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Calculus Volume 2

5.6 Ratio and Root Tests

Calculus Volume 25.6 Ratio and Root Tests
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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

  • 5.6.1. Use the ratio test to determine absolute convergence of a series.
  • 5.6.2. Use the root test to determine absolute convergence of a series.
  • 5.6.3. Describe a strategy for testing the convergence of a given series.

In this section, we prove the last two series convergence tests: the ratio test and the root test. These tests are particularly nice because they do not require us to find a comparable series. The ratio test will be especially useful in the discussion of power series in the next chapter.

Throughout this chapter, we have seen that no single convergence test works for all series. Therefore, at the end of this section we discuss a strategy for choosing which convergence test to use for a given series.

Ratio Test

Consider a series n=1an.n=1an. From our earlier discussion and examples, we know that limnan=0limnan=0 is not a sufficient condition for the series to converge. Not only do we need an0,an0, but we need an0an0 quickly enough. For example, consider the series n=11/nn=11/n and the series n=11/n2.n=11/n2. We know that 1/n01/n0 and 1/n20.1/n20. However, only the series n=11/n2n=11/n2 converges. The series n=11/nn=11/n diverges because the terms in the sequence {1/n}{1/n} do not approach zero fast enough as n.n. Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.

Theorem 5.16

Ratio Test

Let n=1ann=1an be a series with nonzero terms. Let

ρ=limn|an+1an|.ρ=limn|an+1an|.
  1. If 0ρ<1,0ρ<1, then n=1ann=1an converges absolutely.
  2. If ρ>1ρ>1 or ρ=,ρ=, then n=1ann=1an diverges.
  3. If ρ=1,ρ=1, the test does not provide any information.

Proof

Let n=1ann=1an be a series with nonzero terms.

We begin with the proof of part i. In this case, ρ=limn|an+1an|<1.ρ=limn|an+1an|<1. Since 0ρ<1,0ρ<1, there exists RR such that 0ρ<R<1.0ρ<R<1. Let ε=Rρ>0.ε=Rρ>0. By the definition of limit of a sequence, there exists some integer NN such that

||an+1an|ρ|<εfor allnN.||an+1an|ρ|<εfor allnN.

Therefore,

|an+1an|<ρ+ε=Rfor allnN|an+1an|<ρ+ε=Rfor allnN

and, thus,

|aN+1|<R|aN||aN+2|<R|aN+1|<R2|aN||aN+3|<R|aN+2|<R2|aN+1|<R3|aN||aN+4|<R|aN+3|<R2|aN+2|<R3|aN+1|<R4|aN|.|aN+1|<R|aN||aN+2|<R|aN+1|<R2|aN||aN+3|<R|aN+2|<R2|aN+1|<R3|aN||aN+4|<R|aN+3|<R2|aN+2|<R3|aN+1|<R4|aN|.

Since R<1,R<1, the geometric series

R|aN|+R2|aN|+R3|aN|+R|aN|+R2|aN|+R3|aN|+

converges. Given the inequalities above, we can apply the comparison test and conclude that the series

|aN+1|+|aN+2|+|aN+3|+|aN+4|+|aN+1|+|aN+2|+|aN+3|+|aN+4|+

converges. Therefore, since

n=1|an|=n=1N|an|+n=N+1|an|n=1|an|=n=1N|an|+n=N+1|an|

where n=1N|an|n=1N|an| is a finite sum and n=N+1|an|n=N+1|an| converges, we conclude that n=1|an|n=1|an| converges.

For part ii.

ρ=limn|an+1an|>1.ρ=limn|an+1an|>1.

Since ρ>1,ρ>1, there exists RR such that ρ>R>1.ρ>R>1. Let ε=ρR>0.ε=ρR>0. By the definition of the limit of a sequence, there exists an integer NN such that

||an+1an|ρ|<εfor allnN.||an+1an|ρ|<εfor allnN.

Therefore,

R=ρε<|an+1an|for allnN,R=ρε<|an+1an|for allnN,

and, thus,

|aN+1|>R|aN||aN+2|>R|aN+1|>R2|aN||aN+3|>R|aN+2|>R2|aN+1|>R3|aN||aN+4|>R|aN+3|>R2|aN+2|>R3|aN+1|>R4|aN|.|aN+1|>R|aN||aN+2|>R|aN+1|>R2|aN||aN+3|>R|aN+2|>R2|aN+1|>R3|aN||aN+4|>R|aN+3|>R2|aN+2|>R3|aN+1|>R4|aN|.

