Math 115 Spring 2014 Written Homework 5-SOLUTIONS Due ...

Math 115 Spring 2014 Written Homework 5-SOLUTIONS

Due Friday, March 7

Instructions: Write complete solutions on separate paper (not spiral bound). If multiple pieces of paper are used, THEY MUST BE STAPLED with your name and lecture written on each page. Please review the Course Information document for more complete instructions.

9 n-1

1. Consider the series -2

10

n=1

(a) Determine the first 3 partial sums of this series. (Simplify completely.)

Solution:

80 S1 = -3 10 = -3

80

81

24

S2 = -3 10

+ (-3) 10

= -3 - = -3 10

1-

82 10

1

-

8 10

54 =-

10

S3 = -3

80 + (-3)

10

81 + (-3)

10

82

24 192

= -3 - - = -3

10

10 100

1-

83 10

1

-

8 10

(b) Determine the N th partial sum, SN , of this series. Solution:

732 =-

100

SN

=

-3

8 10

0 + -3

8 10

1 + -3

8 10

2 + . . . + -3

8 N -1 10

= -3

1-

8N 10

1

-

8 10

(c) Using the formula for SN you determined in part (b), evaluate lim SN . N

Solution:

S = lim SN N

= lim -3 N

1-

8N 10

1

-

8 10

-3

8N

=

1

-

8 10

lim

N

1-

10

by limit law 5

8N

= -30 lim 1 - lim

by limit law 2

N

N 10

8N

= -30 1 - lim

since 1 is a constant sequence

N 10

= -30 [1 - 0] since = -30

8N is a geometric sequence with |r| < 1

10

8 n-1

(d) What is the value of -3

?

10

n=1

Solution: By the definition of an infinite series,

8 n-1

-3 10

= lim SN = -30 N

n=1

2. A pile driver is pounding a piling into the ground. On the first impact, the piling moves 100 cm into the ground. On the second impact, it moves 96 cm further into the ground. (a) Assume the distances the piling moves with each impact form an arithmetic sequence. How far will the piling move on the 10th impact? How far into the ground will it be after 10 impacts? Solution: Let D1 = 100 and D2 = 96. Using these two terms and the assumption that the sequence is arithmetic, we get the common difference of d = 96 - 100 = -4. Then the associated sequence is generated by D(n) := 100 + (n - 1)(-4).

On the 10-th move, the pile will move D(10) = 100 + (10 - 1)(-4) = 64 cm.

The sum of the 10-term arithmetic sequence is the total distance the pile has been driven

S10 = 10

D1 + D10 2

= 10

100 + 64 2

= 820 cm

(b) Assume the distances the piling moves with each impact form a geometric sequence. How far will it move on the 10th impact? How far into the ground will it be after 10 impacts? What is the farthest the piling can be driven into the ground (assuming infinite time)?

Solution: Let D1 = 100 and D2 = 96. Using these two terms and the assumption that the

sequence

is

geometric,

we

get

the

common

difference

of

r

=

96 100

=

0.96.

Then

the

associated

sequence is generated by D(n) := 100(0.96)n-1.

On

the

10-th

move,

the

pile

will

move

D(10)

=

100(0.96)10-1

=

969 1008

69.3

cm.

The sum of the 10-term geometric sequence is the total the pile has been driven

1 - r10

1 - 0.9610

S10 = D1 1 - r = 100 1 - 0.96 837.9cm

Assuming we drive the pile forever, the sum of the infinite geometric series is

S

=

a1 1-r

=

100

1

-

96 100

= 2500 cm

3. Assume the following series are either arithmetic or geometric. (i) Determine the generat-

ing function for the associated sequence. (ii)Determine the limit of the associated sequence.

(iii) Evaluate the series. If the series is divergent, you must clearly justify your answer.

5 1

(a) 5 + + 5 + . . .

33

Solution: (i) This is a geometric series as the associated sequence is geometric.

r

=

5

3

=

1 3

5

=

1 .

5

5

3

3

1 n-1

The associated sequence is generated by the function a(n) := 5

.

3

(ii) Since r = 1/ 3 is between (-1, 1), the limit of a(n) is 0.

(iii) Since the value of r is between (-1, 1), we know that the infinite series converges and

the sum is represented by the formula

S

=

a1 1-r

=

5 1 - 1

3

=

15 3-1

(b) 4 + 6 + 9 + . . .

Solution: (i) This is a geometric series as the associated sequence is geometric.

693 r= = =

462

3 n-1

The associated sequence is generated by the function a(n) := 4

.

2

3 (ii) Since r = 2 is greater than 1 and a1 = 4 is positive, the limit of a(n) is .

(iii) Since the value of |r| 1, we know that the geometric series is divergent. Equivalently, since the limit of the associated sequence is not 0, we know that the series is divergent.

(c) 3 + 3(10)-1 + 3(10)-2 + . . .

Solution: (i) This is a geometric series as the associated sequence is geometric.

