the next line –b4 is drawn back to S4, and the last line is drawn out to S5. It seems to be converging to S, which is in between S2, S4 and S5, S3, and S1. The odd terms are decreasing and bounded below. The even terms are increasing and bounded above.
Figure 5.18 For an alternating series b1b2+b3b1b2+b3 in which b1>b2>b3>,b1>b2>b3>, the odd terms S2k+1S2k+1 in the sequence of partial sums are decreasing and bounded below. The even terms S2kS2k are increasing and bounded above.

Theorem 5.13

Alternating Series Test

An alternating series of the form

n=1(−1)n+1bnorn=1(−1)nbnn=1(−1)n+1bnorn=1(−1)nbn

converges if

  1. 0bn+1bn0bn+1bn for all n1n1 and
  2. limnbn=0.limnbn=0.

This is known as the alternating series test.

We remark that this theorem is true more generally as long as there exists some integer NN such that 0bn+1bn0bn+1bn for all nN.nN.

Example 5.19

Convergence of Alternating Series

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

  1. n=1(−1)n+1/n2n=1(−1)n+1/n2
  2. n=1(−1)n+1n/(n+1)n=1(−1)n+1n/(n+1)

Solution

  1. Since
    1(n+1)2<1n2and1n20,1(n+1)2<1n2and1n20,
    the series converges.
  2. Since n/(n+1)0n/(n+1)0 as n,n, we cannot apply the alternating series test. Instead, we use the nth term test for divergence. Since
    limn(−1)n+1nn+10,limn(−1)n+1nn+10,
    the series diverges.

Checkpoint 5.18

Determine whether the series n=1(−1)n+1n/2nn=1(−1)n+1n/2n converges or diverges.

Remainder of an Alternating Series

It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series

n=1(−1)n+1bnn=1(−1)n+1bn

satisfying the hypotheses of the alternating series test. Let SS denote the sum of this series and {Sk}{Sk} be the corresponding sequence of partial sums. From Figure 5.18, we see that for any integer N1,N1, the remainder RNRN satisfies

|RN|=|SSN||SN+1SN|=bn+1.|RN|=|SSN||SN+1SN|=bn+1.
Theorem 5.14

Remainders in Alternating Series

Consider an alternating series of the form

n=1(−1)n+1bnorn=1(−1)nbnn=1(−1)n+1bnorn=1(−1)nbn

that satisfies the hypotheses of the alternating series test. Let SS denote the sum of the series and SNSN denote the NthNth partial sum. For any integer N1,N1, the remainder RN=SSNRN=SSN satisfies

|RN|bN+1.|RN|bN+1.

In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the NthNth partial sum SNSN is in magnitude at most the size of the next term bN+1.bN+1.

Example 5.20

Estimating the Remainder of an Alternating Series

Consider the alternating series

n=1(−1)n+1n2.n=1(−1)n+1n2.

Use the remainder estimate to determine a bound on the error R10R10 if we approximate the sum of the series by the partial sum S10.S10.

Solution

From the theorem stated above,

|R10|b11=11120.008265.|R10|b11=11120.008265.

Checkpoint 5.19

Find a bound for R20R20 when approximating n=1(−1)n+1/nn=1(−1)n+1/n by S20.S20.

Absolute and Conditional Convergence

Consider a series n=1ann=1an and the related series n=1|an|.n=1|an|. Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series n=1(−1)n+1/n.n=1(−1)n+1/n. The series whose terms are the absolute value of these terms is the harmonic series, since n=1|(−1)n+1/n|=n=11/n.n=1|(−1)n+1/n|=n=11/n. Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.

By comparison, consider the series n=1(−1)n+1/n2.n=1(−1)n+1/n2. The series whose terms are the absolute values of the terms of this series is the series n=11/n2.n=11/n2. Since both of these series converge, we say the series n=1(−1)n+1/n2n=1(−1)n+1/n2 exhibits absolute convergence.

Definition

A series n=1ann=1an exhibits absolute convergence if n=1|an|n=1|an| converges. A series n=1ann=1an exhibits conditional convergence if n=1ann=1an converges but n=1|an|n=1|an| diverges.

