Chapter 8 Infinite Sequences and Serieslballou/chap8.pdf · 1 Chapter 8 Sequences and Infinite Series. Section 8.1 Overview. An infinite sequence is a list of numbers in a . definite

1

Chapter 8 Sequences and Infinite Series Section 8.1 Overview An infinite sequence is a list of numbers in a definite order

1 2 3, , , , ,na a a a For example, consider the infinite sequence

2,4,8, , 2 ,n

the first term 1 2a = , the second term 2 4a = , and the nth term is 2nna = . Note the integer n is

called the index of the sequence and indicates the position where na occurs in the list. Definition An infinite sequence of numbers is a function whose domain is the set of positive integers. Notation for sequences: List terms: { }1 2, , , ,na a a

Write nth term of sequence { }na

Write nth term of sequence with an index { } 1n na ∞

= (note we can start with any number n

we choose. Examples:

2

1 3 4 5 1, , , , ,1 1 2 3 1n

n nn n

∞

=

+ + = − −

1 1 1 1 1 1, , , , , ,! 1 2 6 24 !n n

=

( ){ } ( ){ }0

1 1, 1,1, 1, , 1 ,n n

n

∞

=− = − − −

Some sequences do not have a defining equation. For example, The sequence of the population size on Jan 1st of each year. Decimal values of π .

{ }1,4,1,5,9,

Fibonacci sequence { }na is defined recursively

1 2 1 21, 1, for 3n n na a a a a n− −= = = + ≥

{ }1,1,2,3,5,8,13,21,

Example: Suppose the list of term of a sequence is 2 4 81, , , ,3 9 27

− −

, write an expression for

the nth term, na .

2

8.2 Sequences From section 8.1: Definition If the terms of a sequence { }na approach a unique number L as n increases, then we say lim nn

a L→∞

= exists, and the sequence converges to L. If the terms of the sequence do not approach

a single number as n increases, the sequence has no limit and the sequence diverges. Theorem 8.1 Limits of Sequences from Limits of Functions Suppose f is a function such that ( ) nf n a= for all positive integers n. If the ( )lim

xf x L

→∞= ,

then the limit of the sequence { }na is also L. Theorem 8.2 Properties of Limits of Sequences Suppose that c is a constant and the limits lim nn

a→∞

and lim nnb

→∞ exist

1. Constant Multiple Rule: lim limn nn nca c a

→∞ →∞=

2. Sum/Difference Rule: [ ]lim lim limn n n nn n na b a b

→∞ →∞ →∞± = ±

3. Product Rule: [ ]lim lim limn n n nn n na b a b

→∞ →∞ →∞⋅ = ⋅

4. Quotient Rule: lim

limlim

nn n

nn nn

aab b

→∞

→∞→∞

=

provided lim 0nn

b→∞

≠

Examples: Determine if the sequence{ }na converges, if it converges find the limit.

1. 2

2

5n

nan n3+

=+

2. 2

21n

nan

=+

3. 3

21nna

n=

+

4. lnln 2n

nan

=

3

5. 1sinna nn

=

6. 21n

nan

= +

Definition A sequence is called non-decreasing if 1n na a+ ≥ for all n. A sequence { }na is called non-increasing if 1n na a+ ≤ for all n. A sequence that is either non-increasing or non-decreasing is said to be monotonic.

Example: Show 12 3na

n=

+ is non-increasing.

Example: Show 1na n

n= + is non-decreasing:

Definition: A sequence { }na is bounded if there exists a number M such that na M≤ for all n. Theorem 8.5 Bounded Monotonic Sequence A bounded monotonic sequence converges Theorem 8.3 Geometric Sequence Let r be a real number. Then

0 1lim 1 1

1 1

n

n

if rr if r

DNE if r or r→∞

<= = ≤ − >

If 0r > , the { }nr converges or diverges monotonically . If 0r < , then { }nr converges or diverges by oscillation. Example:

4

Theorem 8.4 Squeeze Theorem for Sequences Let { } { } { }, ,n n na b and c be sequences with n n na b c≤ ≤ for all integers n greater than some N. If lim limn nn n

a c L→∞ →∞

= = then lim nnb L

→∞= .