Since R>1,R>1, the geometric series

R|aN|+R2|aN|+R3|aN|+R|aN|+R2|aN|+R3|aN|+

diverges. Applying the comparison test, we conclude that the series

|aN+1|+|aN+2|+|aN+3|+|aN+1|+|aN+2|+|aN+3|+

diverges, and therefore the series n=1|an|n=1|an| diverges.

For part iii. we show that the test does not provide any information if ρ=1ρ=1 by considering the pseriespseries n=11/np.n=11/np. For any real number p,p,

ρ=limn1/(n+1)p1/np=limnnp(n+1)p=1.ρ=limn1/(n+1)p1/np=limnnp(n+1)p=1.

However, we know that if p1,p1, the pseriespseries n=11/npn=11/np diverges, whereas n=11/npn=11/np converges if p>1.p>1.

The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.

Example 5.23

Using the Ratio Test

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

  1. n=12nn!n=12nn!
  2. n=1nnn!n=1nnn! n=1(−1)n(n!)2(2n)!n=1(−1)n(n!)2(2n)!
  3. n=1(−1)n(n!)2(2n)!n=1(−1)n(n!)2(2n)!

Solution

  1. From the ratio test, we can see that
    ρ=limn2n+1/(n+1)!2n/n!=limn2n+1(n+1)!·n!2n.ρ=limn2n+1/(n+1)!2n/n!=limn2n+1(n+1)!·n!2n.

    Since (n+1)!=(n+1)·n!,(n+1)!=(n+1)·n!,
    ρ=limn2n+1=0.ρ=limn2n+1=0.

    Since ρ<1,ρ<1, the series converges.
  2. We can see that
    ρ=limn(n+1)n+1/(n+1)!nn/n!=limn(n+1)n+1(n+1)!·n!nn=limn(n+1n)n=limn(1+1n)n=e.ρ=limn(n+1)n+1/(n+1)!nn/n!=limn(n+1)n+1(n+1)!·n!nn=limn(n+1n)n=limn(1+1n)n=e.

    Since ρ>1,ρ>1, the series diverges.
  3. Since
    |(−1)n+1((n+1)!)2/(2(n+1))!(−1)n(n!)2/(2n)!|=(n+1)!(n+1)!(2n+2)!·(2n)!n!n!=(n+1)(n+1)(2n+2)(2n+1)|(−1)n+1((n+1)!)2/(2(n+1))!(−1)n(n!)2/(2n)!|=(n+1)!(n+1)!(2n+2)!·(2n)!n!n!=(n+1)(n+1)(2n+2)(2n+1)

    we see that
    ρ=limn(n+1)(n+1)(2n+2)(2n+1)=14.ρ=limn(n+1)(n+1)(2n+2)(2n+1)=14.

    Since ρ<1,ρ<1, the series converges.

Checkpoint 5.21

Use the ratio test to determine whether the series n=1n33nn=1n33n converges or diverges.

Root Test

The approach of the root test is similar to that of the ratio test. Consider a series n=1ann=1an such that limn|an|n=ρlimn|an|n=ρ for some real number ρ.ρ. Then for NN sufficiently large, |aN|ρN.|aN|ρN. Therefore, we can approximate n=N|an|n=N|an| by writing

|aN|+|aN+1|+|aN+2|+ρN+ρN+1+ρN+2+.|aN|+|aN+1|+|aN+2|+ρN+ρN+1+ρN+2+.

The expression on the right-hand side is a geometric series. As in the ratio test, the series n=1ann=1an converges absolutely if 0ρ<10ρ<1 and the series diverges if ρ1.ρ1. If ρ=1,ρ=1, the test does not provide any information. For example, for any p-series, n=11/np,n=11/np, we see that

ρ=limn|1np|n=limn1np/n.ρ=limn|1np|n=limn1np/n.

To evaluate this limit, we use the natural logarithm function. Doing so, we see that

lnρ=ln(limn1np/n)=limnln(1n)p/n=limnpn·ln(1n)=limnpln(1/n)n.lnρ=ln(limn1np/n)=limnln(1n)p/n=limnpn·ln(1n)=limnpln(1/n)n.

Using L’Hôpital’s rule, it follows that lnρ=0,lnρ=0, and therefore ρ=1ρ=1 for all p.p. However, we know that the p-series only converges if p>1p>1 and diverges if p<1.p<1.

