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Calculus Volume 2

5.4 Comparison Tests

Calculus Volume 25.4 Comparison 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

  • 5.4.1. Use the comparison test to test a series for convergence.
  • 5.4.2. Use the limit comparison test to determine convergence of a series.

We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. In this section, we show how to use comparison tests to determine the convergence or divergence of a series by comparing it to a series whose convergence or divergence is known. Typically these tests are used to determine convergence of series that are similar to geometric series or p-series.

Comparison Test

In the preceding two sections, we discussed two large classes of series: geometric series and p-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test.

For example, consider the series

n=11n2+1.n=11n2+1.

This series looks similar to the convergent series

n=11n2.n=11n2.

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since

0<1n2+1<1n20<1n2+1<1n2

for all positive integers n,n, the kthkth partial sum SkSk of n=11n2+1n=11n2+1 satisfies

Sk=n=1k1n2+1<n=1k1n2<n=11n2.Sk=n=1k1n2+1<n=1k1n2<n=11n2.

(See Figure 5.16(a) and Table 5.1.) Since the series on the right converges, the sequence {Sk}{Sk} is bounded above. We conclude that {Sk}{Sk} is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, {Sk}{Sk} converges, and thus

n=11n2+1n=11n2+1

converges.

Similarly, consider the series

n=11n1/2.n=11n1/2.

This series looks similar to the divergent series

n=11n.n=11n.

The sequence of partial sums for each series is monotone increasing and

1n1/2>1n>01n1/2>1n>0

for every positive integer n.n. Therefore, the kthkth partial sum SkSk of n=11n1/2n=11n1/2 satisfies

Sk=n=1k1n1/2>n=1k1n.Sk=n=1k1n1/2>n=1k1n.

(See Figure 5.16(b) and Table 5.2.) Since the series n=11/nn=11/n diverges to infinity, the sequence of partial sums n=1k1/nn=1k1/n is unbounded. Consequently, {Sk}{Sk} is an unbounded sequence, and therefore diverges. We conclude that

n=11n1/2n=11n1/2

diverges.

This shows two graphs side by side. The first shows plotted points for the partial sums for the sum of 1/n^2 and the sum 1/(n^2 + 1). Each of the partial sums for the latter is less than the corresponding partial sum for the former. The second shows plotted points for the partial sums for the sum of 1/(n - 0.5) and the sum 1/n. Each of the partial sums for the latter is less than the corresponding partial sum for the former.
Figure 5.16 (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging pseries.pseries. (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.
kk 11 22 33 44 55 66 77 88
n=1k1n2+1n=1k1n2+1 0.50.5 0.70.7 0.80.8 0.85880.8588 0.89730.8973 0.92430.9243 0.94430.9443 0.95970.9597
n=1k1n2n=1k1n2 11 1.251.25 1.36111.3611 1.42361.4236 1.46361.4636 1.49141.4914 1.51181.5118 1.52741.5274
Table 5.1 Comparing a series with a p-series (p = 2)
kk 11 22 33 44 55 66 77 88
n=1k1n1/2n=1k1n1/2 22 2.66672.6667 3.06673.0667 3.35243.3524 3.57463.5746 3.75643.7564 3.91033.9103 4.04364.0436
n=1k1nn=1k1n 11 1.51.5 1.83331.8333 2.09332.0933 2.28332.2833 2.452.45 2.59292.5929 2.71792.7179
Table 5.2 Comparing a series with the harmonic series
Theorem 5.11

Comparison Test

  1. Suppose there exists an integer NN such that 0anbn0anbn for all nN.nN. If n=1bnn=1bn converges, then n=1ann=1an converges.
  2. Suppose there exists an integer NN such that anbn0anbn0 for all nN.nN. If n=1bnn=1bn diverges, then n=1ann=1an diverges.

Proof

We prove part i. The proof of part ii. is the contrapositive of part i. Let {Sk}{Sk} be the sequence of partial sums associated with n=1an,n=1an, and let L=n=1bn.L=n=1bn. Since the terms an0,an0,

Sk=a1+a2++aka1+a2++ak+ak+1=Sk+1.Sk=a1+a2++aka1+a2++ak+ak+1=Sk+1.

