9.2 Series and Convergence

Riverfront Park, Spokane, WA

This is an example of an This is an example of an infinite infinite seriesseries..

1

1

Start with a square one unit by one Start with a square one unit by one unit:unit:

1

21

21

4

1

4

1

81

8

1

161

16

1

32 1

64

1

32

1

64 1

This series This series convergesconverges (approaches a limiting value.) (approaches a limiting value.)

Many series do not converge:Many series do not converge:1 1 1 1 1

1 2 3 4 5

1 2 31

n nn

a a a a a

The infinite series for a sequence is written as:

1 1S a

2 1 2S a a

3 1 2 3S a a a ...

1 2 3n nS a a a a

In an infinite series:In an infinite series:1 2 3

1n k

k

a a a a a

aa11, a, a22,…,… are are termsterms of the series. of the series. aann is the is the nnthth term term..

Partial sums:Partial sums:1 1S a

2 1 2S a a

3 1 2 3S a a a

1

n

n kk

S a

nnthth partial sum partial sum

If If SSnn has a limit as , then the series has a limit as , then the series convergesconverges, otherwise it , otherwise it

divergesdiverges..n

Ex. 1 Find the nth partial sum of the series:1

1

2nn

1S 1

1

2a 2S 1 1

2 4

3S 1 1 1

2 4 8

2 1

2

n

n nS

1 1

1 1

2 2

n

nn n

What’s the pattern for each sum??

4

1 1 1 1

2 4 8 16S

3

4

7

8 15

16

The denoms are all and the numerators?? 2n

1lim lim 1

2n nn nS

So the series converges to 1.

1

Then what is the infinite sum of the series??

This is called a geometric series.

Geometric Series:

In a In a geometric seriesgeometric series, each term is found by multiplying the preceding term , each term is found by multiplying the preceding term

by the same number, by the same number, r, r, called the called the common ratiocommon ratio..

2 3

0

n n

n

a ar ar ar ar ar

This infinite sum converges to if , and diverges if .This infinite sum converges to if , and diverges if .1

a

r1r 1r

1 1r is the is the interval of convergenceinterval of convergence..

Example 2:Example 2:

3 3 3 3

10 100 1000 10000

.3 .03 .003 .0003 .333...

1

3

310

11

10

aa

rr

3109

10

3

9

1

3

1 1 11

2 4 8

11

12

11

12

13

2

2

3

aa

rr

Example 3:Example 3:

1

1

n

n

a rS

r

The nth partial sum of a geometric series is:The nth partial sum of a geometric series is:

This is a handy formula to know, as you may be asked to find annth partial sum of a geometric series!

From this, you can see where formula for the infinite sum comes from:

1lim

1

n

nn

a rS S

r

0

1

a

r

Ex. 4 Find the partial sum of the series:

1

1 1

1n n n

Write out a few terms:1 1 1 1 1 1 1

1 ...2 2 3 3 4 1n n

11

1nSn

Since the middle terms zero out, we are left with:

1lim lim 1

1nn nS

n

The series converges to 1.

The last problem is an example of a telescoping series:

1 2 2 3 3 4 4 5( ) ( ) ( ) ( ) ...b b b b b b b b

Since the middle terms will always cancel out, we can find thepartial sum of a telescoping series by:

1 1n nS b b

11

n nn

b b

Ex. 5 Find the sum of the series:2

1

2

4 1n n

This can be written in telescoping form using partial fractions:

2

2 2

4 1 (2 1)(2 1)nan n n

1 1

2 1 2 1n n

So the nth partial sum is:1 1 1 1 1 1

...1 3 3 5 2 1 2 1nS

n n

11

2 1n

1lim lim 1

2 1nn n

Sn

21

2

4 1n n

1

Ex. 6 a) Find the sum (if it exists) of the series: 0

3

2nn

Re-write:

0

13

2

n

n

13,

2a r

Since we know that the series converges to the sum of: 0 1r

1n

aS

r

3

61 1 2

b) Find the sum (if it exists) of the series:

0

3

2

n

n

Since the series diverges. 3

12

r

Ex. 7 Use a geometric series to express as the ratio of two integers. (as a rational number)

0.080808

Note that we can write as:2 4 6 8

8 8 8 80.080808... ...

10 10 10 10

2 20

8 1

10 10

n

n

2

8

10a

2

1

10r

1

aS

r

2

2

8 10

1 (1/10 )

8

99

nth Term Test for Divergence

The first requirement for convergence of a series is that the terms of the sequence must approach zero. That is,

1n

n

a

If converges, then lim 0n

na

The contrapositive of this fact leads us to a useful test for divergence:

If , then diverges.lim 0nn

a

1

nn

a

So how could we use this test? Let’s take a look…

(Notice we are talking about the limit of the sequence, not thepartial sum!)

Ex. 8 Determine if the series diverges:

a)0

2n

n

Let’s look at the sequence: lim 2n

n

Since the limit of the sequence is not 0, by the nth term test, the seriesdiverges!

b)

1

!

2 ! 1n

n

n

!

lim2 ! 1n

n

n

1

2

Since the limit of the sequence is not 0, by the nth term test, the seriesdiverges!

c)

1

1

n n

1

limn n

0The limit of the nth term is 0, so…

the test fails! We can’t draw aconclusion about

convergence or divergence.

Bouncing Ball Problem

A ball is dropped from a height of 6 feet and begins bouncing.The height of each bounce is three-fourths the height of its

previous bounce. Find the total vertical distance traveled by the ball.

Let D1 be the initial distance the ball travels:

1 6D Let D2 be the distance traveled up and down:

2 6 3 4 6(3/ 4) 12(3/ 4)D 2

3 6(3 4)(3 / 4) 6(3/ 4)(3 / 4) 12(3/ 4)D

2 36 12(3/ 4) 12(3/ 4) 12(3/ 4) ...D 1

0

36 12

4

n

n

D

1

0

36 12

4

n

n

D

0

3 36 12

4 4

n

n

Notice this is ageometric series!

16 9

1 3 4

42 feet