Example : Find the limit of sin 21n

nan

=+

Theorem 8.6 Growth Rate of Sequences The following sequences are ordered according to increasing growth rates as n →∞ ; that is, if

{ }na appears before { }nb in the list, then lim nn

n

ab→∞

=0:

{ } { } { } { } { } { } { }ln ln !q p p r p s n nn n n n b n n+

The ordering applies for positive real numbers p, q, r, and s and 1b > .

Example: Find the limit of !n n

nan

=

Definition A sequence { }na has limit L and we write

lim nna L

→∞= or as na L n→ →∞

if we can make the terms na as close to L as we like by taking n sufficiently large.[Formally, if for every 0ε > , there exists a N such that na L ε− < whenever n N> .] If lim nn

a→∞

exists, we say that the sequence converges (or is convergent) and converges to L.

Otherwise, we say that the sequence diverges.

5

Section 8.3 Infinite Series Definition An infinite series is an expression of the form

1 21

n nn

a a a a∞

=

= + + + +∑

Where { }na is an infinite sequence of real numbers.

Can an infinite series have a sum? Look at 1

1 1 1 1 12 2 4 8 16n

n

∞

=

= + + + +∑

Definition The nth partial sum of a series is

1 21

n

n n ii

s a a a a=

= + + + =∑

So for our example, 112

s = , 21 1 32 4 4

s = + = , 31 1 1 72 4 8 8

s = + + = , 41 1 1 152 4 8 16 16

s 1= + + + = , notice

that the partial sums are approaching one. Thus, we say that the series 1

12n

n

∞

=∑ converges and

1

1 12n

n

∞

=

=∑ .

Note: The partial sums of a series will form a sequence!

Some series will diverge, i.e. will not have a sum. For instance, consider 1nn

∞

=∑ , it is a divergent

series, the nth partial sum goes to infinity as n goes to infinity. Definition

The series 1

nn

a∞

=∑ is said to converge with sum S if the sequence of partial sums { }ns converges

to S .

1 1lim lim

n

k k nn nk ka a s S

∞

→∞ →∞= =

= = =∑ ∑

If the sequence { }ns does not converge, the series diverges and has no sum.

Example: Is the series ( )1

1 n

n

∞

=

−∑ convergent or divergent?

6

Example: Telescoping Series Show that 21

1n n n

∞

= +∑ converges and find it sum.

Definition: Geometric Series

The series 1

nn

a∞

=∑ is said to be a geometric series if each term after the first term is a fixed

multiple of the term immediately before it 2 3

0

n n

na ar ar ar ar ar

∞

=

+ + + + + + =∑

The geometric series 0

n

nar

∞

=∑ with 0a ≠ diverges if 1r ≥ and converges if 1r < with sum

0 1n

n

aS arr

∞

=

= =−∑

Proof: If 1r = , then ns a a a na= + + + = → ±∞ , since lim nn

s→∞

does not exist, then the geometric series

diverges. If 1r ≠ we have

2 1nns a ar ar ar −= + + + +

And 2 3 n

nrs ar ar ar ar= + + + + Subtracting theses equations we get

nn ns rs a ar− = −

( )11

n

n

a rs

r−

=−

If 1r < , ( )1

lim lim lim1 1 1 1

nn

nn n n

a r a a as rr r r r→∞ →∞ →∞

−= = − =

− − − −

Thus when 1r < , the geometric series is convergent with sum 1

ar−

and if 1r ≥ the series is

divergent.

Example: Does 0

37 2

n

nn

∞

= ⋅∑ converge or diverge? If it converges, what is its sum?

7

Example: Does 0

43n

n

∞

=∑ converge or diverge? If it converges, what is its sum?

Example: Does 2 1

1

25

n

nn

+∞

=∑ converge or diverge? If it converges, what is its sum?

Example: Write 15.423 as a rational number.

8

Section 8.4 The Divergence and Integral Test Theorem 8.8 Properties of Convergent Series 1. Suppose ka∑ converges to A and let c be a real number. The series kca∑ converges and

k kca c a cA= =∑ ∑ .