Theorem 5.17

Root Test

Consider the series n=1an.n=1an. Let

ρ=limn|an|n.ρ=limn|an|n.
  1. If 0ρ<1,0ρ<1, then n=1ann=1an converges absolutely.
  2. If ρ>1ρ>1 or ρ=,ρ=, then n=1ann=1an diverges.
  3. If ρ=1,ρ=1, the test does not provide any information.

The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms anan satisfy |an|=bnn,|an|=bnn, then |an|n=bn|an|n=bn and we need only evaluate limnbn.limnbn.

Example 5.24

Using the Root Test

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

  1. n=1(n2+3n)n(4n2+5)nn=1(n2+3n)n(4n2+5)n
  2. n=1nn(ln(n))nn=1nn(ln(n))n

Solution

  1. To apply the root test, we compute
    ρ=limn(n2+3n)n/(4n2+5)nn=limnn2+3n4n2+5=14.ρ=limn(n2+3n)n/(4n2+5)nn=limnn2+3n4n2+5=14.

    Since ρ<1,ρ<1, the series converges absolutely.
  2. We have
    ρ=limnnn/(lnn)nn=limnnlnn=by L’Hôpital’s rule.ρ=limnnn/(lnn)nn=limnnlnn=by L’Hôpital’s rule.

    Since ρ=,ρ=, the series diverges.

Checkpoint 5.22

Use the root test to determine whether the series n=11/nnn=11/nn converges or diverges.

Choosing a Convergence Test

At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.

Problem-Solving Strategy: Choosing a Convergence Test for a Series

Consider a series n=1an.n=1an. In the steps below, we outline a strategy for determining whether the series converges.

  1. Is n=1ann=1an a familiar series? For example, is it the harmonic series (which diverges) or the alternating harmonic series (which converges)? Is it a pseriespseries or geometric series? If so, check the power pp or the ratio rr to determine if the series converges.
  2. Is it an alternating series? Are we interested in absolute convergence or just convergence? If we are just interested in whether the series converges, apply the alternating series test. If we are interested in absolute convergence, proceed to step 3,3, considering the series of absolute values n=1|an|.n=1|an|.
  3. Is the series similar to a pseriespseries or geometric series? If so, try the comparison test or limit comparison test.
  4. Do the terms in the series contain a factorial or power? If the terms are powers such that an=bnn,an=bnn, try the root test first. Otherwise, try the ratio test first.
  5. Use the divergence test. If this test does not provide any information, try the integral test.

Media

Visit this website for more information on testing series for convergence, plus general information on sequences and series.

Example 5.25

Using Convergence Tests

For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.

  1. n=1n2+2nn3+3n2+1n=1n2+2nn3+3n2+1
  2. n=1(−1)n+1(3n+1)n!n=1(−1)n+1(3n+1)n!
  3. n=1enn3n=1enn3
  4. n=13n(n+1)nn=13n(n+1)n

Solution

  1. Step 1. The series is not a pseriespseries or geometric series.
    Step 2. The series is not alternating.
    Step 3. For large values of n,n, we approximate the series by the expression
    n2+2nn3+3n2+1n2n3=1n.n2+2nn3+3n2+1n2n3=1n.

    Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series n=11/n.n=11/n. Using the limit comparison test, we see that
    limn(n2+2n)/(n3+3n2+1)1/n=limnn3+2n2n3+3n2+1=1.limn(n2+2n)/(n3+3n2+1)1/n=limnn3+2n2n3+3n2+1=1.

    Since the series n=11/nn=11/n diverges, this series diverges as well.
  2. Step 1.The series is not a familiar series.
    Step 2. The series is alternating. Since we are interested in absolute convergence, consider the series
    n=13n(n+1)!.n=13n(n+1)!.

    Step 3. The series is not similar to a p-series or geometric series.
    Step 4. Since each term contains a factorial, apply the ratio test. We see that
    limn(3(n+1))/(n+1)!(3n+1)/n!=limn3n+3(n+1)!·n!3n+1=limn3n+3(n+1)(3n+1)=0.limn(3(n+1))/(n+1)!(3n+1)/n!=limn3n+3(n+1)!·n!3n+1=limn3n+3(n+1)(3n+1)=0.

    Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.
  3. Step 1. The series is not a familiar series.
    Step 2. It is not an alternating series.
    Step 3. There is no obvious series with which to compare this series.
    Step 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.
    Step 5. To apply the divergence test, we calculate that
    limnenn3=.limnenn3=.

