01 Infinite Sequences and Series Slide

8/6/2019 01 Infinite Sequences and Series Slide

1/72

CALCULUS II MA 1123

Infinite Sequences andSeries

Telkom Institute of Technology

Bandung - Indonesia


2/72

Infinite Se uencesInfinite Se uences

Definition

Sequences or infinite sequence is a function whosedomain is the set of positive integers.

Notation: f : N R

= nThis function is also sequence of real number. We

denote a sequence by a1, a2, a3, ...., by or{ }

=0nna n n .

Formula of sequence:

1. An explicit formula for the nth term2. The first few terms of the sequence

3. A recursion formula1 111 naaa == 1

8/11/2008 2Calculus II (MA 1123) Faculty of Science Telkom Institute of Technology

n n ...,

4,

3,

2,

n

n

a++

1


3/72

Conver ence of Se uenceConver ence of Se uence

Definisi:e say a sequence an converges o or as m

L and write

La =lim

If for each positive number , there is acorres ondin ositive number N such that

n

A sequence that fails to converge to any finite


4/72

Notes:Notes:

Here and is many sequence problem, it is convenient touse the followin almost obvious fact.

Lxfx

=

)(limIf Lnfn

=

)(lim, then

This is convenient because we can apply lHopitals Rule

to the continuous variable problem.



5/72

Pro erties of Limit Se uencePro erties of Limit Se uence

Let {an} be convergent to L and {bn} be convergent toM, then

1. MLblimalimbalim ==nnn

2.( ) ( ) ( )

M.Lblim.alimb.alimnnnnnnn

==

3.( )

M

L

blim

alim

b

alim

n

nn

n

n

n==

, untuk M 0



6/72

Monotone Se uenceMonotone Se uence

Definition:

A sequence {an} is called

(i) increasing if an < an+1

(ii) nondecreasing if an an+1

(iii) decreasing if an > an+1

(iv) nonincreasing if an an+1



7/72

Exam leExam le

Determine if the sequence coverges or diverges :

1n2na n =

1.

Answer:

We has )( =xxf , in particular, I Hopital Rule,

2

1

12lim)(lim =

=

x

xxf

xx

that is, an converges to .

2

1

12lim =

nn

n

,



8/72

Exam le Cont.Exam le Cont.

2.

n

n

11a

+=

Answer:We has

x

xxf

+= 11)( , In particular, IHopitals Rule,

+= x

xx

11ln.limexp

+=

x

x 1

1ln

limexpx

x x

+

11lim

ee

x

x

x==

+

=

1

1

limexp

+

=

2

2

1

1.

1

limexp x

x

xx

en

n

n=

+

11lim

Thus, x


that is, an converges to e.


9/72

Problem SetProblem Set

2

Determine if the sequence coverges or diverges :

3n2n

a2n

+

=

2n32 +

=

...4

,3

,2

,1

an+1 =

2

(an +

na

) , a1=21. 7.

1nn +

na n =

...9

5,

7

4,

5

3,

3

2,1

.

9.3.

( )n

n

n4

a

=

...

4

31

1,

3

21

1,

2

11

1,110.4.

na n =

...

15

4,

14

3,

13

2,

12

1an+1 = 1 +

2

1an , a1=1 11.6.

5.


5432


10/72


11/72

Se uence of Partial SumsSe uence of Partial Sums

Let Sn denote the sum of the first n terms of the series

, then

=1i

ia

S1 = a1

.

.

.Sn = a1 + a2 + a3 + a4 + + an =

=iia

1

n =1i

i

Sequence {Sn} is called the sequence of partial sums



12/72

Definition: Convergent of Infinite SeriesDefinition: Convergent of Infinite Series

The infinite series

ia conver es and has sum S if the=1i

sequence of partial sums {Sn} converges to S, if {Sn}

diverges, then the series diverges.



13/72

Geometric SeriesGeometric Series

A series of the form

=

1n

1nar =a +ar +a r

2

+ ... + a r

n-1

+ ...

w a .

The partial sum of geometric series isnSn =

=

1i

1iar = a +ar +a r + ... + a rn-

r1an

nr1

, .



