Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 100 pdf

EXERCISE 30.9 Use the Limit Comparison Test to determine whether the following series are convergent or divergent.. Its proof relies on our knowledge of geometric series, series for whic

L i m i t C o m p a r i s o n T e s t

Suppose∞k=1akand∞k=1bkare series whose terms are positive for all k≥ N for some constant N

i If limk→∞ak

b k = L for 0 < L < ∞, then both series converge or both series diverge

ii If limk→∞ ak

b k = 0 and∞k =1bkconverges, then∞k=1anconverges as well iii If limk →∞ abkk = ∞ and∞k =1bkdiverges, then∞k=1akdiverges as well

Proof

i Choose positive constants α and β with 0 < α < L < β Then for k sufficiently large

α <ak

bk

< β and αbk< ak< βbk

Suppose∞k=1bkdiverges Then∞k=1αbkdiverges as well.∞k=1akdiverges by the Comparison Test

Suppose∞k=1bk converges Then∞k=1βbk converges as well.∞k=1ak con-verges by the Comparison Test

The proofs of parts (ii) and (iii) are left as an exercise for the reader

EXAMPLE 30.29 Determine whether∞k=2 2k2+3k

k 4 −k+1converges or diverges

SOLUTION For k large the terms of the series “look like”2k2

k 4 =k22.∞k=2 2

k 2 = 2∞k =2 k12 converges (It’s a p-series with p > 1.) We use the Limit Comparison Test to show∞k=2 2k2+3k

k 4 −k+1 converges as well

lim

k →∞

ak

bk = lim

k →∞

2k2+3k

k 4 −k+1 2

k 2

= lim

k →∞

2k2+ 3k

k4− k + 1·

k2

2 = lim

k →∞

2k4+ 3k3 2k4− 2k + 2= 1

Therefore∞k=2 2k2+3k

k 4 −k+1converges

EXERCISE 30.9 Use the Limit Comparison Test to determine whether the following series are convergent

or divergent

(a)∞k=1 10k4

−2 (b)∞k=2

√

k 3 +1 k(k 2 −1)

Answers

(a) Convergent: Compare with∞k=1 1044 or∞k=1 101k, convergent geometric series (b) Convergent: “Looks like”∞k=2 kk3/23 =∞k =2 k 3/21 , a convergent p-series

In the next example we tackle a more challenging problem

EXAMPLE 30.30 Determine whether∞k=2 ln kk2 converges or diverges.

SOLUTION There are several options available for determining convergence

One option is to use the Integral Test To do this, verify that f (x)=ln xx2 is positive, decreasing, and continuous on [2,∞) and then compute2∞ ln xx2 dx This improper integral requires integration by parts As an exercise, verify that2∞ ln xx2 dx=ln 22+1 and conclude that∞k=2 ln kk2 converges

An alternative is to use the Limit Comparison Test The challenge is to make a good choice for comparison

1 < ln k < k for x≥ 3 so 1

k2<ln k

k2 < k

k2=1k for x≥ 3 The terms are greater than those of a known convergent series and less than those of a known divergent series! This doesn’t clarify the convergence issues If we try to use limit comparison by comparing with∞k=3 1

k 2 we get

lim

k →∞

ln k

k 2

1

k 2

= lim

k →∞ln k= ∞, which is inconclusive

On the other hand, using limit comparison by comparing with∞k=3 1k we get

lim

k →∞

ln k

k 2

1 k

= lim

k →∞

ln k

k2 · k = lim

k →∞

ln k

k = 0, which is inconclusive

Let’s compare with∞k=3 1

k 1.5 This is a convergent p-series

lim

k →∞

ln k

k 2

1

k1.5

= lim

k →∞

ln k

k2 · k1.5= lim

k →∞

ln k

k0.5

We can compute this limit by computing limx→∞ln x

x

1 using L’Hˆopital’s Rule.17

lim

x →∞

ln x

x1 = limx→∞

1 x 1

2 √1 x

= limx→∞1x · 2

√ x

1 = limx→∞√2x = 0

Therefore,

lim

k →∞

ln k

k0.5 = 0 and

∞



k =2

ln k

k2 converges

The Ratio and the Root Test

In order for a series to converge its terms must be going toward zero sufficiently rapidly Therefore, it is reasonable to think that valuable information can be obtained by looking at the ratio of successive terms, ak+1/ak, for k very large The following convergence test, the