Therefore, the sequence of partial sums is increasing. Further, since anbnanbn for all nN,nN, then

n=Nkann=Nkbnn=1bn=L.n=Nkann=Nkbnn=1bn=L.

Therefore, for all k1,k1,

Sk=(a1+a2++aN1)+n=Nkan(a1+a2++aN1)+L.Sk=(a1+a2++aN1)+n=Nkan(a1+a2++aN1)+L.

Since a1+a2++aN1a1+a2++aN1 is a finite number, we conclude that the sequence {Sk}{Sk} is bounded above. Therefore, {Sk}{Sk} is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that {Sk}{Sk} converges, and therefore the series n=1ann=1an converges.

To use the comparison test to determine the convergence or divergence of a series n=1an,n=1an, it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and p-series, these series are often used. If there exists an integer NN such that for all nN,nN, each term anan is less than each corresponding term of a known convergent series, then n=1ann=1an converges. Similarly, if there exists an integer NN such that for all nN,nN, each term anan is greater than each corresponding term of a known divergent series, then n=1ann=1an diverges.

Example 5.17

Using the Comparison Test

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

  1. n=11n3+3n+1n=11n3+3n+1
  2. n=112n+1n=112n+1
  3. n=21ln(n)n=21ln(n)

Solution

  1. Compare to n=11n3n=11n3 Since n=11n3n=11n3 is a p-series with p=3,p=3, it converges. Further,
    1n3+3n+1<1n31n3+3n+1<1n3

    for every positive integer n.n. Therefore, we can conclude that n=11n3+3n+1n=11n3+3n+1 converges.
  2. Compare to n=1(12)n.n=1(12)n. Since n=1(12)nn=1(12)n is a geometric series with r=1/2r=1/2 and |1/2|<1,|1/2|<1, it converges. Also,
    12n+1<12n12n+1<12n

    for every positive integer n.n. Therefore, we see that n=112n+1n=112n+1 converges.
  3. Compare to n=21n.n=21n. Since
    1ln(n)>1n1ln(n)>1n

    for every integer n2n2 and n=21/nn=21/n diverges, we have that n=21ln(n)n=21ln(n) diverges.

Checkpoint 5.16

Use the comparison test to determine if the series n=1nn3+n+1n=1nn3+n+1 converges or diverges.

Limit Comparison Test

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

n=21n21.n=21n21.

It is natural to compare this series with the convergent series

n=21n2.n=21n2.

However, this series does not satisfy the hypothesis necessary to use the comparison test because

1n21>1n21n21>1n2

for all integers n2.n2. Although we could look for a different series with which to compare n=21/(n21),n=21/(n21), instead we show how we can use the limit comparison test to compare

n=21n21andn=21n2.n=21n21andn=21n2.

Let us examine the idea behind the limit comparison test. Consider two series n=1ann=1an and n=1bn.n=1bn. with positive terms anandbnanandbn and evaluate

limnanbn.limnanbn.

If

limnanbn=L0,limnanbn=L0,

then, for nn sufficiently large, anLbn.anLbn. Therefore, either both series converge or both series diverge. For the series n=21/(n21)n=21/(n21) and n=21/n2,n=21/n2, we see that

limn1/(n21)1/n2=limnn2n21=1.limn1/(n21)1/n2=limnn2n21=1.

Since n=21/n2n=21/n2 converges, we conclude that

n=21n21n=21n21

converges.

The limit comparison test can be used in two other cases. Suppose

limnanbn=0.limnanbn=0.

In this case, {an/bn}{an/bn} is a bounded sequence. As a result, there exists a constant MM such that anMbn.anMbn. Therefore, if n=1bnn=1bn converges, then n=1ann=1an converges. On the other hand, suppose

limnanbn=.limnanbn=.

In this case, {an/bn}{an/bn} is an unbounded sequence. Therefore, for every constant MM there exists an integer NN such that anMbnanMbn for all nN.nN. Therefore, if n=1bnn=1bn diverges, then n=1ann=1an diverges as well.

