practice solutions

Summer 2006 - Calculus IIMike Limarzi

Practice Problems Solutions

Section 10.2: #11, 13, 17, 29, 35, 36, 37, 42

11. Determine the boundedness and monotonicity of the sequence.

n+ (1)n

n.

Solution:The sequence starts by increasing, but then begins to decrease. Therefore, it is notmonotone.

The sequence {an} is bounded by M = 2. The lub is 32 and the glb is 0.


(0.9)n.

Solution:The sequence {an} is monotone decreasing.

The sequence an = (0.9)n is bounded by M = 1. The lub is .9 and the glb is 0.


4n4n2 + 1

.

Solution:The sequence {an} is monotone increasing.

The sequence {an} is bounded by M = 2. The lub is 2 the glb is 45 .

1


2n 12n

.

Solution:The sequence {an} is monotone increasing.

The sequence {an} is bounded by M = 2. The lub is 1 and the glb is 12 .


1k 1k + 1


The sequence {an} is bounded by M = 12 . The lub is12 and the glb is 0.


cosn.

Solution:Lets look at the first few terms:

a1 = cos = 1

a2 = cos 2 = 1

a3 = cos 3 = 1

So, the sequence starts by increasing, but then begins to decrease. Therefore, it is notmonotone.

To show the sequence is bounded, we must show there exists someM such that |an| Mfor all n.

2

|an| = | cosn| 1

This is true since cosx is bounded by 1.So, {an} is bounded by M = 2. The lub is 1 and the glb is -1.


lnn+ 2n+ 2

.


The sequence {an} is bounded by M = 1. The lub is 13 ln 3 and the glb is 0.

42. Let M be a positive integer. Show that an =Mn

n!decreases for n M .

Solution:To show the sequence decreases, we must show an > an+1 for all n M .

n M

n+ 1 M

Multiply through by Mn > 0:

Mn(n+ 1) > Mn+1

Divide through by (n+ 1)! > 0:

Mn

n!>

Mn+1

(n+ 1)!

an > an+1

So, we have shown that {an} is decreasing when n M .

3

Section 10.3: #17, 20, 21, 24, 30

17. State whether the sequence converges, and, if it does, find the limit.

(2n+ 1)2

(3n 1)2.

Solution:The sequence {an} converges to 49 .


n4 + 1n4 + n+ 6

Solution:We look at this sequence as n approaches .

limn

n4 + 1n4 + n+ 6

Multiply top and bottom by n4 to get:

limn

1 + 1n4

1 + 1n3

+ 6n4

Since we know 1np 0 when p > 0, we get:

limn

1 + 1n4

1 + 1n3

+ 6n4

=11= 1

So, the sequence converges to 1.


cosn

Solution:The sequence {an} oscillates between 0 and 1. Thus, the sequence diverges.

4

24. State whether the sequence converges, and, if it does, find the limit.4 1

n


limn

4 1

n

Since4 x is a continuous function for x < 4, and 1n is bounded by 1, we can bring

the limit inside, to get:

limn

4 1

n

=

4 lim

n

1n

=4 0 = 2


30. State whether the sequence converges, and, if it does, find the limit.(1 +

1n

)n/2


limn

(1 +

1n

)n/2We rewrite this as:

limn

((1 +

1n

)n)1/2

= limn

(1 +

1n

)n

5

Again, we knowx is a continuous function for x > 0, and 1+ 1n > 0, we can bring the

limit inside, to get:

=

lim

n

(1 +

1n

)nWe recognize the inside as e, so we get:

=e

So, the sequence converges toe.

Section 10.4: #6, 16, 20, 24, 25, 28, 31

6. State whether the sequence converges as n; if it does, find the limit.

3n

4n


limn

3n

4n

Using properties of exponents, we can group the top and the bottom to get:

limn

(34

)nSince |34 | < 1, we can use Proposition 10.4.2 to see that

limn

(34

)n= 0

So, this sequence converges to 0.


23n1

7n+2

6


limn

23n1

7n+2

Simplifying, we get:

= limn

12(2

3)n

49 7n

= limn

198

8n

7n

= limn

198

(87

)nBut, since |87 | > 1, this sequence diverges.


n2 sinn

Solution:Look at the first few terms:

a1 = 1 sin = 0

a2 = 4 sin 2 = 0

a3 = 9 sin 3 = 0

In fact, for every n, sinn = 0. So, this means that the sequence converges to 0.


n!2n

7


limn

n!2n

Simplify:

limn

(n 1)!2

= limn

12(n 1)!

And, since (n 1)!, the sequence diverges.


5n+1

42n1

Solution:The sequence converges to 0.