2. Suppose ka∑ converges to A and kb∑ converges to B. The series ( )k ka b±∑ converges

and ( )k k k ka b a b A B± = ± = ±∑ ∑ ∑ . 3. Whether a series converges does not depend on a finite number of terms added to or removed

from the series. Specifically, if M is a positive integer, then 1

kk

a∞

=∑ and k

k Ma

∞

=∑ both converge

or both diverge. However, the value fo the convergent series does change if the nonzero terms are added or deleted.

Example: Find the sum of 1

1 5 3 73 6 5 9

k k

k

∞

=

+

∑

Theorem 8.9 Divergence Test If ka∑ converges, then lim 0kk

a→∞

= . Equivalently, if lim 0kka

→∞≠ , then the series diverges.

Proof: Let 1 2 1k k ks a a a a−= + + + + , so 1n n na s s −= − .

Since 1

kk

a∞

=∑ is convergent then the sequence { }ks of the partial sums is convergent and

lim kks S

→∞= so ( )1lim lim 0k k kk k

a s s S S−→∞ →∞= − = − = .

Example: Show 1arctan

nn

∞

=∑ is divergent.

Example: Show 2

21

3 5n

nn n

∞

=

++∑ is divergent.

Note: If lim 0nna

→∞= , we do not know if the series

1n

na

∞

=∑ converges or diverges.

9

Theorem 8.11 The Integral Test Suppose f is a continuous, positive, decreasing function for 1x ≥ and let ( )ka f k= for

1,2,3k = . Then

1k

ka

∞

=∑ and ( )

1

f x dx∞

∫

Either both converge or both diverge. In the case of convergence, the value of the integral is not, in general, equal the value of the series. Consider the following:

(i) If ( )1

f x dx∞

∫ is convergent, then

( ) ( )2 1 1

nn

ii

a f x dx f x dx∞

=

≤ ≤∑ ∫ ∫ since ( ) 0f x ≥

na

2a

3a

4a

( )2 1

nn

ii

a f x dx=

≤∑ ∫

( )1

11

nn

ii

f x dx a−

=

≤⌠⌡

∑

1na −

1a

2a

3a

10

Therefore, ( )1 12 1

n

n ii

S a a a f x dx M∞

=

= + ≤ + =∑ ∫ . Since nS M≤ for all n then the sequence { }nS

is bounded from above. Also 1 1n n n nS S a S+ +≤ + = since ( )1 1 0na f n+ = + ≥ Thus { }nS is a non-

decreasing bounded from above sequence so it is convergent. Thus 1

nn

a∞

=∑ is convergent.

(ii) If ( )1

f x dx∞

∫ is divergent, then

( )1

as n

f x dx n→∞ →∞∫ because ( ) 0f x ≥ . But ( ) 11

n

nf x dx S −≤∫ so 1nS − →∞ which implies

that nS →∞ and so 1

nn

a∞

=∑ is divergent.

Example: Show 1

k

ke

∞−

=∑ is convergent with the integral test.

Example: Show the Harmonic Series, 1

1k k

∞

=∑ is divergent with the integral test.

Example: For what values of p are the p-series 1

1p

k k

∞

=∑ is convergent.

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Theorem 8.12 Convergence of the p-Series

The p-series 1

1p

k k

∞

=∑ converges when 1p > and diverges when 1p ≤ .

To estimate the sum of a series, we add the first n terms together. Now the error of estimation is the difference between the actual sum, S, and the partial sum, nS . We will call the difference the remainder, nR . Now

1 2n n n nR S S a a+ += − = + + We can bound the remainder by comparing the sum of the areas of rectangles with area under the curve ( )y f x= for x n≥ . We see that

( )nn

R f x dx∞

≤ ∫ and ( )1

nn

R f x dx∞

+

≥ ∫

So

( ) ( )1

nn n

f x dx R f x dx∞ ∞

+

≤ ≤∫ ∫

which implies that ( ) ( )1

n nn n

S f x dx S S f x dx∞ ∞

+

+ ≤ ≤ +∫ ∫

Example: Estimate the sum of the series 21

1n n

∞

=∑ using the inequality above and 10n = .

