IntroRealAnal-ch09

8/11/2019 IntroRealAnal-ch09

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Chapter 9

Sequences of Functions

9.1 Pointwise Convergence

We have accumulated much experience working with sequences of numbers. Thenext level of complexity is sequences of functions. This chapter explores severalways that sequences of functions can converge to another function. The basicstarting point is contained in the following denitions.

Denition 9.1.1. Suppose S R and for each n N there is a functionf n : S R . The collection {f n : nN } is a sequence of functions dened on S .For each xed x S , f n (x) is a sequence of numbers, and it makes senseto ask whether this sequence converges. If f n (x) converges for each xS , thisdenes a new function f : S

R by

f (x) = limn

f n (x).

The function f is called the pointwise limit of the sequence f n , or, equivalently,it is said f n converges pointwise to f . This is abbreviated f n

S

f , or simplyf n f , if the domain is clear from the context.Example 9.1.1. Let

f n (x) =0, x 0xn , 0 < x < 11, x 1

.

Then f n

f where

f (x) =0, x 11, x 1

.

(See Figure 9.1.) This example shows that a pointwise limit of continuousfunctions need not be continuous.

9-1


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9-2 CHAPTER 9. SEQUENCES OF FUNCTIONS

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

Figure 9.1: The rst ten functions from the sequence of Example 9.1.1.

3 2 1 1 2 3

0.4

0.2

0.2

0.4

Figure 9.2: The rst four functions from the sequence of Example 9.1.2.

Example 9.1.2. For each nN , dene f n : R R byf n (x) =

nx1 + n 2x2

.

(See Figure 9.2.) Clearly, each f n is an odd function and lim | x |f n (x) = 0. Abit of calculus shows that f n (1/n ) = 1 / 2 and f n (1/n ) = 1/ 2 are the extremevalues of f n . Finally, if x = 0,

|f n (x)| = nx1 + n2x2 < nxn2x2 = 1nx

implies f n 0. This example shows that functions can remain bounded awayfrom 0 and still converge pointwise to 0.

May 1, 2009 http://math.louisville.edu/ lee/ira


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Pointwise and Uniform Convergence 9-3

0.2 0.4 0.6 0.8 1.0

10

20

30

40

50

60

Figure 9.3: The rst four functions from the sequence of Example 9.1.3.

Example 9.1.3. Dene f n : R R by

f (x) =22n +4 x 2n +3 , 12 n +1 < x < 32n +222n +4 x + 2 n +4 , 32 n +2 x < 12n0, otherwise

To gure out what this looks like, it might help to look at Figure 9.3.The graph of f n is a piecewise linear function supported on [1 / 2n +1 , 1/ 2n ]

and the area under the isoceles triangle of the graph over this interval is 1.Therefore,

10 f n = 1 for all n .

If x > 0, then whenever x > 1/ 2n , we have f n (x) = 0. From this it easily

follows that f n 0.The lesson to be learned from this example is that it may not be true thatlimn

10 f n =

10 limn f n .

Example 9.1.4. Dene f n : R R by

f n (x) =n2 x

2 + 12n , |x | 1n|x |, |x | > 1n

.

(See Figure 9.4.) The parabolic section in the center was chosen so f n ( 1/n ) =1/n and f n ( 1/n ) = 1. This splices the sections together at ( 1/n, 1/n ) sof n is diff erentiable everywhere. Its clear f n |x |, which is not di ff erentiableat 0.

This example shows that the limit of di ff erentiable functions need not bediff erentiable.

The examples given above show that continuity, integrability and di ff eren-tiability are not preserved in the pointwise limit of a sequence of functions. Tohave any hope of preserving these properties, a stronger form of convergence isneeded.



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1.0 0.5 0.5 1 .0

0.2

0.4

0.6

0.8

1.0

Figure 9.4: The rst ten functions f from Example 9.1.4.