14/72

Pro erties ofPro erties of Geometric SeriesGeometric Series

n .

n

nrlim

= 0, then the series converges to

r1a

2. if then sequence {rn} diverges because

r > 1 nn

rlim

= ,



15/72

Exam leExam le

...11111

+++++1. Show converges

Answer:Consider the partial sums

S1 =2

= 1 -2

S2 =42

+ =4

= 1 (2

)2

S3 = 111 ++ = 7 = 1 ( 1 )3

then,

S = 1 1 n

2and

nn

Slim

=n

lim (1 (2

1)n) = 1

so because the se uence of artial sums Sn


converges to 1, then that infinite series converges to 1.


16/72

Exam le 2Exam le 2

1

2.2. (Kolaps Series). Show the infinite series converges.=1i

Answer:Consider the partial sums

)1i(i +=

i-

1i +

then,

and

Sn =

+

/

++

/

/

+

/

/

+

/

1nn

...43322

1 =

+

1n

1

so because the se uence of artial sums Sn

nn

Slim

=n

lim

+

1n

1 = 1


converges to 1, then that infinite series converges to 1.


17/72

Exam le 3Exam le 3

3.3.

= i

1(Show that the harmonic series diverges)

Answer:

We show that Sn grows without bound. Imagine n to beLarge and write

Sn = 1 +n

...8765432

++++++++

Sn = 1 + 1...1111111 ++ ++++ ++n

1 +

n

1...

8

1

8

1

8

1

8

1

4

1

4

1

2

1++

++++

++

It is clear that by taking n sufficiently large, we can

= 1 +n

...2222

+++++


.n

, ,so does the harmonic series.


18/72

Dever ence tests ofDever ence tests ofnnth termth term

If the series

=0nna converges, then n

nalim

= 0, ekivalen if

nn

alim

0 then the series diverges.

2n

= ++1n 4n3n3

That series diverges sincen 2 1 1

4n3n3m

2n ++=

2

n

n

4

n

33

m

++ 3

= (Tidak Nol)

So proved that diverges.

= ++1n2

2

4n3n3

n



19/72

The New ProblemThe New Problem

To prove that a series converges is does not suffice to

s ow t at nn

am

= , s nce t s property may o or

divergent as well as convergent series.

=1n n

1

=1 + n

1

...8

1

7

1

6

1

5

1

4

1

3

1

2

1

++++++++ + . . .It is clear that n

nalim

= 0, but the harmonic series diverges.

.



20/72

Positive Series TestPositive Series Test

1. The integral test

Let function f be continuous, positive, nonincreasing

function the interval [1,)

.

1,

=1n

)n(f converges.

. ,

=1n

)n(f diverges.1



21/72

Exam leExam le

1. Determine the series

n 2en converges or diverges=1n

Answer. We has2

xex)x(f = , thus2 2b

b1 2dxex

1

xexm1b

1bxem

2b

x 2elim1 11lim1 1

= =

= = =1 eebb

e

So because dxex2x

converges, then

n 2en

=n

also converges.



22/72

Exam leExam le

2. Determine the series

1converges or diverges

Answer. We has , thus

=2n

xxxf

ln1)( =

=b

b

xx

dx

xx

dx

22

ln

lim

ln

=

2

ln

)(lnlim

x

xd

b

bb

dx 1,

also diverges.

2 lnxx=2n

nlnn



23/72


24/72


2. Convergence ofp-series

e ser es

=1

1i

piw ere p s a constant, s ca e p-ser es.

By the integral test, we have

1

tp1x

1t p1

xx

m1 pt

=1

t p1 =

p1t

Note the case

1. p = 1 gives the harmonic series, then the seriesdiverges

2. p > 1 givesp1tlim = 0, then the series converges


t


25/72


3. p < 1 givesp1tlim =, then the series diverges.

4. p < 0, the nth term of =n

i Pi1

1, namely,

doesnt evenPn

1

Tend toward 0. So the series diver es b nth term test.