12

Section 8.5 The Ratio, Root and Comparison Tests Theorem 8.14 The Ratio Test Let ka∑ be a series with positive terms and suppose that

1lim kk

k

aa

ρ+

→∞=

Then 1. If 1ρ < , the series converges. 2. If 1ρ > or ρ is infinite, the series diverges 3. If 1ρ = , the test is inconclusive.

Comment: This test is very useful if ka contains a factorial or kc where c is some constant. However, this test will be inconclusive if ka looks like a p-series.

Example: Determine if the series 1

!10k

k

k∞

=∑ is convergent or divergent.


32

k

kk k

∞


Theorem 8.15 The Root Test Let ka∑ be a series with positive terms and suppose that

lim kkk

a ρ→∞

=

Then 1. If 1ρ < , the series converges. 2. If 1ρ > or ρ is infinite, the series diverges 3. If 1ρ = , the test is inconclusive.

Comment: This test is useful when it is easy to take the kth root of ka .


31

49 1

k

k

k kk k

∞

=

+ + +

∑ is convergent or divergent.

13

Example: Determine if the series ( )2

1 2

k

kk

k∞



2 k

k

kk

∞

=

−


Theorem 8.16 Direct Comparison Test Suppose that ka∑ and kb∑ are series with positive terms

1. If 0 k ka b< ≤ and kb∑ is convergent then ka∑ is convergent.

2. If 0k ka b≥ > and kb∑ is divergent then ka∑ is divergent.


23 1

k

kk

∞

= +∑ is convergent or divergent.

Example: Determine if the series 2 1k

kk

∞

= −∑ is convergent or divergent.

14

Theorem 8.17 Limit Comparison Test Suppose that ka∑ and kb∑ are series with positive terms and

lim kk

k

a Lb→∞

=

1. If 0 L< < ∞ then ka∑ and kb∑ both converge or both diverge.

2. If 0L = and kb∑ converges then ka∑ converges.

3. If L = ∞ and kb∑ diverges, then ka∑ diverges.

Example: Determine if the series 3 7 2

1

5k

kk k

∞

=

+

+∑ is convergent or divergent.


31

51k

k kk k

∞

=

−+ −∑ is convergent or divergent.

Example: Determine if the series 1.11

arctank

kk

∞



11 lnk k

∞

= +∑ is convergent or divergent.

15

Section 8.6 Alternating Series, Absolute and Conditional Convergence A series where the terms alternate between positive and negative is called an alternating series:

( ) ( ) 1

1 11 or 1k k

k kk n

a a∞ ∞

+

= =

− −∑ ∑

Where ka is a positive number.

For instance, the alternating harmonic series is ( ) 1

1

11 k

k k

∞+

=

−∑

( ) 1

1

1 1 1 11 12 3 4

k

k k

∞+

=

− = − + − +∑

Theorem 8.18 Alternating Series Test The series

( ) 11 2 3 4

11 k

kk

a a a a a∞

+

=

− = − + − +∑

Converges if the following conditions are satisfied 1. 1 0 k ka a k N+ ≥ > ∀ > (i.e. the terms in the series are non-increasing). 2. lim 0kk

a→∞

=

Example: Determine if the alternating harmonic series ( ) 1

1

11 k

k k

∞+

=

−∑ is convergent or divergent.

Example: Determine if the series ( ) 21

211

k

k

kk

∞

=

−+∑ is convergent or divergent.


1 cosk

k kπ∞

=

−


16

Theorem 8.20 Remainder in Alternating Series

If ( ) 1

11 k

kk

S a∞

+

=

= −∑ is the sum of an alternating series that satisfies the conditions of the

Alternating Series Test then 1n n nR S S a += − ≤

Example: Find the sum of the series ( )0

1!

n

n n

∞

=

−∑ correct to three decimal places.

Definition Absolute and Conditional Convergence Assume the infinite series ka∑ converges. The series ka∑ converges absolutely if the series

ka∑ converges. Otherwise, the series ka∑ converges conditionally.