9.2 Uniform Convergence

Denition 9.2.1. The sequence f n : S R converges uniformly to f : S Ron S , if for each > 0 there is an N N so that whenever n N and x S ,then |f n (x) f (x)| < .In this case, we write f n S f , or simply f n f , if the set S is clear fromthe context.

a b

f ( x)

f ( x) + !

f ( x !

f n( x)

Figure 9.5: |f n (x) f (x)| < on [a, b], as in Denition 9.2.1.The di ff erence between pointwise and uniform convergence is that with point-

wise convergence, the convergence of f n to f can vary in speed at each point of S . With uniform convergence, the speed of convergence is roughly the same allacross S . Uniform convergence is a stronger condition to place on the sequencef n than pointwise convergence in the sense of the following theorem.

Theorem 9.2.1. If f n S f , then f n

S

f .May 1, 2009 http://math.louisville.edu/ lee/ira


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9.3. METRIC PROPERTIES OF UNIFORM CONVERGENCE 9-5

Proof. Let x0 S and > 0. There is an N N such that when n N ,then |f (x)

f n (x)| < for all x

S . In particular, |f (x0)

f n (x0)| < when

n N . This shows f n (x0) f (x0). Since x0 S is arbitrary, it follows thatf n f .The rst three examples given above show the converse to Theorem 9.2.1

is false. There is, however, one interesting and useful case in which a partialconverse is true.

Denition 9.2.2. If f nS

f and f n (x) f (x) for all xS , then f n increases to f on S . If f n S f and f n (x) f (x) for all xS , then f n decreases to f onS . In either case, f n is said to converge to f monotonically .The functions of Example 9.1.4 decrease to |x |. Notice that in this case,

the convergence is also happens to be uniform. The following theorem showsExample 9.1.4 to be an instance of a more general phenomenon.

Theorem 9.2.2 (Dinis Theorem) . If

(a) S is compact,

(b) f nS

f monotonically,(c) f n C (S ) for all nN , and (d) f C (S ),

then f n f .Proof. There is no loss of generality in assuming f n f , for otherwise we con-sider f n and f . With this assumption, if gn = f n f , then gn is a sequenceof continuous functions decreasing to 0. It su ffi ces to show gn 0.To do so, let > 0. Using continuity and pointwise convergence, for eachx S nd an open set Gx containing x and an N x N such that gN x (y) < for all y Gx . Notice that the monotonicity condition guarantees gn (y) < forevery y Gx and n N x .

The collection {Gx : xS } is an open cover for S , so it must contain a nitesubcover {Gx i : 1 i n}. Let N = max {N x i : 1 i n} and choose m N .If x S , then x Gx i for some i, and 0 gm (x) gN (x) gN i (x) < . Itfollows that gn 0.

9.3 Metric Properties of Uniform Convergence

If S R , let B(S ) = {f : S R : f is bounded }. For f B(S ), denef S = lub {|f (x)| : xS }. (It is abbreviated to f , if the domain S is clearfrom the context.) Apparently, f 0, f = 0 f 0 and, if g B (S ),May 1, 2009 http://math.louisville.edu/ lee/ira


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then f g = g f . Moreover, if hB (S ), thenf g = lub {|f (x) g(x)| : xS } lub {|f (x) h(x)| + |h(x) g(x)| : xS }

lub {|f (x) h(x)| : xS } + lub {|h(x) g(x)| : xS }= f h + h g

Combining all this, it follows that f g is a metric 1 on B (S ).The denition of uniform convergence implies that for a sequence of boundedfunctions f n : S R ,

f n f f n f 0.Because of this, the metric f g is often called the uniform metric or the sup-metric . Many ideas developed using the metric properties of R can be carriedover into this setting. In particular, there is a Cauchy criterion for uniformconvergence.