So we has, namely:

-.2. The p-series diverges if 0 p 1



26/72

Exam leExam le

Does the following series converge or diverge?

1.

=1

001,1

nn

-

1

=1,

n n

because p=1,001 > 1

1.

By the p-series test, the series diverges

=1 2n n

= 21

1

because p= < 1



27/72


28/72

Exam leExam le

Does the following series converge or diverge:

n.

= 3n2 5n

Answer:

1

n

n n

n=

5n 2 we know that

,

=1n n

1

is harmonic series andn

1

5n2 n

, so ecause t e ser es=1n

n

2

n

iverges, t en

the series diverges.= 2n n



29/72

Exam leExam le

1

Answer:

= +1n 5n

We will comparing the series with bn= and an=

we know that the series is p-series with p = 2 > 1

5n2 +2n

21

, so because the series

2n

1

5n

12 +

=1n

and

=1n2n

1converges

then the series converges.= +1n

2 5n



30/72


Does the following series converge or diverge

= +1n2

5n ( )= 3n22n

4.1.

= 3n2

5n

1 = 1n 1n2

12. 5.

= +1nn 12

13.



31/72

4. Limit Comparison Test

Suppose an and bn positive series andn

n

n b

a

lim = L

.= `1n

n

= `1nn

diverge together

2. i L = 0 an= `1n

n= `1n

naconverges, t en converges.



32/72

Exam leExam le

Does the following series converge or diverge :

= +1n23 7n5n

1.

Jawab:

If we note the series, the nth term is like bn=

Thus,

2n

32n +

n

n

n b

alim

= 2

1

2

23

175lim

n

nnn

+= 75

lim23 +

+=

nn

nn

n

+

23

3n2the series converges.

So because L=2 and the series=1

2n n

converges, then


=n


33/72

Exam leExam le

12.

Answer:

= +1n 2 4n

If we note the series, the nth term is like bn =

thus,

n1

n

n

n b

alim

= 1

n1

4nlim

2

n

+ 4n

nlim

2

2

n += =

the series diverges.

So because L=1 and the series diverges, then

= +1n 2 4n

1 =1nn



34/72


Does the following series converge or diverge:

= ++1n

2

3n2n

= 1n3

4n

4.1.

= +1n 1nn

=1n

2n2.5.

=

+

1n2n

3n23.



35/72

5. Ratio Test

=1k kaa

Let be a series of positive term and

= kk a

m

=ka1. if < 1, the series converges.

suppose

=1kka diverges.2. if > 1, the series

= , ..



36/72

Exam leExam le

Does the following series converge or diverge: 3n

.=1 !n n

n 1+nAnswer:

Suppose nth term is an =!n, then (n+1)th is an+1=

thus,( )!1+n

1+n31n+

3

n

( )1lim

+=

nn0=

( )!13lim

+=

nnn!

3lim

n

nn=

n

n

n a1lim +

o ecause = < , en e ser es=1 !n n

converges.



37/72


38/72


Does the following series converge or deverge:

=1n

n

n

=1 !n n 3

4.1.

( )

=1 !2n n

n

( )=1 !2n n2. 5.

=

+

1 !

4

n

n

n

n3.



39/72

6. The Root Test

=1k kaLet be a series with a positive terms and suppose

ka1. if a < 1, the series converges

kk

=

=1kka diverges

=

2. if a > 1, then the series

3. if a



40/72

Exam leExam le

Does the following series converge and diverge: + 22

n

n.

= 1 1n n

Answer: n

Suppose nth is an =n

1

, thus

22 +n1

=

= nn

nn

So because a = 2 (> 1), then the series

=

+

1

22n

n

n

diverges.



41/72

Exam leExam le

+ 2n

n.

= 1 12n n

Answer: n

Suppose nth term is an =n

12

, thus

12+nn212 nnnn

n

So because a = (< 1), then the series =

1 1n n

converges.



42/72


Does the following series converge or diverge: n n

=

1 lnn n =

+1 2n n

3.1.

=

+

1 12

23

n

n

n

n

=

+1 23n n

n2. 4.