17 See Appendix F for a discussion of L’Hˆopital’s Rule.

Ratio Test, does this Its proof relies on our knowledge of geometric series, series for which the ratio of successive terms is fixed

T h e R a t i o T e s t

Let∞k=1akbe a series such that ak>0 for k sufficiently large Suppose

lim

k →∞

ak+1

ak = L, where L is a real number or∞

i If L < 1, then∞k=1akconverges

ii If L > 1, then∞k=1akdiverges

iii If L= 1, the test is inconclusive

Proof

i Suppose L < 1 We will compare our series with a convergent geometric series Since

0≤ L < 1, we can choose a constant r such that 0 ≤ L < r < 1.∞k =1rkconverges

lim

k →∞

ak+1

ak = L < r, so for k sufficiently large,aka+1

k

< r

In other words, there exists an integer N such that

ak+1

ak < rfor all k≥ N

Therefore, for k≥ N,

ak+1< akr, and

ak+2< ak+1r < (akr)· r = akr2

ak +3< ak +2r < (akr2)· r = akr3

The series∞k=1aN+k= aN +1+ aN +2+ aN +3+ · · · converges by comparison to

∞



k =1

aNrk= aNr+ aNr2+ aNr3+ · · ·

This latter series is a geometric series with|r| < 1 Each of its terms is larger than the corresponding term in∞k=1aN+k

aN+i< aNri for i= 1, 2, 3, Because∞k=N+1akconverges we conclude that∞k=1akconverges

ii Suppose limk→∞ak +1

a k = L, L > 1, or limk →∞aka+1k = ∞ Then there exists a number

Nsuch thatak+1

ak >1 for all k≥ N; that is, 0 < ak< ak+1for all k≥ N But then

lim

k →∞ak= 0;

the series∞k=1akdiverges by the nth Term Test for Divergence

iii To show that limk →∞ aka+1k = 1 is inconclusive, we need only produce both convergent and divergent series for which the limit of the ratio of successive terms is 1

∞



k =1

1

k diverges lim

k →∞

ak+1

ak = lim

k →∞

1

k +1 1 k

= lim

k →∞

k

k+ 1= 1.

∞



k =1

1

k2 converges lim

k →∞

ak+1

ak = lim

k →∞

1 (k +1) 2

1

k 2

= lim

k →∞

k2 (k+ 1)2= 1

As demonstrated, the Ratio Test is not useful for p-series or series whose terms look like polynomials It is a great test to use for series whose terms involve factorials and/or exponentials

EXAMPLE 30.31 Test the following series for convergence

(a)∞k=1 10k!k (b) ∞k=1 kk!k

SOLUTION Both series have positive terms We use the Ratio Test

(a)

ak+1

ak =

10k+1 (k +1)!

10 k

k!

= 10

k +1 (k+ 1)!·

k!

10k = 10

k· 10 (k+ 1)k!·

k!

10k =k10

+ 1

Compute: limk→∞ak +1

a k = limk →∞ k10+1 = 0 < 1 Therefore,∞k =1 10

k

k! converges (b)

ak+1

ak =

(k +1)k+1 (k +1)!

k k

k!

=(k+ 1)

k +1 (k+ 1)! ·

k!

kk =(k+ 1)

k(k+ 1) · k!