Denition 9.3.1. Let S R . A sequence of functions f n : S R is a Cauchy sequence under the uniform metric, if given > 0, there is an N N such thatwhen m, n N , then f n f m < .Theorem 9.3.1. Let f n B(S ). There is a function f B(S ) such that f n f i ff f n is a Cauchy sequence in B (S ).Proof. () Let f n f and > 0. There is an N N such that n N implies f n f < / 2. If m N and n N , then

f m f n f m f + f f n <

2 +

2 =

shows f n is a Cauchy sequence.() Suppose f n is a Cauchy sequence in B(S ) and

> 0. Choose N Nso that when f m f n < whenever m N and n N . In particular, fora xed x0 S and m, n N , |f m (x0) f n (x0)| f m f n < shows thesequence f n (x0) is a Cauchy sequence in R and therefore converges. Since x0 isan arbitrary point of S , this denes an f : S R such that f n f .Finally, if m, n N and xS the fact that |f n (x) f m (x)| < gives

|f n (x) f (x)| = limm |f n (x) f m (x)| .

This shows that when n N , then f n f . We conclude that f B (S )and f n

f .

A collection of functions S is said to be complete under uniform convergence,if every Cauchy sequence in S converges to a function in S . Theorem 9.3.1 showsB (S ) is complete under uniform convergence. Well see several other collectionsof functions that are complete under uniform convergence.

1 Denition 2.2.3



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9.4. SERIES OF FUNCTIONS 9-7

9.4 Series of Functions

The denitions of pointwise and uniform convergence are extended in the naturalway to series of functions. If k=1 f k is a series of functions dened on aset S , then the series converges pointwise or uniformly, depending on whetherthe sequence of partial sums, sn =

nk=1 f k converges pointwise or uniformly,

respectively. It is absolutely convergent or absolutely uniformly convergent, if n =1 |f n | is convergent or uniformly convergent on S , respectively.The following theorem is obvious and its proof is left to the reader.

Theorem 9.4.1. Let n =1 f n be a series of functions dened on S . If n =1 f nis absolutely convergent, then it is convergent. If n =1 f n is absolutely uniformly convergent, then it is uniformly convergent.

The following theorem is a restatement of Theorem 9.2.2 for series.Theorem 9.4.2. If

n =1 f n is a series of nonnegative continuous functions

converging pointwise to a continuous function on a compact set S , then n =1 f nconverges uniformly on S .

A simple, but powerful technique for showing uniform convergence of seriesis the following.

Theorem 9.4.3 (Weierstrass M-Test) . If f n : S R is a sequence of functions and M n is a sequence nonnegative numbers such that f n S M n for all nNand n =1 M n converges, then n =1 f n is absolutely uniformly convergent.Proof. Let > 0 and sn be the sequence of partial sums of n =1 |f n |. There isan N N such that when n > m N , then nk = m M k < . So,

sn sm n

k = m +1

|f k | n

k= m +1

f k n

k = m

M k < .

This shows sn is a Cauchy sequence in B(S ) and must converge according toTheorem 9.3.1.

9.5 Continuity and Uniform Convergence

Theorem 9.5.1. If f n : S R such that each f n is continuous at x0 and f n S f , then f is continuous at x0 .

Proof. Let > 0. Since f n f , there is an N N such that whenever n N and xS , then |f n (x) f (x)| < / 3. Because f N is continuous at x0 , there isa > 0 such that x(x0 , x0 + ) S implies |f N (x) f N (x0)| < / 3. Usingthese two estimates, it follows that when x

(x0

, x0 + )

S ,

|f (x) f (x0)| = |f (x) f N (x) + f N (x) f N (x0) + f N (x0) f (x0)| |f (x) f N (x)| + |f N (x) f N (x0)| + |f N (x0) f (x0)|< / 3 + / 3 + / 3 = .

Therefore, f is continuous at x0 .



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The following corollary is immediate from Theorem 9.5.1.

Corollary 9.5.2. If f n is a sequence of continuous functions converging uniformly to f on S , then f is continuous.

Example 9.1.1 shows that continuity is not preserved under pointwise con-vergence. Corollary 9.5.2 is establishes that if S R , then C (S ) is completeunder the uniform metric.

The fact that C ([a, b]) is closed under uniform convergence is often useful be-cause, given a bad function f C ([a, b]), its often possible to nd a sequencef n of good functions in C ([a, b]) converging uniformly to f .Theorem 9.5.3 (Weierstrass Approximation Theorem) . If f C ([a, b]), then there is a sequence of polynomials pn f .