43/72

Alternating SeriesAlternating Series

The series of the form

( ) ...aaaaa1 43211n

n1n ++=

=

+

with an > 0, for all n.

An important example is the alternating harmonic

series, namely;

( ) ...4

1

3

1

2

11

n

11

1n

1n ++=

=

+



44/72


45/72

Exam leExam le

1. Answer (Alternating series test)

1

1

e ave an= n , an an+1 = 1+n

since

, e ser es converges,

1a. 11

11

1

>+==+

=+ nnn

n

an

n an >an+1

1.

nnn

n

So because a dan b proved, the alternating seriesconverges.



46/72

Exam leExam le

2. Answer (alternating series test)

1

1

e ave an=!n, an an+1 =( )!1+n

since

, e ser es converges,

1

a.

( )11

!11

1

>+=+

=+

n

nan

n an >an+1

1.! nn

nn

So because a dan b proved, the series converges.



47/72


Does the following alternating series converge or deverge:

( )=

+

+

1

1

13

1n

n

n

)=

1 3

1

n

n

n

4.1.

( )

= ++

1

21

n

n

nn

n ( )= +

1 )1(

1n

n

nn2. 5.

( )

=

+1

1

!1

n

nn

n

n3.



48/72

Absolute Convergence and ConditionalAbsolute Convergence and Conditional

If the series is said to converge adsolutely if the series

=1nnb

In the other words, a series

converges absolutely, then

=1nnb converges.

of absolute values converges.

nb

nb converges.

And the series is said converge conditionally if the series

diver es but=1n =1n



49/72

Ratio Test for Absolute ConvergenceRatio Test for Absolute Convergence

Let

na with an 0 andna 1lim +

= r, then=1n n

1. if r < 1, the series converges absolutely2. if r > 1, the series diverges

. ,convergence can be drawn from this test.



50/72

Exam leExam le

Does the following series conditional converge, absoluteconver e or diver es:

1.( )

=

+1

1

!

21

n

nn

n

We have an= ( )!

21 1

n

nn+ , and an+1 =( )

( )!1

21

12

+

+

+

n

nn

thus,

( )

( ) 21

!11

limlim1

1 n

a

ar

nn

n

nn

n

n +

+

+

+==

( )!12!2

lim1

+=

+

n

nn

n

n 1

2lim

+=

nn0=

We conclude from the Absolute Ratio Test, the seriesabsolute converges.



51/72

Exam leExam le

2. ( )

+ 11

1 n

=n

Answer:

+ 11

1 n =1n n

diverges, since it is a p-series

=

=

=1 1

1

n n

n

n

a

However,

( )

+ 11

1n

So the series

with p = .

conditional converges.=n



52/72


Does the following series conditional converge,

( )

1n

n n ( )

11 n1. 4.

,

=1n

)4(n

=1n

+

1

)1(n

=12

n n

n

=1 lnn nn

+

1n

. .

= +1 23n n = +1 1n nn3. 6.



53/72

Power SeriesPower Series

The power series has two form:

1. A power series in x has the form

nxa = 2

=0n

2. A power series in (x b) has the form

( )

=

0n

nn bxa = a0 + a1 (x-b) + a2 (x-b)

2 + . . .

For this time, we talk about convergence set / for what xsdoes the power series converges.



54/72

Radius of Conver enceRadius of Conver ence

Radius of conver ence found with the ratio test forabsolutely converges :

Let ( )

nn bxa and nn

n bxaL)(

lim1

1 =+

+

=0n n

1. If L < 1, the series converges.. , ,tried.



55/72

nx


56/72

AnswerAnswer nx

=0n

. .

n

n

n

n

n

xxL :lim

1

1

= ++

x=

)1(lim +=

nx

So the series converges absolutely if L< 1, namelyIf x = 2 or x = -2 , the ratio fails. However, When x = 2

( ) ( )

=

= +=

+ 11 11

21

2

nnn

n

nn


, .


57/72

AnswerAnswer

When x = 2

The series is the alternatin harmonic series which

( ) ( )==

+

=+

11 121 nnn nn

We conclude that the convergence set is the interval

converges.