As shown by the alternating harmonic series, a series n=1ann=1an may converge, but n=1|an|n=1|an| may diverge. In the following theorem, however, we show that if n=1|an|n=1|an| converges, then n=1ann=1an converges.

Theorem 5.15

Absolute Convergence Implies Convergence

If n=1|an|n=1|an| converges, then n=1ann=1an converges.

Proof

Suppose that n=1|an|n=1|an| converges. We show this by using the fact that an=|an|an=|an| or an=|an|an=|an| and therefore |an|+an=2|an||an|+an=2|an| or |an|+an=0.|an|+an=0. Therefore, 0|an|+an2|an|.0|an|+an2|an|. Consequently, by the comparison test, since 2n=1|an|2n=1|an| converges, the series

n=1(|an|+an)n=1(|an|+an)

converges. By using the algebraic properties for convergent series, we conclude that

n=1an=n=1(|an|+an)n=1|an|n=1an=n=1(|an|+an)n=1|an|

converges.

Example 5.21

Absolute versus Conditional Convergence

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

  1. n=1(−1)n+1/(3n+1)n=1(−1)n+1/(3n+1)
  2. n=1cos(n)/n2n=1cos(n)/n2

Solution

  1. We can see that
    n=1|(−1)n+13n+1|=n=113n+1n=1|(−1)n+13n+1|=n=113n+1

    diverges by using the limit comparison test with the harmonic series. In fact,
    limn1/(3n+1)1/n=13.limn1/(3n+1)1/n=13.

    Therefore, the series does not converge absolutely. However, since
    13(n+1)+1<13n+1and13n+10,13(n+1)+1<13n+1and13n+10,

    the series converges. We can conclude that n=1(−1)n+1/(3n+1)n=1(−1)n+1/(3n+1) converges conditionally.
  2. Noting that |cosn|1,|cosn|1, to determine whether the series converges absolutely, compare
    n=1|cosnn2|n=1|cosnn2|

    with the series n=11/n2.n=11/n2. Since n=11/n2n=11/n2 converges, by the comparison test, n=1|cosn/n2|n=1|cosn/n2| converges, and therefore n=1cosn/n2n=1cosn/n2 converges absolutely.

Checkpoint 5.20

Determine whether the series n=1(−1)n+1n/(2n3+1)n=1(−1)n+1n/(2n3+1) converges absolutely, converges conditionally, or diverges.

To see the difference between absolute and conditional convergence, look at what happens when we rearrange the terms of the alternating harmonic series n=1(−1)n+1/n.n=1(−1)n+1/n. We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.

Begin by adding enough of the positive terms to produce a sum that is larger than some real number M>0.M>0. For example, let M=10,M=10, and find an integer kk such that

1+13+15++12k1>10.1+13+15++12k1>10.

(We can do this because the series n=11/(2n1)n=11/(2n1) diverges to infinity.) Then subtract 1/2.1/2. Then add more positive terms until the sum reaches 100. That is, find another integer j>kj>k such that

1+13++12k112+12k+1++12j+1>100.1+13++12k112+12k+1++12j+1>100.

Then subtract 1/4.1/4. Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.

The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In Example 5.22, we show how to rearrange the terms to create a new series that converges to 3ln(2)/2.3ln(2)/2. We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number r;r; however, the proof of that fact is beyond the scope of this text.

In general, any series n=1ann=1an that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series n=1ann=1an that converges absolutely, the value of n=1ann=1an is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.

Example 5.22

Rearranging Series

Use the fact that

112+1314+15=ln2112+1314+15=ln2

to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is 3ln(2)/2.3ln(2)/2.

Solution

Let

n=1an=112+1314+1516+1718+.n=1an=112+1314+1516+1718+.

Since n=1an=ln(2),n=1an=ln(2), by the algebraic properties of convergent series,

n=112an=1214+1618+=12n=1an=ln22.n=112an=1214+1618+=12n=1an=ln22.

Now introduce the series n=1bnn=1bn such that for all n1,n1, b2n1=0b2n1=0 and b2n=an/2.b2n=an/2. Then

n=1bn=0+12+014+0+16+018+=ln22.n=1bn=0+12+014+0+16+018+=ln22.