    Therefore, by the divergence test, the series diverges.
  4. Step 1. This series is not a familiar series.
    Step 2. It is not an alternating series.
    Step 3. There is no obvious series with which to compare this series.
    Step 4. Since each term is a power of n,n, we can apply the root test. Since
    limn(3n+1)nn=limn3n+1=0,limn(3n+1)nn=limn3n+1=0,

    by the root test, we conclude that the series converges.
Checkpoint 5.23

For the series n=12n3n+n,n=12n3n+n, determine which convergence test is the best to use and explain why.

In Table 5.3, we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series n=1ann=1an to have nonnegative terms, if n=1ann=1an has negative terms, these tests can be applied to n=1|an|n=1|an| to test for absolute convergence.

Series or Test Conclusions Comments
Divergence Test
For any series n=1an,n=1an, evaluate limnan.limnan.
If limnan=0,limnan=0, the test is inconclusive. This test cannot prove convergence of a series.
If limnan0,limnan0, the series diverges.
Geometric Series
n=1arn1n=1arn1
If |r|<1,|r|<1, the series converges to
a/(1r).a/(1r).
Any geometric series can be reindexed to be written in the form a+ar+ar2+,a+ar+ar2+, where aa is the initial term and rr is the ratio.
If |r|1,|r|1, the series diverges.
p-Series
n=11npn=11np
If p>1,p>1, the series converges. For p=1,p=1, we have the harmonic series n=11/n.n=11/n.
If p1,p1, the series diverges.
Comparison Test
For n=1ann=1an with nonnegative terms, compare with a known series n=1bn.n=1bn.
If anbnanbn for all nNnN and n=1bnn=1bn converges, then n=1ann=1an converges. Typically used for a series similar to a geometric or pp-series. It can sometimes be difficult to find an appropriate series.
If anbnanbn for all nNnN and n=1bnn=1bn diverges, then n=1ann=1an diverges.
Limit Comparison Test
For n=1ann=1an with positive terms, compare with a series n=1bnn=1bn by evaluating
L=limnanbn.L=limnanbn.
If LL is a real number and L0,L0, then n=1ann=1an and n=1bnn=1bn both converge or both diverge. Typically used for a series similar to a geometric or pp-series. Often easier to apply than the comparison test.
If L=0L=0 and n=1bnn=1bn converges, then n=1ann=1an converges.
If L=L= and n=1bnn=1bn diverges, then n=1ann=1an diverges.
Integral Test
If there exists a positive, continuous, decreasing function ff such that an=f(n)an=f(n) for all nN,nN, evaluate Nf(x)dx.Nf(x)dx.
Nf(x)dxNf(x)dx and n=1ann=1an both converge or both diverge. Limited to those series for which the corresponding function ff can be easily integrated.
Alternating Series
n=1(−1)n+1bnorn=1(−1)nbnn=1(−1)n+1bnorn=1(−1)nbn
If bn+1bnbn+1bn for all n1n1 and bn0,bn0, then the series converges. Only applies to alternating series.
Ratio Test
For any series n=1ann=1an with nonzero terms, let
ρ=limn|an+1an|.ρ=limn|an+1an|.
If 0ρ<1,0ρ<1, the series converges absolutely. Often used for series involving factorials or exponentials.
If ρ>1orρ=,ρ>1orρ=, the series diverges.
If ρ=1,ρ=1, the test is inconclusive.
Root Test
For any series n=1an,n=1an, let
ρ=limn|an|n.ρ=limn|an|n.
If 0ρ<1,0ρ<1, the series converges absolutely. Often used for series where |an|=bnn.|an|=bnn.
If ρ>1orρ=,ρ>1orρ=, the series diverges.
If ρ=1,ρ=1, the test is inconclusive.
Table 5.3 Summary of Convergence Tests

Student Project

Series Converging to ππ and 1/π1/π

Dozens of series exist that converge to ππ or an algebraic expression containing π.π. Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of ππ in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.