(k+ 1)k!kk = k + 1

k

k

=



1+1k

k

Compute: limk→∞ 1+1kk= e > 1 Therefore,∞k =1 k

k

k! diverges

REMARKIn Example 30.31 part (b) we showed that for the series∞k=1 kk!k, limk →∞ ak+1ak = e It follows that for the series ∞k =1 k!k k, limk →∞ ak+1ak =1e <1 Thus

∞

k =1 k!k k converges and its terms must tend toward zero We can conclude that kkgrows much more rapidly than does k! From part (a) of Example 30.31 we can deduce that k! in turn grows much more rapidly than does bkfor any constant b The Ratio Test gives us an alternative way of proving the fact that limk→∞bk!k = 0 for every real number b

The following test gives us a convenient way of testing for convergence when the terms

of a series are raised to the kth power

T h e R o o t T e s t

Let∞k=1akbe a series such that ak>0 for k sufficiently large Suppose that

lim

k →∞

k

√a

k= L, where L is a real number or ∞

i If L < 1, then∞k=1akconverges

ii If L > 1, then∞k=1akdiverges

iii If L= 1, the test is inconclusive

The proof of the Root Test runs along the same lines as that of the Ratio Test, so it is omitted

EXERCISE 30.10 Use the Root Test to show that∞k=1 2n3n33+3n+1

n converges

Extending Our Results

In this section we’ve presented several tests that can be used to determine the convergence

or divergence of a series: The nth Term Test for Divergence, the Comparison and Limit Comparison Tests, the Integral Test, the Ratio Test, and the Root Test Of these, all but the nth Term Test require that eventually all the terms of the series being tested must be positive If, instead, all the terms of the series are negative, or eventually the terms are all negative, then we can factor out a (−1) and the tests can be applied Otherwise, we can look at∞k=1|ak| If ∞k =1|ak| converges, not only does∞k =1ak converge, but it

converges absolutely If∞k=1|ak| diverges, then∞k =1akwill either diverge or converge conditionally.18If the series is alternating we can try to apply the Alternating Series Test The Ratio Test and Root Test can be easily generalized to apply to arbitrary series

Generalized Ratio Test

Let∞k=1akbe a series with ak= 0 for k sufficiently large Suppose

lim

k →∞

|ak +1|

|ak| = L, where L is a real number or ∞.

i If L < 1, then∞k=1anconverges absolutely (and therefore converges)

ii If L > 1, then∞k=1andiverges

iii If L= 1, then the test is inconclusive

The proof is left as a problem at the end of the section The Root Test can be similarly generalized

The generalized Ratio Test can be applied to a power series in order to find its radius

of convergence and to show that in the interior of its interval of convergence (not including endpoints) the series converges absolutely

Let ∞k=0ak(x− b)k be a power series centered at x= b Let wk= ak(x− b)k Compute

18 Conditional convergence, defined in Section 30.4, meansa k converges but  |a k | diverges.

k →∞

|wk +1|

|wk| = limk →∞

|ak +1(x− b)k +1|

|ak(x− b)k| = limk →∞

|ak +1|

|ak| |x − b| =

 lim

k →∞

|ak +1|

|ak|



|x − b|

If limk→∞|ak +1 |

|a k | = 0 then by the generalized Ratio Test the series converges absolutely for all x

If limk→∞|ak +1 |

|a k | = Q, where Q is finite and nonzero, then, the generalized Ratio Test says the series converges absolutely for all x such that Q|x − b| < 1, i.e., |x − b| <Q1 The series diverges for Q|x − b| > 1, i.e., |x − b| >Q1 Therefore, the radius of convergence is Q1

If limk→∞|ak+1 |

|a k | = ∞, then, by the generalized Ratio Test, the series converges only for x= b

EXAMPLE 30.32 The Bessel function J0(x)=∞k =0(−1)k(k!)x2k2 2 2k is a function defined as a power series.