100!

k

k k

∞

=

−∑ converges absolutely, converges conditional or

diverges.

Example: Determine if the series ( ) 1

41

1 k

k k

+∞

=

−∑ converges absolutely, converges conditional or

diverges.

17

Theorem 8.21 Absolute Convergence Implies Convergence If ka∑ converges, then ka∑ converges. If ka∑ diverges, then ka∑ diverges.


sink

kk k

∞

=∑ converges or diverges.

Rearranging a Series If we rearrange the terms of a finite sum, then the value of the sum remains unchanged. If we rearrange the terms of an infinite series, na∑ , we will change the order of the terms. For instance, we might have

1 2 6 3 4 12 5a a a a a a a+ + + + + + +

The Rearrangement Theorem for Absolutely Convergent Series If na∑ is absolutely convergent with sum S then any rearrangement of na∑ has the same sum S.

However, any conditionally convergent series can be rearranged to give a different sum. Consider the alternating harmonic series

( ) 1

1

1 1 1 1 1 1 1 11 1 ln 22 3 4 5 6 7 8

n

n n

∞+

=

− = − + − + − + − + =∑

Now multiply the series by 12

then

1 1 1 1 1 ln 22 4 6 8 2− + − + =

Now insert a zeros in between the terms of this series, 1 1 1 1 10 0 0 0 ln 22 4 6 8 2

+ + − + + + − + =

Now 1ln 2 ln 22

+ is

1 1 1 1 1 31 ln 23 2 5 7 4 2

+ − + + − + =

Notice we have the same terms but in different order and we have changed the sum of the series. Riemann proved that

If na∑ is a conditionally convergent series and r is any real number whatsoever, then

there is a rearrangement of na∑ that has sum equal to r.

18

Summary for Determining the convergence or divergence of ∑ na 1. Look at the form of the series.

a. Is it a p-series, 1pn∑ , the series converges if 1p > ; it diverges if 1p ≤ .

b. Is it a geometric series nar∑ , the series converges if 1r < ; it diverges if 1r ≥ .

(The sum of a convergent geometric series is ( )/ 1S a r= − .) c. Is it a telescoping series? If it is find the nth partial sum, nS then if lim nn

S→∞

exists, the

series converges and has sum lim nnS S

→∞= .

d. Is it an alternating series, if it is use the Alternating Series Test. 2. Is it a positive term series? If it is,

a. And the terms are similar to those of a geometric series or p-series, try the Direct Comparison Test or the Limit Comparison Test.

b. And if ( )na f n= and ( )1

f x dx∞

∫ is “easy” to evaluate use the Integral Test,

remember to verify that the conditions of the integral test are satisfied. c. And if the terms of the series contain factorials or nc (c is a positive constant) try the

Ratio Test. d. And if it is easy to take the nth root of the terms, try the Root Test.

3. If the series has some negative terms. Check if na∑ converges; if it does then

na∑ converges since absolute convergence implies convergence. 4. If you can easily determine that lim 0nn

a→∞

≠ , then the nth Term Test for Divergence

indicates that the series na∑ diverges.

Test the series for convergence or divergence

1. 2

1

3!

n

n

nn

∞

=∑

2. 2

31

11n

nn

∞

=

++∑

3. ( ) 1

31

32

n

nn

+∞

=

−∑

19

4. 1sin

nn

∞

=∑

5. ( )2

1

2 n

nn

nn

∞

=∑

6. ( )3

1

ln1n

n nn

∞

= +∑

7. 1

arctann

nn n

∞

=∑

8. ( )1

ln1 n

n

nn

∞

=

−∑

9. ( )2

1

cos / 24n

nn n

∞

= +∑

10. 2

1 1

n

n

nn

∞

=

+

∑

To determine if a series∑ na converges absolutely, converges conditionally or diverges.

1. Look at na∑ if it converges then the series converges absolutely.

2. If na∑ diverges then check na∑ using the process above; if it converges then the series is conditionally convergent and if it diverges then the series is divergent.

