1

January 13,15 Geometric series

1.1 The geometric series 1.2 Definitions and notation1.3 Applications of series

Infinite sequence: An ordered set of infinite number of quantities. Examples:

Chapter 1 Infinite Series, Power Series

,, , , , , )4

,111111)3

,2

1 , ,

8

1 ,

4

1 ,

2

1 1, )2

,2 , 8, 6, 4, 2, )1

1321

1

n

n

aaaaa

, , , , ,

n

Limit of an infinite sequence:

.01

lim :Example

. allfor such that integer an exists thereis,it smallhowever ,0any For

:means lim

2

n

NnlaN

la

n

n

nn

Examples p5.1-3; Problems 2.7, 8.

2

Infinite series: The sum of an infinite sequence of numbers. Examples:

1321

1

1

111

1

)4

)1( )1(1)1(1)3

2

1

2

1

8

1

4

1

2

1 2)1

2 2 8 6 4 2 )1

nnn

n

n

nnn

n

aaaaa

nn

Note:1)The sum of an infinite series may not be finite.2)Even when the sum is finite, we still cannot do it by adding the terms one by one.

Partial sum of an infinite series:

Sum of an infinite series is defined as:

n

n

iin aaaaaS

3211

nn

SS

lim

3

Geometric series: A geometric series has the general term of and1 nn ara

1

21

n

n araraar

Partial sum of a geometric series:

Sum of a geometric series:

r

raarararaarS

nn

i

nin

1

)1(

1

121

.1 ifonly and if ,1

lim

rr

aSS n

n

.3

32

1

1

27

8

9

4

3

21 :Example

Application: Change recurring decimals into fractions. Examples:

12

7

9

3

4

13.025.030.58

99

2525.0

27

5

999

185185.0

.999101

10)1010(.0.0

3

363

abcabc

abcabcabcabc

4

Reading: L’Hospital’s rule:

.)('

)('lim

)(

)(lim

)(

)(lim

)('

)('lim

)(/)('

)(/)('lim

)(/1

)(/1lim

)(

)(lim

then,)(lim)(lim If )4

.)('

)('lim

)/1('

)/1('lim

)/1(')/1(

)/1(')/1(lim

)/1(

)/1(lim

)(

)(lim then ,0)(lim)(lim If )3

.)('

)('lim

)(

)(lim

)(

)(lim

)('

)('lim

)(/)('

)(/)('lim

)(/1

)(/1lim

)(

)(lim

then,)(lim)(lim If )2

.)('

)('lim

))((')(

))((')(lim

)(

)(lim then ,0)(lim)(lim If 1)

:Proof

exsists.limit theuntil )(

)(lim

)(''

)(''lim

)('

)('lim

)(

)(lim ,Repeatedly

.)('

)('lim

)(

)(lim then exists,

)('

)('lim and ,)(or 0)(lim)(lim If

2

2

2

2

0

2

2

00

2

2

2

2

000

000

)(

)(

)or ()or ()or ()or ()or (

0000000

00

00000

0000

00000

xg

xf

xg

xf

xg

xf

xf

xg

xfxf

xgxg

xf

xg

xg

xf

xgxf

xg

xf

tg

tf

tgt

tft

tg

tf

xg

xfxgxf

xg

xf

xg

xf

xg

xf

xf

xg

xfxf

xgxg

xf

xg

xg

xf

xgxf

xg

xf

xxxgxg

xxxfxf

xg

xfxgxf

xg

xf

xg

xf

xg

xf

xg

xf

xg

xf

xg

xf

xg

xfxgxf

xxxxxxx

xx

xt

ttxxx

xxxxxxxxxxxxxx

xxxx

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n

n

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5

Read: Chapter 1: 1-3Homework:1.1.2,8;1.2.1,7.Due: January 24

6

January 17 Convergence of series

1.4 Convergent and divergent series

Convergence of series: If for an infinite series we have , where S is a finite

number, the series is said to be convergent. Otherwise it is divergent.Note: 1)Whether or not a series is convergent is of essential interest for the series.2)For a divergent series, Sn either approaches infinity or is oscillatory.3)Adding or removing a finite number of terms from an infinite series will not affect whether or not it converges.

SSnn

lim

Problems 4.3, 6.

divergent. is )1( )1(1)1(1 )2

divergent. is 2

12

16

18

8

14

4

12

2

11

16

1

10

1

9

1

8

1

7

1

6

1

5

1

4

1

3

1

2

11

4

1

3

1

2

11 series Harmonic 1)

:examples More

1

1

n

n

pp

7

1.5 Testing series for convergence; the preliminary testPreliminary test:If then the series is divergent.,0lim

nn

a

.0limlim then,convergent is series theIf

:Proof

1 SSSSa nn

nn

n

Note: 1)The requirement that is a necessary condition for convergence of a series,

but is not sufficient. E.g., the harmonic series.2)If , then further test is needed.

0lim n

na

0lim n

na

Problems 5.3, 9.