Then using the algebraic limit properties of convergent series, since n=1ann=1an and n=1bnn=1bn converge, the series n=1(an+bn)n=1(an+bn) converges and

n=1(an+bn)=n=1an+n=1bn=ln2+ln22=3ln22.n=1(an+bn)=n=1an+n=1bn=ln2+ln22=3ln22.

Now adding the corresponding terms, anan and bn,bn, we see that

n=1(an+bn)=(1+0)+(12+12)+(13+0)+(1414)+(15+0)+(16+16)+(17+0)+(1818)+=1+1312+15+1714+.n=1(an+bn)=(1+0)+(12+12)+(13+0)+(1414)+(15+0)+(16+16)+(17+0)+(1818)+=1+1312+15+1714+.

We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since n=1(an+bn)=3ln(2)/2,n=1(an+bn)=3ln(2)/2, we conclude that

1+1312+15+1714+=3ln(2)2.1+1312+15+1714+=3ln(2)2.

Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.

Section 5.5 Exercises

State whether each of the following series converges absolutely, conditionally, or not at all.

250.

n=1(−1)n+1nn+3n=1(−1)n+1nn+3

251.

n=1(−1)n+1n+1n+3n=1(−1)n+1n+1n+3

252.

n=1(−1)n+11n+3n=1(−1)n+11n+3

253.

n=1(−1)n+1n+3nn=1(−1)n+1n+3n

254.

n=1(−1)n+11n!n=1(−1)n+11n!

255.

n=1(−1)n+13nn!n=1(−1)n+13nn!

256.

n=1(−1)n+1(n1n)nn=1(−1)n+1(n1n)n

257.

n=1(−1)n+1(n+1n)nn=1(−1)n+1(n+1n)n

258.

n=1(−1)n+1sin2nn=1(−1)n+1sin2n

259.

n=1(−1)n+1cos2nn=1(−1)n+1cos2n

260.

n=1(−1)n+1sin2(1/n)n=1(−1)n+1sin2(1/n)

261.

n=1(−1)n+1cos2(1/n)n=1(−1)n+1cos2(1/n)

262.

n=1(−1)n+1ln(1/n)n=1(−1)n+1ln(1/n)

263.

n=1(−1)n+1ln(1+1n)n=1(−1)n+1ln(1+1n)

264.

n=1(−1)n+1n21+n4n=1(−1)n+1n21+n4

265.

n=1(−1)n+1ne1+nπn=1(−1)n+1ne1+nπ

266.

n=1(−1)n+121/nn=1(−1)n+121/n

267.

n=1(−1)n+1n1/nn=1(−1)n+1n1/n

268.

n=1(−1)n(1n1/n)n=1(−1)n(1n1/n) (Hint: n1/n1+ln(n)/nn1/n1+ln(n)/n for large n.)n.)

269.

n=1(−1)n+1n(1cos(1n))n=1(−1)n+1n(1cos(1n)) (Hint: cos(1/n)11/n2cos(1/n)11/n2 for large n.)n.)

270.

n=1(−1)n+1(n+1n)n=1(−1)n+1(n+1n) (Hint: Rationalize the numerator.)

271.

n=1(−1)n+1(1n1n+1)n=1(−1)n+1(1n1n+1) (Hint: Find common denominator then rationalize numerator.)

272.

n=1(−1)n+1(ln(n+1)lnn)n=1(−1)n+1(ln(n+1)lnn)

273.

n=1(−1)n+1n(tan−1(n+1)tan−1n)n=1(−1)n+1n(tan−1(n+1)tan−1n) (Hint: Use Mean Value Theorem.)

274.

n=1(−1)n+1((n+1)2n2)n=1(−1)n+1((n+1)2n2)

275.

n=1(−1)n+1(1n1n+1)n=1(−1)n+1(1n1n+1)

276.

n=1cos(nπ)nn=1cos(nπ)n

277.

n=1cos(nπ)n1/nn=1cos(nπ)n1/n

278.

n=11nsin(nπ2)n=11nsin(nπ2)

279.

n=1sin(nπ/2)sin(1/n)n=1sin(nπ/2)sin(1/n)

In each of the following problems, use the estimate |RN|bN+1|RN|bN+1 to find a value of NN that guarantees that the sum of the first NN terms of the alternating series n=1(−1)n+1bnn=1(−1)n+1bn differs from the infinite sum by at most the given error. Calculate the partial sum SNSN for this N.N.