20

Chapter 9 Power Series Section 9.1 Approximating Functions with Polynomials Definition: Taylor Polynomial Definition Let f be a function with derivatives of order k for 1, 2, ,k N= in some interval containing a as an interior point. Then for any integer n from 0 through N , the Taylor polynomial of order n generated by f at x a= is the polynomial

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )2"

'2! !

nn

n

f a f aP x f a f a x a x a x a

n= + − + − + + −

Example: Find the Taylor series for ( )f x x= at 4a = . Also find the Taylor polynomial of order 4. Taylor’s Theorem If f and its first n derivatives ( )', ", , nf f f are continuous on the closed interval between a and b and ( )nf is differentiable on the open interval between a and b, then there exists a number c between a and b such that

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( )( ) ( )

12 1"

'2! ! 1 !

n nn nf a f a f c

f b f a f a b a b a b a b an n

++= + − + − + + − + −

+

Taylor’s Formula If f has derivates of all order on an open interval I containing a, then for each positive integer n and for each x in I.

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )2"

'2! !

nn

n

f a f af x f a f a x a x a x a R x

n= + − + − + + − +

Where ( )( ) ( )( ) ( )

11

1 !

nn

n

f cR x x a

n

++= −

+ for some c between a and b.

Notice

• ( ) ( ) ( )n nf x P x R x= + for x I∈

• ( )nP x is the polynomial approximation of order n

• ( )nR x is called the remainder of order n or the error term for the approximation of f by

( )nP x

• If ( ) 0 as nR x n→ →∞ for all x I∈ , then we say that the Taylor series generated by f at x a= converges to f on I.

21

The Remainder Estimation Theorem If there is a positive constant M such that ( ) ( )1nf t M+ ≤ for all t between x and a, inclusive, then

the remainder term ( )nR x in Taylor’s Theorem satisfies the inequality

( ) ( )

1

1 !

n

n

x aR x M

n

+−≤

+

If this condition holds for every n and the other conditions of Taylor’s Theorem are satisfied by f, then the series converges to ( )f x . Example: Show that cos x is equal to the sum of it Maclaurin series. Section 9.2 Properties of Power Series Definition A power series about 0x = is a series of the form

20 1 2

0

k kk k

kc x c c x c x c x

∞

=

= + + + + +∑

A power series about x a= is a series of the form

( ) ( ) ( ) ( )20 1 2

0

k kk k

kc x a c c x a c x a c x a

∞

=

− = + − + − + + − +∑

In which the center a and the coefficients 0 1, , , ,nc c c are constants.

Consider the geometric series, 0

k

kr

∞

=∑ , it converges if 1r < , with sum 1

1 r−. So

Similarly, we can show that

0

1 and converges for 11

n

nx x

x

∞

=

= <−∑

When considering a power series in general, a question we must answer is: For what values of x does the power series converge? To find the values of x consider na∑ and use the Ratio Test (or the Root Test).

22

Example: 51 5

k

kk

xk

∞

=∑

Example: ( ) ( )2

01

2 !

kk

k

xk

∞

=

−∑

Example: 1

k k

kk x

∞

=∑

Theorem9,3: Convergence of Power Series

A power series ( )0

kk

kc x a

∞

=

−∑ centered at a converges n one of three ways:

1. The series converges absolutely for all x in which case the interval of convergence is ( ),−∞ ∞ and the radius of convergence is R = ∞ .

2. There is a real number 0R > such that the series converges absolutely for x a R− < and

diverges for x a R− > , in which case the radius of convergence is R . (Check the endpoints.).

3. The series converges only at a, in which case the interval of convergence is 0R = . Note

• R is called the radius of convergence of the power series. • The interval of radius R centered at x a= is called the interval of convergence

( ) ( ] [ ) [ ]{ }, , , , , , ,a R a R a R a R a R a R a R a R− + − + − + − +

Example: Find the radius and interval of convergence for ( )0

12 2

k

k

xk

∞

=

++∑

23

Theorem 9.4 Combining Power Series Suppose the power series k

kc x∑ and kkd x∑ converge absolutely to ( )f x and ( )g x ,

respectively, on the interval I. 1. Sum and Difference: The power series ( ) k

k kc d x±∑ converges absolutely to ( ) ( )f x g x± on I.