  1. The series
    π=4n=1(−1)n+12n1=443+4547+49π=4n=1(−1)n+12n1=443+4547+49

    was discovered by Gregory and Leibniz in the late 1600s.1600s. This result follows from the Maclaurin series for f(x)=tan−1x.f(x)=tan−1x. We will discuss this series in the next chapter.
    1. Prove that this series converges.
    2. Evaluate the partial sums SnSn for n=10,20,50,100.n=10,20,50,100.
    3. Use the remainder estimate for alternating series to get a bound on the error Rn.Rn.
    4. What is the smallest value of NN that guarantees |RN|<0.01?|RN|<0.01? Evaluate SN.SN.
  2. The series
    π=6n=0(2n)!24n+1(n!)2(2n+1)=6(12+12·3(12)3+1·32·4·5·(12)5+1·3·52·4·6·7(12)7+)π=6n=0(2n)!24n+1(n!)2(2n+1)=6(12+12·3(12)3+1·32·4·5·(12)5+1·3·52·4·6·7(12)7+)

    has been attributed to Newton in the late 1600s.1600s. The proof of this result uses the Maclaurin series for f(x)=sin−1x.f(x)=sin−1x.
    1. Prove that the series converges.
    2. Evaluate the partial sums SnSn for n=5,10,20.n=5,10,20.
    3. Compare SnSn to ππ for n=5,10,20n=5,10,20 and discuss the number of correct decimal places.
  3. The series
    1π=89801n=0(4n)!(1103+26390n)(n!)43964n1π=89801n=0(4n)!(1103+26390n)(n!)43964n

    was discovered by Ramanujan in the early 1900s.1900s. William Gosper, Jr., used this series to calculate ππ to an accuracy of more than 1717 million digits in the mid-1980s.mid-1980s. At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for ππ and 1/π.1/π.
    1. Prove that this series converges.
    2. Evaluate the first term in this series. Compare this number with the value of ππ from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?
    3. Investigate the life of Srinivasa Ramanujan (18871920)(18871920) and write a brief summary. Ramanujan is one of the most fascinating stories in the history of mathematics. He was basically self-taught, with no formal training in mathematics, yet he contributed in highly original ways to many advanced areas of mathematics.

Section 5.6 Exercises

Use the ratio test to determine whether n=1ann=1an converges, where anan is given in the following problems. State if the ratio test is inconclusive.

317.

an=1/n!an=1/n!

318.

an=10n/n!an=10n/n!

319.

an=n2/2nan=n2/2n

320.

an=n10/2nan=n10/2n

321.

n=1(n!)3(3n!)n=1(n!)3(3n!)

322.

n=123n(n!)3(3n)!n=123n(n!)3(3n)!

323.

n=1(2n)!n2nn=1(2n)!n2n

324.

n=1(2n)!(2n)nn=1(2n)!(2n)n

325.

n=1n!(n/e)nn=1n!(n/e)n

326.

n=1(2n)!(n/e)2nn=1(2n)!(n/e)2n

327.

n=1(2nn!)2(2n)2nn=1(2nn!)2(2n)2n

Use the root test to determine whether n=1ann=1an converges, where anan is as follows.

328.

ak=(k12k+3)kak=(k12k+3)k

329.

ak=(2k21k2+3)kak=(2k21k2+3)k

330.

an=(lnn)2nnnan=(lnn)2nnn

331.

an=n/2nan=n/2n

332.

an=n/enan=n/en

333.

ak=keekak=keek

334.

ak=πkkπak=πkkπ

335.

an=(1e+1n)nan=(1e+1n)n

336.

ak=1(1+lnk)kak=1(1+lnk)k

337.

an=(ln(1+lnn))n(lnn)nan=(ln(1+lnn))n(lnn)n

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series k=1akk=1ak with given terms akak converges, or state if the test is inconclusive.

338.

ak=k!1·3·5(2k1)ak=k!1·3·5(2k1)

339.

ak=2·4·62k(2k)!ak=2·4·62k(2k)!

340.

ak=1·4·7(3k2)3kk!ak=1·4·7(3k2)3kk!

341.

an=(11n)n2an=(11n)n2

342.

ak=(1k+1+1k+2++12k)kak=(1k+1+1k+2++12k)k (Hint: Compare ak1/kak1/k to k2kdtt.)k2kdtt.)

343.

ak=(1k+1+1k+2++13k)kak=(1k+1+1k+2++13k)k

344.

an=(n1/n1)nan=(n1/n1)n

Use the ratio test to determine whether n=1ann=1an converges, or state if the ratio test is inconclusive.

345.

n=13n22n3n=13n22n3

346.

n=12n2nnn!n=12n2nnn!

Use the root and limit comparison tests to determine whether n=1ann=1an converges.

347.

an=1/xnnan=1/xnn where xn+1=12xn+1xn,xn+1=12xn+1xn, x1=1x1=1 (Hint: Find limit of {xn}.){xn}.)

In the following exercises, use an appropriate test to determine whether the series converges.