Theorem 5.12

Limit Comparison Test

Let an,bn0an,bn0 for all n1.n1.

  1. If limnan/bn=L0,limnan/bn=L0, then n=1ann=1an and n=1bnn=1bn both converge or both diverge.
  2. If limnan/bn=0limnan/bn=0 and n=1bnn=1bn converges, then n=1ann=1an converges.
  3. If limnan/bn=limnan/bn= and n=1bnn=1bn diverges, then n=1ann=1an diverges.

Note that if an/bn0an/bn0 and n=1bnn=1bn diverges, the limit comparison test gives no information. Similarly, if an/bnan/bn and n=1bnn=1bn converges, the test also provides no information. For example, consider the two series n=11/nn=11/n and n=11/n2.n=11/n2. These series are both p-series with p=1/2p=1/2 and p=2,p=2, respectively. Since p=1/2>1,p=1/2>1, the series n=11/nn=11/n diverges. On the other hand, since p=2<1,p=2<1, the series n=11/n2n=11/n2 converges. However, suppose we attempted to apply the limit comparison test, using the convergent pseriespseries n=11/n3n=11/n3 as our comparison series. First, we see that

1/n1/n3=n3n=n5/2asn.1/n1/n3=n3n=n5/2asn.

Similarly, we see that

1/n21/n3=nasn.1/n21/n3=nasn.

Therefore, if an/bnan/bn when n=1bnn=1bn converges, we do not gain any information on the convergence or divergence of n=1an.n=1an.

Example 5.18

Using the Limit Comparison Test

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

  1. n=11n+1n=11n+1
  2. n=12n+13nn=12n+13n
  3. n=1ln(n)n2n=1ln(n)n2

Solution

  1. Compare this series to n=11n.n=11n. Calculate
    limn1/(n+1)1/n=limnnn+1=limn1/n1+1/n=1.limn1/(n+1)1/n=limnnn+1=limn1/n1+1/n=1.
    By the limit comparison test, since n=11nn=11n diverges, then n=11n+1n=11n+1 diverges.
  2. Compare this series to n=1(23)n.n=1(23)n. We see that
    limn(2n+1)/3n2n/3n=limn2n+13n·3n2n=limn2n+12n=limn[1+(12)n]=1.limn(2n+1)/3n2n/3n=limn2n+13n·3n2n=limn2n+12n=limn[1+(12)n]=1.

    Therefore,
    limn(2n+1)/3n2n/3n=1.limn(2n+1)/3n2n/3n=1.

    Since n=1(23)nn=1(23)n converges, we conclude that n=12n+13nn=12n+13n converges.
  3. Since lnn<n,lnn<n, compare with n=11n.n=11n. We see that
    limnlnn/n21/n=limnlnnn2·n1=limnlnnn.limnlnn/n21/n=limnlnnn2·n1=limnlnnn.

    In order to evaluate limnlnn/n,limnlnn/n, evaluate the limit as xx of the real-valued function ln(x)/x.ln(x)/x. These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain
    limxlnxx=limx1x=0.limxlnxx=limx1x=0.

    Therefore, limnlnn/n=0,limnlnn/n=0, and, consequently,
    limnlnn/n21/n=0.limnlnn/n21/n=0.

    Since the limit is 00 but n=11nn=11n diverges, the limit comparison test does not provide any information.
    Compare with n=11n2n=11n2 instead. In this case,
    limnlnn/n21/n2=limnlnnn2·n21=limnlnn=.limnlnn/n21/n2=limnlnnn2·n21=limnlnn=.

    Since the limit is but n=11n2n=11n2 converges, the test still does not provide any information.
    So now we try a series between the two we already tried. Choosing the series n=11n3/2,n=11n3/2, we see that
    limnlnn/n21/n3/2=limnlnnn2·n3/21=limnlnnn.limnlnn/n21/n3/2=limnlnnn2·n3/21=limnlnnn.