2. Multiplication by a power: The power series m k k mk kx c x c x +=∑ ∑ converges absolutely to

( )mx f x on I, provided m is an integer such that 0k m+ ≥ for all terms in the series.

3. Composition: If ( ) mh x bx= , where m is a positive integer and b is a real number, the power

series ( )( )kkc h x∑ converges absolutely tot eh composite function ( )( )f h x for all x such

that ( )h x is in I.

Example: Given the power series for 0

1 and converges for 11

k

kx x

x

∞

=

= <−∑ find the power series

for:

1. 11 x+

2. 11 5x−

3. 3

1 5x

x−

Theorem 9.5 Differentiating and Integrating Power Series Let the function f be defined by the power series ( )k

kc x a−∑ on its interval of convergence I . 1. f is a continuous function on I. 2. The power series may be differentiated or integrated term by term, and the resulting power

series converges to ( )'f x or ( )f x dx C+∫ , respectively, at all points in the interior of I, where C is an arbitrary constant.

24

Example: Recall that

( ) 2 3 4

0

11 11

n n

nx x x x x

x

∞

=

− = − + + − + =+∑

Find 1. ( )'f x 2. ( )"f x

3. 11

dxx+

⌠⌡

Example: The series

2 3 4 5

12! 3! 4! 5!

x x x x xe x= + + + + + +

Converges to xe for all x.

a. Find the series for ( )xd edx

. Do you get the series for xe ? Explain your answer.

b. Find the series for xe dx∫ . Do you get the series for xe ? Explain your answer.

c. Replace x by x− in the series for xe to find the series that converges to xe− for all x. Then multiply the series for xe and xe− to find the first six terms of a series for x xe e−⋅ .

25

Section 9.3 Taylor Series Definition Taylor/Maclaurin Series for a Function Suppose the function f has derivatives of all orders on an interval containing the point a. The Taylor Series generated by f at x a= is

( )( ) ( ) ( )

( ) ( )( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

0

32 3

!

"'

2! 3! !

kk

k

nn

f af x x a

k

f a f a f af a f a x a x a x a x a

n

∞

=

= −

= + − + − + − + + − +

∑

The Maclaurin Series is the Taylor series generated by f at 0x =

( )( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )0

32 3

0!

" 0 0 00 ' 0

2! 3! !

kk

k

nn

ff x x

k

f f ff f x x x x

n

∞

=

=

= + + + + + +

∑

Example: Find the Maclaurin series for ( ) 11

f xx

=+

Example: Find the Maclaurin series for ( ) cosf x x= Example: Find the Maclaurin series for ( ) xf x e=

26

Example: Find the Taylor series of ( ) 12 1

f xx

=−

for 1a = .

Example: Find the Taylor series of ( ) cosf x x= at a π= . Theorem 9.7 Convergence of Taylor Series Let f have derivatives of all orders on an open interval I containing a. The Taylor series for f centered at a converges to f for all x in I if and only if ( )lim 0nn

R x→∞

= for all x in I where

( )( ) ( )( ) ( )

11

1 !

nn

n

f cR x x a

n

++= −

+

Is the remainder at x (with c between x and a).

27

Section 9.4 Working with Taylor Series Applying Taylor Series Taylor series can added subtracted, and multiplied by constants and the resulting series is a Taylor series with interval of converges the intersection of their respective intervals of convergence. Example: Find the Taylor series at 0x = of ( ) 5xf x e−= Example: Find the Taylor series at 0x = of ( ) ( )cosf x x xπ=

Example: Express 1xe dxx−⌠

⌡

as a power series.

Example: Express

2xe dx∫ as a power series.

Example: Use series to evaluate 0

limx x

x

e ex

−

→

−

Chapter 8 Infinite Sequences and Serieslballou/chap8.pdf · 1 Chapter 8 Sequences and Infinite Series. Section 8.1 Overview. An infinite sequence is a list of numbers in a . definite

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