To prove this theorem, we rst need a lemma.

Lemma 9.5.4. For nN let cn =

1

1(1 t2)n dt

1and

kn (t) =cn (1 t2)n , |t | 10, |t | > 1

.

(See Figure 9.6.) Then

(a) kn (t) 0 on [1, 1] for all nN ;(b)

1

1 kn = 1 for all nN ; and,

(c) if 0 <

< 1, then kn 0 on [1, ][

, 1].

1.0 0.5 0.5 1 .0

0.2

0.4

0.6

0.8

1.0

1.2

Figure 9.6: Here are the graphs of kn (t) for n = 1 , 2, 3, 4, 5.



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9.5. CONTINUITY AND UNIFORM CONVERGENCE 9-9

Proof. Parts (a) and (b) follow easily from the denition of kn .To prove (c) rst note that

1 = 1

1kn

1/ n

1/ ncn (1 t2)n dt cn

2 n 1

1n

n

.

Since 1 1nn

1e , it follow that there is an > 0 such that cn < n.2Letting (0, 1) and t 1,kn (t) kn ( ) n(1 2)n 0

by LHospitals Rule. Since kn is an even function, this establishes (c).

A sequence of functions satisfying conditions such as those in Lemma 9.5.4is called a convolution kernel or a Dirac sequence . Several such kernels play akey role in the study of Fourier series.

We now turn to the proof of the theorem.

Proof. There is no generality lost in assuming [ a, b] = [0, 1], for otherwise weconsider the linear change of variables g(x) = f ((b a)x + a). Similarly, wecan assume f (0) = f (1) = 0, for otherwise we consider g(x) = f (x) (( f (1) f (0)) x+ f (0), which is a polynomial added to f . We can further assume f (x) = 0when x /[0, 1].

Set

pn (x) = 1

1f (x + t)kn (x) dt. (9.1)

To see pn is a polynomial, change variables in the integral using u = x + t toarrive at

pn (x) = x+1

x1f (u)kn (u x) du =

1

0f (u)kn (x u) du,

because f (x) = 0 when x / [0, 1]. Notice that kn (x u) is a polynomial in uwith coeffi cients being polynomials in x, so integrating f (u)kn (x u) yields apolynomial in x. (Just try it for a small value of n and a simple function f !)2 It is interesting to note that with a bit of work using complex variables one can prove

cn = (n + 3 / 2) (n + 1) =

n + 1 / 2n

n 1/ 2n 1

n 3/ 2n 2

3/ 2

1 .



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Use (9.1) and Lemma 9.5.4(b) to see for (0, 1) that

| pn (x) f (x)| = 1

1f (x + t)kn (t) dt f (x)

= 1

1(f (x + t) f (x))kn (t) dt

1

1|f (x + t) f (x)|kn (t) dt

=

|f (x + t) f (x)|kn (t) dt + < | t |1 |f (x + t) f (x)|kn (t) dt. (9.2)

Well handle each of the nal integrals in turn.Let > 0 and use the uniform continuity of f to choose a

(0, 1) such

that when |t | < , then |f (x + t) f (x)| < / 2. Then, using Lemma 9.5.4(b)again,

|f (x + t) f (x)|kn (t) dt m , then hm /p n = 4 n m 1

N , so sn (x hm )

sn (x) = 0 and

f (x hm ) f (x) hm

=m

k=0

sk (x hm ) sk (x) hm

. (9.5)

On the other hand, if n < m , then a worst-case estimate is that

sn (x hm ) sn (x)hm

34

n

/ 14n

= 3 n .

This gives

m 1

k=0

sk (x hm ) sk (x) h

m

m 1

k =0

sk (x hm ) sk (x) h

m

3m 1

3 1 3m 3m

2 =

3m

2 .

Since 3m / 2

, it is apparent f (x) does not exist.