Find the set of all x for which the series converges absolutely

SOLUTION Apply the generalized Ratio Test

|wk +1|

|wk| =







(−1)k +1x2(k +1) (k+ 1)!(k + 1)!22(k +1)





·







(k!)(k!)22k (−1)kx2k





2k +2|22k

|x2k|(k + 1)(k + 1)22k+2

2| (k+ 1)222 =x

2

4

1 (k+ 1)2

lim

k →∞

x2

(k+ 1)2= 0 < 1

The Bessel function J0(x)converges absolutely for all x

EXAMPLE 30.33 Find the interval of convergence of the power series∞k=2 (xln k+3)k

SOLUTION Begin by applying the generalized Ratio Test Compute|wk+1 |

|w k |



 (x +3) k +1

ln(k +1)







 (x +3) k

ln k





=|x + 3|| ln k|

| ln(k + 1)|

lim

k →∞

|wk +1|

|wk| = limk →∞|x + 3| ln k

ln(k+ 1)= |x + 3|

The series converges for|x + 3| < 1 and diverges for |x + 3| > 1

We must check the endpoints of the interval of convergence, x= −4 and x = −2, independently

When x= 4 we have∞k =2

( −4+3) k

ln k =∞k =2

( −1) k

ln k This converges by the Alternating Series Test, because the terms are decreasing in magnitude and tending toward zero When x= −2 we have∞k =2

( −2+3) k

ln k =∞k =2 ln k1 This series diverges by comparison with the harmonic series (See Example 30.28 for details.)

The interval of convergence is centered at x= −3, the center of the series The interval

of convergence is [−4, −2) The series converges absolutely on (−4, −2) and converges conditionally at x= −4

Summary of Convergence19 Criteria for  ak

Geometric Series:ark  converges for |r| < 1

diverges for|r| ≥ 1 p-series:k1p

 converges for p > 1 diverges for p≤ 1

nth Term Test for Divergence If limk→∞ak= 0, then the series diverges

Comparison Test

terms all positive

If ak≤ bkfor all k andbkconverges, so doesak

If bk≤ akfor all k andbkdiverges, so doesak

Limit Comparison Test

terms all positive

Suppose limk→∞ak

b k = L, L possibly infinite

If 0 < L <∞,akandbkeither both converge or both diverge

If L= 0, andbkconverges, so doesak

If L= ∞, andbkdiverges, so doesak

Integral Test

terms positive

If f (k)= akand f (x) is positive, continuous, and decreasing on [1,∞), then ∞

1 f (x) dxandf (k)either both converge or both diverge

Ratio Test

particularly useful with factorials

and exponents

Compute limk→∞ |ak+1 |

|a k | = L, L possibly infinite

If L < 1,akconverges absolutely

If L > 1,akdiverges

If L= 1, the test is inconclusive

Root Test

particularly useful when aklooks

like (−)k

Compute limk →∞√k

|ak| = L, L possibly infinite

If L < 1,akconverges absolutely

If L > 1,akdiverges

If L= 1, the test is inconclusive

Alternating Series Test

terms alternate sign

If the series is alternating, the terms are decreasing in magnitude, and the terms tend toward zero, then the series converges

P R O B L E M S F O R S E C T I O N 3 0- 5

For Problems 1 and 2, write out the first three terms of the series and then answer the question posed.

1 For what values of c will the series∞k=1ckconverge? Explain

2 For what values of w will the series∞k=1kwconverge? Explain

3 Suppose 0≤ ak≤ bk≤ ckfor all k Consider∞k=1ak,∞k=1bk and∞k=1ck What conclusions can be drawn if you know that∞k=1bk

(a) converges (b) diverges

19 Here we usea k to mean∞k=1a k The starting value of k is irrelevant since convergence is determined by the tail of the series.

In Problems 4 through 19, determine whether the series converges or diverges It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and

Root Tests.

4.∞k=2 √3

k

5.∞k=10 10

k √ k

6.∞k=1 ln kk

7.∞n=5n−9/10

8.∞k=1 ekk

9.∞k=22k−10/9

10.∞k=1e−2k

11.∞k=1 k+2

3k 2

12.∞k=12e−0.1k

13.∞k=2 kln k1

14.∞k=1ke−k2

15.∞k=1 k

k 3 +k+1

16.∞k=1 2

3 k +1

17.∞k=2 k−0.55

18.∞n=1 2n3−1n

19.∞n=1 en1+e

20 Suppose that ak= f (k) for k = 1, 2, 3, , where f (x) is positive, decreasing, and continuous on [1,∞) Put the following expressions in order, from smallest to largest Explain your reasoning with a picture or two

n −1

k =2ak, nk=3ak, 2nf (x) dx

21 Explain why the hypothesis that f (x) is decreasing is important in the Integral Test

22 Use your knowledge of improper integrals to give an upper and lower bound for

∞

k =1 k12

23 Let ∞k=1ak be a series and Sk= a1+ a2+ · · · + ak its kth partial sum, where

k= 1, 2, 3, Let L be a constant, 0 < L < 1

(a) If limk →∞ak= L, what can you conclude about∞k =1ak?