28. State whether the sequence converges as n; if it does, find the limit. 11/n

dxx


limn

11/n

dxx

To solve this, we integrate:

limn

2x

11/n

= limn

21 2

1/n

8

= limn

2 21/n

= 2 20 = 2



nn

2n2


Section 10.5: #13, 15, 26, 27, 28, 34, 35

13. Calculate the indicated limit.

limx

x+ sinxx sinx

Solution:

limx

x+ sinxx sinx

=1 + 1


limx

ex + ex 21 cos 2x

9

Solution:

limx

ex + ex 21 cos 2x

=12


limn0+

x

x+ sinx

Solution:Verify that we can use LHopitals rule:

0

0 + sin0=

00

So, we apply LHopitals rule:

limn0+

12x1/2

12x1/2 + cos

x 12x1/2

Cancelling out, we get:

limn0+

11 + cos

x=

12

So, the limit is 12 .


limx

cosx cos 3xsin(x2)

Solution:

limx

cosx cos 3xsin(x2)

= 4

10


limn0

a+ x

a x

x


a

a

0=

00


limn0

12(a+ x)

1/2 + 12(a x)1/2

1

Letting x 0, we get:

=12a1/2 + 12a

1/2

1

=a1/2

1=

1a

So, this limit is1a.

34. Find the limit of the sequence.

limn

ln(1 1/n)sin(1/n)


ln(1 0)sin(0)

=00


limn

1/n2

11/n

cos(1/n) (1) 1n2

11

Cancelling, we get:

limn

1

11/n

cos(1/n)

Let n:

11

cos(0)= 1

So, the sequence converges to -1.

We could solve this a different way. We can make the substitution: x = 1n . So, whenn, we get that x 0+. Making this substitution simplifies our derivatives.

limn

ln(1 1/n)sin(1/n)

Make the switch x = 1n :

limx0+

ln(1 x)sin(x)

Apply LHopitals rule:

limx0+

11xcos(x)

Let x 0+:11

cos(0)= 1

So, we do get the same answer, but the derivatives in the second way are easier tocompute.

35. Find the limit of the sequence.

limx

1n[ln(n+ 1) lnn]


Section 10.6: #5, 19, 25, 27, 30

12


limx

x2 sin1x.

Solution:

limx

x2 sin1x=


limx

(cos

1x

)x.

Solution:

limx

(cos

1x

)x= 1


limx

x2 + 2x x.

Solution:

limx

x2 + 2x x = 1


limx

(x3 + 1)1

ln x .

13

Solution:

limx

(x3 + 1)1

ln x = e3


limx

(1 +

1x

)3x

Solution:From our knowledge of limits, we guess that the answer will be e3.To solve this using LHopitals rule, we look at eln:

limx

eln(1+ 1

x

)3xex is continuous everywhere, so bring limit inside:

elimx ln(1+ 1

x

)3xNow, look just at the ln term:

limx

ln(1 +

1x

)3xSlap down the exponent:

= limx

3x ln(1 +

1x

)Bring out 3 and rewrite as quotient:

= 3 limx

ln(1 + 1x

)1x

Make the substitution n = 1x :

= 3 limn0+

ln(1 + n)n

Verify we can use LHopitals rule:

= 3ln(1)0

=00

14

Apply LHopitals rule:

= 3 limn0+

11+n

1

Let n 0+:= 3

11

1= 3

Plugging back in we get:

elimx ln

(1+ 1

x

)3x

= e3

So, the limit is e3, as we suspected.

Section 10.7: #6, 14, 22, 23, 30

6. Evaluate the improper integrals that converge. 10

dxx.

Solution:We note that the problem occurs as b 0+. So, we look at:

limb0+

1b

dxx.

Integrate:

limb0+

2x

1b

= limb0+

21 2

b

= limb0+

2 2b

Letting b 0:= 2 2

0 = 2.

15

So, the integral converges to 2.

14. Evaluate the improper integrals that converge. e

dx

x lnx.

Solution:We look at this integral as b:

limb

be

dx

x lnx.

Integrate with u substitution:Let u = lnx.Let du = 1x .Bounds change from e to 1 and from b to b:

= limb

b1

du

u.

Integrate:

= limb

lnub1

Plug back in for x:

= limb

ln(lnx)be

= limb

ln(ln b) ln(ln e)

But, we note thatlim

bln(ln b) =

Therefore, this integral diverges.


xexdx.

16

Solution:We look at this integral as b :

limb

0bxexdx.

Integrate by parts:u = x v = ex

du = dx dv = exdx.

= limb

xex 0

bexdx.