348.

n=1(n+1)n3+n2+n+1n=1(n+1)n3+n2+n+1

349.

n=1(−1)n+1(n+1)n3+3n2+3n+1n=1(−1)n+1(n+1)n3+3n2+3n+1

350.

n=1(n+1)2n3+(1.1)nn=1(n+1)2n3+(1.1)n

351.

n=1(n1)n(n+1)nn=1(n1)n(n+1)n

352.

an=(1+1n2)nan=(1+1n2)n (Hint: (1+1n2)n2e.)(1+1n2)n2e.)

353.

ak=1/2sin2kak=1/2sin2k

354.

ak=2sin(1/k)ak=2sin(1/k)

355.

an=1/(n+2n)an=1/(n+2n) where (nk)=n!k!(nk)!(nk)=n!k!(nk)!

356.

ak=1/(2kk)ak=1/(2kk)

357.

ak=2k/(3kk)ak=2k/(3kk)

358.

ak=(kk+lnk)kak=(kk+lnk)k (Hint: ak=(1+lnkk)(k/lnk)lnkelnk.)ak=(1+lnkk)(k/lnk)lnkelnk.)

359.

ak=(kk+lnk)2kak=(kk+lnk)2k (Hint: ak=(1+lnkk)(k/lnk)lnk2.)ak=(1+lnkk)(k/lnk)lnk2.)

The following series converge by the ratio test. Use summation by parts, k=1nak(bk+1bk)=[an+1bn+1a1b1]k=1nbk+1(ak+1ak),k=1nak(bk+1bk)=[an+1bn+1a1b1]k=1nbk+1(ak+1ak), to find the sum of the given series.

360.

k=1k2kk=1k2k (Hint: Take ak=kak=k and bk=21k.)bk=21k.)

361.

k=1kck,k=1kck, where c>1c>1 (Hint: Take ak=kak=k and bk=c1k/(c1).)bk=c1k/(c1).)

362.

n=1n22nn=1n22n

363.

n=1(n+1)22nn=1(n+1)22n

The kth term of each of the following series has a factor xk.xk. Find the range of xx for which the ratio test implies that the series converges.

364.

k=1xkk2k=1xkk2

365.

k=1x2kk2k=1x2kk2

366.

k=1x2k3kk=1x2k3k

367.

k=1xkk!k=1xkk!

368.

Does there exist a number pp such that n=12nnpn=12nnp converges?

369.

Let 0<r<1.0<r<1. For which real numbers pp does n=1nprnn=1nprn converge?

370.

Suppose that limn|an+1an|=p.limn|an+1an|=p. For which values of pp must n=12nann=12nan converge?

371.

Suppose that limn|an+1an|=p.limn|an+1an|=p. For which values of r>0r>0 is n=1rnann=1rnan guaranteed to converge?

372.

Suppose that |an+1an|(n+1)p|an+1an|(n+1)p for all n=1,2,…n=1,2,… where pp is a fixed real number. For which values of pp is n=1n!ann=1n!an guaranteed to converge?

373.

For which values of r>0,r>0, if any, does n=1rnn=1rn converge? (Hint: n=1an=k=1n=k2(k+1)21an.)n=1an=k=1n=k2(k+1)21an.)

374.

Suppose that |an+2an|r<1|an+2an|r<1 for all n.n. Can you conclude that n=1ann=1an converges?

375.

Let an=2[n/2]an=2[n/2] where [x][x] is the greatest integer less than or equal to x.x. Determine whether n=1ann=1an converges and justify your answer.

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if limna2nan<1/2,limna2nan<1/2, then anan converges, while if limna2n+1an>1/2,limna2n+1an>1/2, then anan diverges.

376.

Let an=1436582n12n+2=1·3·5(2n1)2n(n+1)!.an=1436582n12n+2=1·3·5(2n1)2n(n+1)!. Explain why the ratio test cannot determine convergence of n=1an.n=1an. Use the fact that 11/(4k)11/(4k) is increasing kk to estimate limna2nan.limna2nan.

377.

Let an=11+x22+xnn+x1n=(n1)!(1+x)(2+x)(n+x).an=11+x22+xnn+x1n=(n1)!(1+x)(2+x)(n+x). Show that a2n/anex/2/2.a2n/anex/2/2. For which x>0x>0 does the generalized ratio test imply convergence of n=1an?n=1an? (Hint: Write 2a2n/an2a2n/an as a product of nn factors each smaller than 1/(1+x/(2n)).)1/(1+x/(2n)).)

378.

Let an=nlnn(lnn)n.an=nlnn(lnn)n. Show that a2nan0a2nan0 as n.n.

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