8

1.6 Convergence test for series of positive terms; absolute convergenceA. The comparison test:

limit. a has therefore, and

,increasinglly monotonica is then , ,let converges, If

:Proof

diverges. then diverges, If 2)

converges. thenconverges, If 1)

then, allfor 0 If

111

11

11

anban

anan

n

iib

nn

nn

nn

nn

nn

nn

nn

SSS

SSaSbb

ba

ab

nba

Problems 6.4, 5.

.convergent is !

1 therefore,convergent is )2(

2

1

.1for 2

1

)1(21

1

!

1

:examples More

111

1

nnn

n

n

nnnn

9

Reading: Monotone convergence theorem:

Let S be a set of real numbers. A real number x is called an upper bound for S if x ≥ s for all s ∈ S. A real number x is the least upper bound for S if x is an upper bound for S, and x ≤ y for every upper bound y of S.Least-upper-bound property: Any set of real numbers that has an upper bound must have a least upper bound.

Monotone convergence theorem:A bounded monotonic sequence of real numbers has a finite limit.

.lim means which , Now . of boundupper least theis

fact that heion with tcontradictin is which bound,upper an be will exist thennot does

thisif because is This . have we allfor that so exists there0,number

givenany for then , of boundupper least theis where, and Suppose

:Proof

1

cacacaac

c

NcaNnN

accaaa

nn

nnn

n

nnnn

10

Read: Chapter 1: 4-6Homework:1.4.7;1.5.3,9;1.6.4,5.Due: January 24

11

January 22, 24 Convergence test

1.6 Convergence test for series of positive terms; absolute convergenceB. The integral test:

Example p12; Problems 6.12.

figures. twofollowing with theunderstand easy to is proof The :Proof

converges. )( ifonly and if converges

then,decreasinglly monotonica and positive is )( and ),( If

1

dxxfa

xfnfa

nn

n

Note: The lower limit in the integral is not necessary. Using x = 0 or x =1 may cause problems.

E.g.,

120 2

.converges1

but infinite, is 1

n ndx

n

12

.1 if diverges

,1 if converges

1

1

1lim

1limlim

1 :series- The:examples More

1

1

1

1

11

1

p

p

np

b

p

x

x

dx

x

dx

np

np

p

b

bp

b

b

pbp

np

The p-series:

13

C. The ratio test:

diverge.must .0lim,1 , allfor Then .1 that so take,1 If 2)

converge.must proved.) be To

.convergent is series convergent absolutely An . thus(and converges, Since

. then,for Form

. , allfor Then .1 that so take,1 If 1)

:Proof

needed. isst further te 1, If 3)

diverges. ,1 If 2)

converges, ,1 If )1

then,lim If

1

1

111

1

1

1

1

nnn

nn

n

nn

nn

nn

nnNNn

n

n

n

nn

nn

n

n

n

aaa

aNn

aab

baNnab

a

aNn

a

a

a

a

Example p14.1,2; Problems 6.19,21.

14

D. Limit comparison test (a special comparison test):

diverges. then diverges if Therefore .1 have we largefor then , 3)If

converges. then converges if Therefore .1 have we largefor ,0 2)If

n test.comparisio the toaccording divergeboth or

convergeboth either and Therefore . largefor

and ,0

satisfy which , and exist e then ther,0 If 1)

:Proof

diverges. then diverges, and , If )3

converges. then converges, and ,0 If )2

diverge.both or convergeboth either and then ,0 If )1

then,lim and positive,both are and If

2121

21

21

nnnnn

n

nnnnn

n

nnnnn

n

n

nn

nn

nn

n

n

nnn

ababb

anl

abbab

anl

bablablnl

b

al

lll

lll

abl

abl

bal

lb

aba

Example p15.1; Problems 6.31,35.

15

Read: Chapter 1:6Homework:1.6.12,13,21,26,28,32,35.Due: January 31

16

January 29 Alternating series

1.7 Alternating series1.8 Conditionally convergent series

Absolute convergence: A series is said to be absolutely convergent if

is convergent.

Theorem: An absolutely convergent series is convergent.

1nna

1

||n

na

converges. also ||Then converges. Therefore

.||2||by above from bounded is and

decreasing-non is of sum partial The .||let converges, || Suppose

:Proof

11 11

111

11

nn

n nnn

nn

nn

nnn

nn

nnnnn

nn

abab

aaab

baaba

Conditional convergence: A series is said to be conditionally convergent if

is convergent but is not convergent.

1nna

1nna

1

||n

na

17Problems 7.4,6,7.

Alternating series: An alternating series is a series whose successive terms alternate in

sign. E.g.,

Test for alternating series (Leibnitz’s alternating series theorem):

An alternating series converges if the absolute values of the terms decreases steadily to 0.

That is

4

1

3

1

2

11

.0lim and 1 n

nnn aaa

.lim that shows facts twoThese

.limlim Therefore ,0limlimNow

.lim Let exists. lim Therefore

above. from bounded is )()()( Also

.decreasing-non is and 0)()()(

sequence the then,0 Suppose

:Proof

21212212

22

121222543212

21243212

1

LS

LSSaSS

LSS

aaaaaaaaaS

aaaaaaS

a

ii

nn

nn

nn

nnn

nn

nn

nnnn

nnn

ln2).( converges 4

1

3

1

2

11

)1(

:Example

1n

n

n

18

1.9 Useful facts about series1)The convergence or divergence of a series is not affected by multiplying each term by the same none-zero constant.2)The sum of two convergent series is convergent.