280.

[T] bn=1/n,bn=1/n, error <10−5<10−5

281.

[T] bn=1/ln(n),bn=1/ln(n), n2,n2, error <10−1<10−1

282.

[T] bn=1/n,bn=1/n, error <10−3<10−3

283.

[T] bn=1/2n,bn=1/2n, error <10−6<10−6

284.

[T] bn=ln(1+1n),bn=ln(1+1n), error <10−3<10−3

285.

[T] bn=1/n2,bn=1/n2, error <10−6<10−6

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

286.

If bn0bn0 is decreasing and limnbn=0,limnbn=0, then n=1(b2n1b2n)n=1(b2n1b2n) converges absolutely.

287.

If bn0bn0 is decreasing, then n=1(b2n1b2n)n=1(b2n1b2n) converges absolutely.

288.

If bn0bn0 and limnbn=0limnbn=0 then n=1(12(b3n2+b3n1)b3n)n=1(12(b3n2+b3n1)b3n) converges.

289.

If bn0bn0 is decreasing and n=1(b3n2+b3n1b3n)n=1(b3n2+b3n1b3n) converges then n=1b3n2n=1b3n2 converges.

290.

If bn0bn0 is decreasing and n=1(−1)n1bnn=1(−1)n1bn converges conditionally but not absolutely, then bnbn does not tend to zero.

291.

Let an+=anan+=an if an0an0 and an=anan=an if an<0.an<0. (Also, an+=0ifan<0an+=0ifan<0 and an=0ifan0.)an=0ifan0.) If n=1ann=1an converges conditionally but not absolutely, then neither n=1an+n=1an+ nor n=1ann=1an converge.

292.

Suppose that anan is a sequence of positive real numbers and that n=1ann=1an converges.

Suppose that bnbn is an arbitrary sequence of ones and minus ones. Does n=1anbnn=1anbn necessarily converge?

293.

Suppose that anan is a sequence such that n=1anbnn=1anbn converges for every possible sequence bnbn of zeros and ones. Does n=1ann=1an converge absolutely?

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

294.

n=1(−1)n+1sin2nnn=1(−1)n+1sin2nn

295.

n=1(−1)n+1cos2nnn=1(−1)n+1cos2nn

296.

1+121314+15+161718+1+121314+15+161718+

297.

1+1213+14+1516+17+1819+1+1213+14+1516+17+1819+

298.

Show that the alternating series 112+1214+1316+1418+112+1214+1316+1418+ does

not converge. What hypothesis of the alternating series test is not met?

299.

Suppose that anan converges absolutely. Show that the series consisting of the positive terms anan also converges.

300.

Show that the alternating series 2335+4759+2335+4759+ does not converge. What hypothesis of the alternating series test is not met?

301.

The formula cosθ=1θ22!+θ44!θ66!+cosθ=1θ22!+θ44!θ66!+ will be derived in the next chapter. Use the remainder |RN|bN+1|RN|bN+1 to find a bound for the error in estimating cosθcosθ by the fifth partial sum 1θ2/2!+θ4/4!θ6/6!+θ8/8!1θ2/2!+θ4/4!θ6/6!+θ8/8! for θ=1,θ=1, θ=π/6,θ=π/6, and θ=π.θ=π.

302.

The formula sinθ=θθ33!+θ55!θ77!+sinθ=θθ33!+θ55!θ77!+ will be derived in the next chapter. Use the remainder |RN|bN+1|RN|bN+1 to find a bound for the error in estimating sinθsinθ by the fifth partial sum θθ3/3!+θ5/5!θ7/7!+θ9/9!θθ3/3!+θ5/5!θ7/7!+θ9/9! for θ=1,θ=1, θ=π/6,θ=π/6, and θ=π.θ=π.

303.

How many terms in cosθ=1θ22!+θ44!θ66!+cosθ=1θ22!+θ44!θ66!+ are needed to approximate cos1cos1 accurate to an error of at most 0.00001?0.00001?

304.

How many terms in sinθ=θθ33!+θ55!θ77!+sinθ=θθ33!+θ55!θ77!+ are needed to approximate sin1sin1 accurate to an error of at most 0.00001?0.00001?