= limb

xex ex0b

= limb

(x 1)ex0b

= limb

(0 1)e0 (b 1)eb

The first term is -1. We must analyze second term:

limb

(b 1)eb

This is equivalent to:

limb

b+ 1eb

Verify we can use LHopitals rule:

+ 1e

=


limb

1eb

Now, we let b: 1

= 0

So, we getlim

b(b 1)eb = 0

17

This means we get

limb

0bxexdx = 1

We must be careful since the integral gives us a negative result. In this case, on theinterval (, 0], x < 0 and ex > 0, so xex < 0. So, this makes sense. Our integralconverges to -1.


xx2 9

dx.

Solution: 53

xx2 9

dx = 4


dx

x2 5x+ 6

Solution:We first factor the denominator: 4

1

dx

(x 2)(x 3)We note that there are problems at x = 2 and x = 3. But, before we look at theseproblems, we use partial fractions to make the integrand easier for us to integrate.

A

x 2+

B

x 3=

1(x 2)(x 3)

A(x 3) +B(x 2) = 1

3A 2B +AX +Bx = 1 + 0x

Solving, we get:A = B

3B 2B = 1

B = 1, A = 1

So, we get: 41

1x 2

+1

x 3dx

18

We must deal with the problems at 2 and 3, so we get 3 integrals:

= 2

1

1x 2

+1

x 3dx+

32

1x 2

+1

x 3dx+

43

1x 2

+1

x 3dx

If any of these three integrals goes to infinity, the whole integral diverges. So, we lookat the first integral: 2

1

1x 2

+1

x 3dx

= limb2

b1

1x 2

+1

x 3dx

= limb2

lnx 3x 2

b

1

= limb2

lnb 3b 2

ln 1 31 2

Letting b go to 2, we see that:

= ln10

ln 2And since lnx when x , we get that the first of the 3 integral diverges.Therefore, 4

1

dx

x2 5x+ 6diverges.

Section 11.1: #27, 33, 47, 52, 68

27. Find the sum of the series:

k=1

1k(k + 3)

Solution:

k=1

1k(k + 3)

=1118

33. Find the sum of the series:

k=0

2k+3

3k

19

Solution:

k=0

2k+3

3k= 24

47. Find a series expansion for the expression:

x

1 + x2|x| < 1

Solution:

x

1 + x2=

k=0

(1)kx2k+1

52. Show the series diverges:

k=2

kk2

3k

Solution:So, we look at the term we are summing:

kk2

3kWe take the limit:

limk

kk2

3kCancelling out:

limk

kk3

3Now, we see that:

limk

kk3

3=

So, the terms do not go to 0. By the basic divergence test, the sum diverges.

68. Show that:

k=1

(k

k + 1

)kdiverges.

20

Solution:We look at the term we are summing:(

k

k + 1

)kWe take the limit:

limk

(k

k + 1

)kRewriting, we get:

= limk

(1

1 + 1k

)k= lim

k

1(1 + 1k

)kLetting n, we get:

=1e

So, we get that

limk

(k

k + 1

)k=

1e

So, by the basic divergence test, since ak does not go to 0, the sum diverges.

Section 11.2: #1, 2, 3, 10, 14, 17, 19, 25, 31, 32

1. Determine whether the series converges or diverges: kk3 + 1

Solution:The series converges.

2. Determine whether the series converges or diverges: 13k + 2

Solution:This sum looks like

1k , which we know diverges. To show this series diverges, we use

the limit comparison test.

Let ak =1

3k + 2.

Let bk =1k.

Look at:

limk

13k+2

1k

21

Simplifying, we get:

limk

k

3k + 2Multiplying through by 1k , we get:

limk

13 + 2k

Letting k :lim

k

13 + 2k

=13

Since 13 > 0, we know that 1

3k+2 and 1

k have the same convergence. And since 1

kdiverges, we get that 1

3k + 2diverges.

3. Determine whether the series converges or diverges: 1(2k + 1)2

Solution:The series converges.

10. Determine whether the series converges or diverges: ln kk3

Solution:We guess that this series converges, since x3 goes to infinity much faster that lnx. Wewill use the basic comparison test.

Let ak =ln kk3

.

Let bk =1k2

.

We know that x > lnx for all x. So, we get:

k

k3 ln k

k3

So, this means that ln k

k3converges only if

1k2

converges. We know that 1

k2

converges by the p-series test, and also by the integral test. Therefore, by the basiccomparison test: ln k

k3

converges.

22

14. Determine whether the series converges or diverges: 11 + 2 ln k

Solution:Since lnx goes to infinity slowly, we guess that this series diverges. We use the basiccomparison test.

Let ak =1

1 + 2 ln k.

Let bk =1k.

We know that x > lnx for all x, and similarly 1 + 2x > 1 + 2 lnx. This meansthat 11+2x

practice solutions

Documents