    As above, in order to evaluate limnlnn/n,limnlnn/n, evaluate the limit as xx of the real-valued function lnx/x.lnx/x. Using L’Hôpital’s rule,
    limxlnxx=limx2xx=limx2x=0.limxlnxx=limx2xx=limx2x=0.

    Since the limit is 00 and n=11n3/2n=11n3/2 converges, we can conclude that n=1lnnn2n=1lnnn2 converges.
Checkpoint 5.17

Use the limit comparison test to determine whether the series n=15n3n+2n=15n3n+2 converges or diverges.

Section 5.4 Exercises

Use the comparison test to determine whether the following series converge.

194.

n=1ann=1an where an=2n(n+1)an=2n(n+1)

195.

n=1ann=1an where an=1n(n+1/2)an=1n(n+1/2)

196.

n=112(n+1)n=112(n+1)

197.

n=112n1n=112n1

198.

n=21(nlnn)2n=21(nlnn)2

199.

n=1n!(n+2)!n=1n!(n+2)!

200.

n=11n!n=11n!

201.

n=1sin(1/n)nn=1sin(1/n)n

202.

n=1sin2nn2n=1sin2nn2

203.

n=1sin(1/n)nn=1sin(1/n)n

204.

n=1n1.21n2.3+1n=1n1.21n2.3+1

205.

n=1n+1nnn=1n+1nn

206.

n=1n4n4+n23n=1n4n4+n23

Use the limit comparison test to determine whether each of the following series converges or diverges.

207.

n=1(lnnn)2n=1(lnnn)2

208.

n=1(lnnn0.6)2n=1(lnnn0.6)2

209.

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

210.

n=1ln(1+1n2)n=1ln(1+1n2)

211.

n=114n3nn=114n3n

212.

n=11n2nsinnn=11n2nsinn

213.

n=11e(1.1)n3nn=11e(1.1)n3n

214.

n=11e(1.01)n3nn=11e(1.01)n3n

215.

n=11n1+1/nn=11n1+1/n

216.

n=1121+1/nn1+1/nn=1121+1/nn1+1/n

217.

n=1(1nsin(1n))n=1(1nsin(1n))

218.

n=1(1cos(1n))n=1(1cos(1n))

219.

n=11n(tan−1nπ2)n=11n(tan−1nπ2)

220.

n=1(11n)n.nn=1(11n)n.n (Hint:(11n)n1/e.)(11n)n1/e.)

221.

n=1(1e−1/n)n=1(1e−1/n) (Hint:1/e(11/n)n,1/e(11/n)n, so 1e−1/n1/n.)1e−1/n1/n.)

222.

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

223.

Does n=1((lnn)n)pn=1((lnn)n)p converge if pp is large enough? If so, for which p?p?

224.

For which pp does the series n=12pn/3nn=12pn/3n converge?

225.

For which p>0p>0 does the series n=1np2nn=1np2n converge?

226.

For which r>0r>0 does the series n=1rn22nn=1rn22n converge?

227.

For which r>0r>0 does the series n=12nrn2n=12nrn2 converge?

228.

Find all values of pp and qq such that n=1np(n!)qn=1np(n!)q converges.

229.

Does n=1sin2(nr/2)nn=1sin2(nr/2)n converge or diverge? Explain.

230.

Explain why, for each n,n, at least one of {|sinn|,|sin(n+1)|,...,|sinn+6|}{|sinn|,|sin(n+1)|,...,|sinn+6|} is larger than 1/2.1/2. Use this relation to test convergence of n=1|sinn|n.n=1|sinn|n.

231.

Suppose that an0an0 and bn0bn0 and that n=1a2nn=1a2n and n=1b2nn=1b2n converge. Prove that n=1anbnn=1anbn converges and n=1anbn12(n=1an2+n=1bn2).n=1anbn12(n=1an2+n=1bn2).

232.

Does n=12lnlnnn=12lnlnn converge? (Hint: Write 2lnlnn2lnlnn as a power of lnn.)lnn.)

233.

Does n=1(lnn)lnnn=1(lnn)lnn converge? (Hint: Use t=eln(t)t=eln(t) to compare to a pseries.)pseries.)