305.

Sometimes the alternating series n=1(−1)n1bnn=1(−1)n1bn converges to a certain fraction of an absolutely convergent series n=1bnn=1bn at a faster rate. Given that n=11n2=π26,n=11n2=π26, find S=1122+132142+.S=1122+132142+. Which of the series 6n=11n26n=11n2 and Sn=1(−1)n1n2Sn=1(−1)n1n2 gives a better estimation of π2π2 using 10001000 terms?

The following alternating series converge to given multiples of π.π. Find the value of NN predicted by the remainder estimate such that the NthNth partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum NN for which the error bound holds, and give the desired approximate value in each case. Up to 1515 decimals places, π=3.141592653589793.π=3.141592653589793.

306.

[T] π4=n=0(−1)n2n+1,π4=n=0(−1)n2n+1, error <0.0001<0.0001

307.

[T] π12=k=0(−3)k2k+1,π12=k=0(−3)k2k+1, error <0.0001<0.0001

308.

[T] The series n=0sin(x+πn)x+πnn=0sin(x+πn)x+πn plays an important role in signal processing. Show that n=0sin(x+πn)x+πnn=0sin(x+πn)x+πn converges whenever 0<x<π.0<x<π. (Hint: Use the formula for the sine of a sum of angles.)

309.

[T] If n=1N(−1)n11nln2,n=1N(−1)n11nln2, what is 1+13+15121416+17+19+11118110112+?1+13+15121416+17+19+11118110112+?

310.

[T] Plot the series n=1100cos(2πnx)nn=1100cos(2πnx)n for 0x<1.0x<1. Explain why n=1100cos(2πnx)nn=1100cos(2πnx)n diverges when x=0,1.x=0,1. How does the series behave for other x?x?

311.

[T] Plot the series n=1100sin(2πnx)nn=1100sin(2πnx)n for 0x<10x<1 and comment on its behavior

312.

[T] Plot the series n=1100cos(2πnx)n2n=1100cos(2πnx)n2 for 0x<10x<1 and describe its graph.

313.

[T] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form n=1Snn,n=1Snn, where snsn is a randomly generated sequence of ±1's±1's in which the values ±1±1 are equally likely to occur. Use a random number generator to produce 10001000 random ±1s±1s and plot the partial sums SN=n=1NsnnSN=n=1Nsnn of your random harmonic sequence for N=1N=1 to 1000.1000. Compare to a plot of the first 10001000 partial sums of the harmonic series.

314.

[T] Estimates of n=11n2n=11n2 can be accelerated by writing its partial sums as n=1N1n2=n=1N1n(n+1)+n=1N1n2(n+1)n=1N1n2=n=1N1n(n+1)+n=1N1n2(n+1) and recalling that n=1N1n(n+1)=11N+1n=1N1n(n+1)=11N+1 converges to one as N.N. Compare the estimate of π2/6π2/6 using the sums n=110001n2n=110001n2 with the estimate using 1+n=110001n2(n+1).1+n=110001n2(n+1).

315.

[T] The Euler transform rewrites S=n=0(−1)nbnS=n=0(−1)nbn as S=n=0(−1)n2n1m=0n(nm)bnm.S=n=0(−1)n2n1m=0n(nm)bnm. For the alternating harmonic series, it takes the form ln(2)=n=1(−1)n1n=n=11n2n.ln(2)=n=1(−1)n1n=n=11n2n. Compute partial sums of n=11n2nn=11n2n until they approximate ln(2)ln(2) accurate to within 0.0001.0.0001. How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate ln(2).ln(2).

316.

[T] In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If an0an0 is such that an0an0 as nn but n=1ann=1an diverges, then, given any number AA there is a sequence snsn of ±1's±1's such that n=1ansnA.n=1ansnA. Show this for A>0A>0 as follows.

  1. Recursively define snsn by sn=1sn=1 if Sn1=k=1n1aksk<ASn1=k=1n1aksk<A and sn=−1sn=−1 otherwise.
  2. Explain why eventually SnA,SnA, and for any mm larger than this n,n, AamSmA+am.AamSmA+am.
  3. Explain why this implies that SnASnA as n.n.
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