58/72

!)1( n


59/72

Answer 3Answer 3 + !)1(nxn

=

. e app y t e rat o test or a so ute convergence.

( ) nn xnxn

L!1

lim+

+=

( )xn

n2lim +=

==

0,

,

xjika

xa

= .



60/72

Theorem 1Theorem 1

n

nxa is alwa s=0n

an interval of one of the following three types.

1. The single pointx= 0.2. An interval (-c, c), plus possibly one or both endpoints.

3. The whole real line.



61/72

Theorem 2Theorem 2

nbxa is alwa s=0n

an interval of one of the following three types.=. .

2. An interval (-c + b, c + b), plus possibly one or bothendpoints.



62/72


Find the convergence set of the given power series: n

( )= +

02

1n n

1.

...

81.4

4ln2

27.3

3ln2

9.2

2ln2

3

2+

++

++

++

+ xxxx2.

( ) ...!3!2

2 ++

++

++xx

x3.

32n

x nn

nn

( )= +02

41nnn

4.( )= +

0

221n

nn5. ( )= +0

231n

nn6.



63/72

O eration on Power SeriesO eration on Power Series

We have done this for one series, a geometric series.

11,


64/72

Theorem 3Theorem 3

Suppose that S(x) is the sum of a power series oninterval I that is

S(x)=

=0n

n

nxa = a0 + a1x+ a2x2 + a3x3+ . . .

[ ]

=0n

nnxaD

Then,1. S(x) = = D[a0 + a1x+ a2x

2 + a3x3+ . . .]

2

=

1

1

n

nn xna

x x

1 1 1

1 2 3 . . .

=

dttS0

)(=0

0n

n tta

=

+

+1

1

nn xn

a

2 3 4=

=

= a0x+ a1x + a2x a3x+ . . ..



65/72

Exam leExam le

Apply Theorem 3 to the geometric series

x1= 1 +x + x2 +x3 + . . . for -1


66/72

ContohContoh

a. ln (1 x)

Integrating term by term givesx

++++== dttttdttx0

32

0

...11

)1ln(

111111 432432x

...432

...432 0

, -1


67/72


Find the power series representation for f(x):

x

xf

+

=

1

)(

=x

x1

ln1

1. 5. f(x)=tan-1(x)

x 122

+ x1( )21 x+..

1=xx ++ 11

.( )x32 +

.

2

1)(xf =4.

x



68/72

Taylor and MaclaurinTaylor and Maclaurin SeriesSeries

Definition: Let f(x) be function with derivatives of all orderatx=b. then we define the Taylor series for fabout x = b

to be )( )( nn bf

)()('' 2bxbf

=

0

!n

nx = = - . . .

In the special case where b = 0, the taylor series for f

!2

)0(" 2xf

, ,

( ) )(

!

)0( nn

x

n

ff(x) = = f(0) + f(0)(x)+ + . . .

=n



69/72

Exam leExam le

Find the Maclaurin Series for the given function:

=.

Answer:

=

f(x) = cosx

f(x) = - sinx f(0) = 0

f(0) = 1

f(x) = - cosx f(0) = -1

flV (x) = sinx flV(0) = 0

thus,753

xxx +12n

n x


...

!7!5!3 ( )= +=

0 !12n n


70/72

Exam leExam le

2. f(x)= ex

f(x) = ex

fx = ex

f(0) = 1

f(x) = ex f(0) = 1

f(x) = ex f(0) = 1

flV (x) = ex flV(0) = 1

...!4!3!2

1)(432

+++++==xxx

xexf x

=

=0 !n

n

n

x


E lE l


71/72

Exam leExam le

Find the Taylor series about x = 1 for f(x)= ex

f(x) = ex

f(x) = ex =

f(1) = e

f(x) = ex f(1) = e

f(x) = ex

f(1) = eflV (x) = ex flV(1) = e

thus,

( ) ( )...

!3

1

!2

1)1()(

32

+

+

++==x

ex

exeeexf x( )

=

=

0 !

1

n

n

n

xe



72/72

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01 Infinite Sequences and Series Slide