234.

Does n=2(lnn)lnlnnn=2(lnn)lnlnn converge? (Hint: Compare anan to 1/n.)1/n.)

235.

Show that if an0an0 and n=1ann=1an converges, then n=1a2nn=1a2n converges. If n=1a2nn=1a2n converges, does n=1ann=1an necessarily converge?

236.

Suppose that an>0an>0 for all nn and that n=1ann=1an converges. Suppose that bnbn is an arbitrary sequence of zeros and ones. Does n=1anbnn=1anbn necessarily converge?

237.

Suppose that an>0an>0 for all nn and that n=1ann=1an diverges. Suppose that bnbn is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does n=1anbnn=1anbn necessarily diverge?

238.

Complete the details of the following argument: If n=11nn=11n converges to a finite sum s,s, then 12s=12+14+16+12s=12+14+16+ and s12s=1+13+15+.s12s=1+13+15+. Why does this lead to a contradiction?

239.

Show that if an0an0 and n=1a2nn=1a2n converges, then n=1sin2(an)n=1sin2(an) converges.

240.

Suppose that an/bn0an/bn0 in the comparison test, where an0an0 and bn0.bn0. Prove that if bnbn converges, then anan converges.

241.

Let bnbn be an infinite sequence of zeros and ones. What is the largest possible value of x=n=1bn/2n?x=n=1bn/2n?

242.

Let dndn be an infinite sequence of digits, meaning dndn takes values in {0,1,…,9}.{0,1,…,9}. What is the largest possible value of x=n=1dn/10nx=n=1dn/10n that converges?

243.

Explain why, if x>1/2,x>1/2, then xx cannot be written x=n=2bn2n(bn=0or1,b1=0).x=n=2bn2n(bn=0or1,b1=0).

244.

[T] Evelyn has a perfect balancing scale, an unlimited number of 1-kg1-kg weights, and one each of 1/2-kg,1/4-kg,1/8-kg,1/2-kg,1/4-kg,1/8-kg, and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?

245.

[T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of 1-kg1-kg weights, and nine each of 0.1-kg,0.1-kg, 0.01-kg,0.001-kg,0.01-kg,0.001-kg, and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?

246.

The series n=112nn=112n is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which nn is odd. Let m>1m>1 be fixed. Show, more generally, that deleting all terms 1/n1/n where n=mkn=mk for some integer kk also results in a divergent series.

247.

In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from n=11nn=11n by removing any term 1/n1/n if a given digit, say 9,9, appears in the decimal expansion of n.n. Argue that this depleted harmonic series converges by answering the following questions.

  1. How many whole numbers nn have dd digits?
  2. How many d-digitd-digit whole numbers h(d).h(d). do not contain 99 as one or more of their digits?
  3. What is the smallest d-digitd-digit number m(d)?m(d)?
  4. Explain why the deleted harmonic series is bounded by d=1h(d)m(d).d=1h(d)m(d).
  5. Show that d=1h(d)m(d)d=1h(d)m(d) converges.
248.

Suppose that a sequence of numbers an>0an>0 has the property that a1=1a1=1 and an+1=1n+1Sn,an+1=1n+1Sn, where Sn=a1++an.Sn=a1++an. Can you determine whether n=1ann=1an converges? (Hint: SnSn is monotone.)

249.

Suppose that a sequence of numbers an>0an>0 has the property that a1=1a1=1 and an+1=1(n+1)2Sn,an+1=1(n+1)2Sn, where Sn=a1++an.Sn=a1++an. Can you determine whether n=1ann=1an converges? (Hint: S2=a2+a1=a2+S1=a2+1=1+1/4=(1+1/4)S1,S2=a2+a1=a2+S1=a2+1=1+1/4=(1+1/4)S1, S3=132S2+S2=(1+1/9)S2=(1+1/9)(1+1/4)S1,S3=132S2+S2=(1+1/9)S2=(1+1/9)(1+1/4)S1, etc. Look at ln(Sn),ln(Sn), and use ln(1+t)t,ln(1+t)t, t>0.)t>0.)

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