3(10)-1 3(10)-2 1

r=

=

=

3

3(10)-1 10

1 n-1

The associated sequence is generated by the function c(n) := 3

.

10

(ii) Since r = 1/10 is between (-1, 1), the limit of c(n) is 0.

(iii) Since the value of r is between (-1, 1), we know that the infinite series converges and the sum is represented by the formula

S

=

a1 1-r

=

3

1

-

1 10

=

30 9

=

10 3

1131 (d) + + + + . . .

16 8 16 4

Solution: (i) This is an arithmetic series as the associated sequence is arithmetic.

1 1 3 11 3 1 d= - = - = - =

8 16 16 8 4 16 16

11

1

The associated sequence is generated by the function a(n) := +

(n - 1) =

n.

16 16

16

1 (ii) Since d = is positive, the limit of a(n) is .

16

(iii) Since the limit of the associated sequence is not equal to 0, we know that the series is divergent.

4. Find r for the infinite geometric series which has sum S = 30 and a1 = 4. Solution: We are given that the series is geometric. Since it is convergent, the sum of the series is

a1 . Hence, 1-r

4 = 30.

1-r 4 = 30(1 - r)

4 = 30 - 30r

30r = 26 26 13

r= = 30 15

5. An infinite geometric series has the sum 16. If the sum of the first two terms is 15, find

a1 and r. Solution:

Again,

S

=

a1 1-r

=

16.

We

are

also

given

that

a1 + a2

=

15.

Since

the

associated

sequence is geometric, a2 = a1r. Thus, we have two equations in the two unknowns a1 and

r.

a1 1-r

=

16

and

a1

+

a1r

=

15

a1 = 16(1 - r) and a1(1 + r) = 15

16(1 - r)(1 + r) = 15

1 - r2 = 15 16

r2 = 1 - 15 = 1 16 16 1

r=? 4

This results in two different geometric sequences that solve the problem. If r = 1/4, then a1 = 16(1 - 1/4) = 16(3/4) = 12. If r = -1/4, then a1 = 16(1 - (-1/4) = 16(5/4) = 20

6. Suppose, in the process of investigating an infinite series, the following formulas for the nth partial sum, Sn, of the series are found. Determine whether each associated series is convergent or divergent by determining lim Sn. If convergent, state the sum.

n

4 (a) Sn = 1 - 7n

Solution:

4

lim Sn

n

=

lim

n

1- 7n

4

= lim 1 - lim

by limit law 2

n

n 7n

1

= 1 - lim 4

since 1 is a constant sequence

n 7n

1 = 1 - 4 lim

n 7n = 1 - 4(0) since

by limit law 5 1n

is a geometric sequence with |r| < 1 7

=1

By the definition of an infinite series, the total sum S is defined as S = lim Sn, when this n

limit exists. Hence, the infinite series is convergent and the sum of the infinite series is 1.

n+1 (b) Sn = n2 + n + 1 Solution:

n+1

lim Sn

n

=

lim

n

n2

+

n

+

1

n + 1 1/n2

=

lim

n

n2

+

n

+

1

?

1/n2

=

lim

n

1

1 n

+

+

1 n2

1 n

+

1 n2

11

lim

n

n + n2

=

11

lim 1 + +

n

n n2

by limit law 4

1

1

lim + lim

=

n n

n n2 1

1 by limit laws 1 and 2

lim 1 + lim + lim

n

n n n n2

0+0 =

1+0+0

=0

Hence, the associated infinite series is convergent and the sum of the associated infinite series is 0.

7. For each decimal below, (i)express it as a series and (ii) use a series formula to determine the rational form of the number. (a) 17.241 Solution: (i)

1

1

1

24 1

17.241 = 1(10) + 7(1) + 2

+4

+1

or 17 + + +

10

100

1000

10 100 1000

(ii)

2 4 1 17000 + 200 + 40 + 1 17241

17.241 = 17 + 17 + + +

=

=

10 100 1000

1000

1000

(b) 5.44444 . . .

Solution:

(i)

1

1

1

1n

5.44444 ? ? ? = 5 + 4 10 + 4 100 + 4 103 + . . . = 5 +

4 10

n=1

(ii)Since the repeating part of the decimal is an infinite geometric series with r between -1 and 1, we can use the convergent geometric series formula

5.44444 ? ? ? = 5+4

1 10

+4

1 100

+4

1 103

4

+.

.

.

=

5+

1

10

-

1 10

4/10 = 5+

9/10

4 = 5+

9

=

45 4 +

99

=

49 9

(c) 1.24242424 . . . (i)

1

1

1

1n

1.24242424 . . . = 1 + 24

+ 24

+ 24

+ . . . = 1 + 24

100

104

106

100

n=1

(ii)Since the repeating part of the decimal is an infinite geometric series with r between -1 and 1, we can use the convergent geometric series formula

1

1

1

24 100 99 24 123

1.24242424 . . . = 1+24

100

+ 24

104

+ 24

106

+... = 1+

=+=

100 99 99 99 99

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