Problems 9.11,18,19.

19

Read: Chapter 1:7-9Homework:1.7.3,4,6,7;1.9.1,4,9,10,16,20.Due: February 7

20

January 31 Power series1.10 Power series; interval of convergencePower series: A power series is an infinite series of the form

We say that the power series is in x and is about the point a.

A special case is a=0 and

Interval of convergence: The range of x where the power series converges.

2210

0

)()()()( axaaxaaaxaxSn

nn

2210

0

)( xaxaaxaxSn

nn

Ratio test of the convergence of a power series:

.1

and 1

points end for the needed isst Further te )3

.1

or 1

when diverges series The 2)

.) if no 0, if (all 11

when converges series The 1)

.lim)(

)(lim 1

11

lax

lax

lax

lax

lxlxl

axl

a

axlaxa

a

axa

axa

n

n

nnn

nn

n

Example p21.2,3; Problems 10.2,7,17.

21

1.11 Theories about power series1)A power series may be differentiated or integrated term by term. The resultant series converges within the original interval of convergence (but not necessarily at the endpoints).

2)Two power series may be added, subtracted, or multiplied. The resultant series converges at least in the common interval of convergence. Two power series can be divided if the denominator series is not 0, and the resultant series has some interval of convergence.

3)One series can be substituted into another if the value of the substituted series is in the interval of convergence of the other series.4)The power series of a function is unique.

These theories are easy to understand if we treat a power series as a well defined function.

1

1

12

).1,1[in converges )(' .]1,1[in converges )( :Examplen

n

n

n

xn

xxfx

n

xxf

).2/1,2/1(in converges ])2([ :Example0

n

nn xxx

22

1.12 Expanding functions in power seriesTaylor series: A Taylor series expansion of a function f (x) at x=a has the form

.)(2

)("))((')()(

!

)()(

Therefore .!)( and , ,2)(" ,)(' ,)(hat can test t We

.)()()()(

2

0

)(

)(210

0

2210

axaf

axafafaxn

afxf

anafaafaafaaf

axaaxaaxaaxf

n

nn

nn

n

nn

Maclaurin series: The Taylor series expansion of a function f (x) at x=0:

.2

)0(")0(')0(

!

)0()( 2

0

)(

xf

xffxn

fxf

n

nn

23

1.13 Techniques for obtaining power series expansionsSome important Taylor series expansions (printed in head):

).1,1( ,11

1

).1,1( ,11

1

especially ,expansion) (Binomial

)1,1( ,!3

)2)(1(

!2

)1(1)1(

]1,1( ,432

)1()1ln(

!4!3!21

!

!6!4!21

)!2(

)1(cos

!7!5!3)!12(

)1(sin

32

32

3

0

2

432

1

1

43

0

2

64

0

22

75

0

312

xxxxx

xxxxx

xxmmm

xmm

mxxn

mx

xxxx

xn

xx

xxxx

n

xe

xxx

n

xx

xxxx

n

xx

n

nm

n

nn

n

nx

n

nn

n

nn

24

Read: Chapter 1:10-13Homework:1.10.2,7,15,23.Due: February 7

25

February 3 Power series expansion1.13 Techniques for obtaining power series expansionsA. Multiply a series by a polynomial or by another series

Example p26.1,2; Problem 13.5,9.

B. Division of two series or of a series by a polynomial

Example p27.1; Problem 13.22.

C. Binomial series

Example p29.2.

D. Substitution of a polynomial or a series for a variable in another series

Example p29.1; Problem 13.10,27.

E. Combination of methods

Example p30; Problem 13.13.

F. Taylor series using the basic Maclaurin series

Example p30.1; Problems 13.39.

26

1.14 Accuracy of series approximationsError in a convergent alternating series:

.)()(

.0)()(then

]).(consider we[otherwise ,0 Suppose

:Proof

term.neglectedfirst than theless iserror theis,That

.)(

reminder then the,0lim and with series galternatinan is If

154321

4321

11

121

11

nnnnnnn

nnnnn

nnn

nnn

nn

nnn

n

aaaaaaR

aaaaR

aa

aaaaSR

aaaaS

Example p34.1; Problem 14.4.

Note: This rule only applies to an alternating series.

27

Theorem:

1

11

1110

11

0

10

1

:Proof

.1

then,for and ,1for converges If

Nn

NNn

NNn

n

nNn

nn

N

n

nn

NN

N

n

nn

nnn

nn

x

xaxaxaxaxaS

x

xaxaS

NnaaxxaS

Problem 14.6.

28

Read: Chapter 1:13-14Homework:1.13.5,8,14,20,28,42;(Find first 3 terms. No computer work needed.)1.14.6.Due: February 14