(b) If limk →∞Sk= L, what can you conclude about∞k =1ak?

24 Let∞k=1akbe a series and Sn=nk =1akits nth partial sum, where n= 1, 2, 3, For each of the following, decide whether or not enough information is given to assure that∞k=1akconverges M and m are constants Explain your reasoning

(a) ak>0 for all k and Sn> mfor all n

(b) ak>0 for all k and Sn< Mfor all n

(c) ak<0 for all k and Sn> mfor all n

(d) m < Sn< Mfor all n

In Problems 25 through 32, determine whether the series converges or diverges In this set of problems knowledge of the Limit Comparison Test is assumed.

25.∞k=1 2k3

−1

26.∞k=1 1

e k −1

27.∞n=2 2nn−12

−n

28.∞k=1 2k3k24−k

+1

29.∞k=3 k

2k 3 −2

30.∞n=2 √1

n 2 −n

31.∞n=2 n +1

ln n

32.∞k=2 5k2k

−5

33 (a) Let∞k=1akbe a convergent series with 0 < ak<1 for k= 1, 2, 3,

i Show that∞k=1ak2converges

ii Show that∞k=1 a1

k diverges

(b) Let∞k=1bkbe a convergent series with 0 < bk for k= 1, 2, 3, Argue that

∞

k =1b2kconverges (Use the results of part (a))

In Problems 34 through 41, determine whether the series converges or diverges In this set of problems knowledge of all the convergence tests from the chapter is assumed.

34.∞k=1 3kk!

35.∞n=1 n23n

36.∞n=1 32nn

37.∞k=1 kk!3

38.∞k=1 k!

k 3 3 k

39.∞k=2 2

(ln k) k

40.∞k=1 1+1kk

41.∞k=1 k5k2−3k2+1k

42 For what values of n, n a positive integer, does∞k=1kk!n converge?

43 For what values of r does∞k=1 rk

k! converge?

In Problems 44 through 51, determine whether the series converges absolutely, con-verges conditionally, or dicon-verges Explain your reasoning carefully.

44.∞k=1 ((k−1)+1)!kk!

45.∞k=1 (−1)k+1

k √ 2k

46.∞k=1 cos kk3

47.∞k=1 sin(2k)2k

48.∞n=3 (10−1)√n

n

49.∞k=1 (−1)k!k5k

50.∞k=1 (−k)k!k

51.∞k=2 3k(−1)3 k

+3

52 Does the series∞k=2(−1)kkln kconverge absolutely, converge conditionally, or diverge? Explain your reasoning carefully and justify your assertions

53 Prove the following version of the Integral Test (It’s a slightly weaker version than the one stated in this section.)

Let∞k=1akbe a series such that ak= f (k) for k = 1, 2, 3, where the function

f is positive, continuous, and decreasing on [1,∞)

(a) If1∞f (x) dxconverges, then∞k=1akconverges

(b) If1∞f (x) dxdiverges, then∞k=1akdiverges

In Problems 54 through 59, use the Ratio Test or Root Test to find the radius of convergence of the power series given.

54.∞k=1(−1)k (2x)k

k!

The generalized Ratio Test can be applied to a power series in order to find its radius

of convergence and to show... k!k k converges and its terms must tend toward zero We can conclude that kkgrows much more rapidly than does k! From part (a) of Example 30.31 we can deduce that k! in turn... that can be used to determine the convergence

or divergence of a series: The nth Term Test for Divergence, the Comparison and Limit Comparison Tests, the Integral Test, the Ratio Test, and