There are many other constructions of nowhere di ff erentiable continuousfunctions. The rst was published by Weierstrass [6] in 1872, although it wasknown in the folklore sense among mathematicians earlier than this. (There isan English translation of Weierstrass paper in [2].) In fact, it is now known in atechnical sense that the typical continuous function is nowhere di ff erentiable.



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9.6. INTEGRATION AND UNIFORM CONVERGENCE 9-13

9.6 Integration and Uniform Convergence

Theorem 9.6.1. If f n : [a, b] R such that ba f n exists for each n and f n f on [a, b], then

b

af = lim

n b

af n

.

Proof. Some care must be taken in this proof, because there are actually twothings to prove. Before the equality can be shown, it must be proved that f isintegrable.

To show that f is integrable, let > 0 and N N such that f f N


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shows that

ba f n

ba f .

Corollary 9.6.2. If n =1 f n is a series of integrable functions converging uniformly on [a, b], then

b

a

n =1f n =

n =1 b

af n

Combining Theorem 9.6.1 with Dinis Theorem, gives the following.

Corollary 9.6.3. If f n is a sequence of continuous functions converging mono-tonically to a continuous function f on [a, b], then

ba f n

ba f .

9.7 Uniform Convergence and Di ff erentiation

The relationship between uniform convergence and di ff erentiation is somewhatmore complex than those weve already examined. First, because there are twosequences involved, f n and f n , either of which may converge or diverge at apoint; and second, because di ff erentiation is more delicate than continuity orintegration.

Example 9.1.4 is an explicit example of a sequence of di ff erentiable functionsconverging uniformly to a function which is not di ff erentiable at a point. Thederivatives of the functions from that example converge pointwise to a functionthat is not a derivative. The Weierstrass Approximation Theorem and Example9.5.1 push this to the extreme by showing the existence of a sequence of poly-nomials converging uniformly to a continuous nowhere di ff erentiable function.

The following theorem starts to shed some light on the situation.

Theorem 9.7.1. If f n is a sequence of derivatives dened on [a, b] and f n

f , then f is a derivative. Moreover, if the sequence F n C ([a, b]) satises F n = f n and F n (a) = 0 , then F n F where F = f .Proof. For each n, let F n be an antiderivative of f n . By considering F n (x) F n (a), if necessary, there is no generality lost with the assumption that F n (a) =0.

Let > 0. There is an N N such thatm, n N = f m f n 0 the series converges at x when |c x | < R and diverges at xwhen |c x | > R .(c) If R(0, ), then the series may converge at none, one or both of c Rand c + R.



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9.8.2 Uniform Convergence of Power Series

The partial sums of a power series are a sequence of polynomials convergingpointwise on the domain of the series. As has been seen, pointwise convergenceis not enough to say much about the behavior of the power series. The followingtheorem opens the door to a lot more.

Theorem 9.8.2. A power series converges absolutely and uniformly on compact subsets of its interval of convergence.

Proof. There is no generality lost in assuming the series has the form of (9.13)with c = 0. Let the radius of convergence R > 0 and K be a compact sbset of (R, R ). Choose r (lub {|x | : xK }, R ). If xK , then |an xn | < |an r n | fornN . Since n =0 |an r

n | converges, the Weierstrass M -test shows n =0 an xn

is absolutely and uniformly convergent on K .

The following two corollaries are immediate consequences of Corollary 9.5.2and Theorem 9.6.1, respectively.

Corollary 9.8.3. A power series is continuous on its interval of convergence.

Corollary 9.8.4. If [a, b] is an interval contained in the interval of convergence for the power series n =0 an (x c)

n , then

b

a

n =0an (x c)n =

n =0an

b

a(x c)n .

The next question is: What about di ff erentiability?Notice that the continuity of the exponential function and LH ospitals Rule

give

limn

n1/n = limn

exp ln nn

= exp limn

ln nn

= exp(0) = 1 .

Therefore, for any sequence an ,

limsup( na n )1/n = limsup n1/n a1/nn = limsup a1/nn . (9.16)

Now, suppose the power series n =0 an xn has a nontrivial interval of con-

vergence, I . Formally di ff erentiating the power series term-by-term gives a newpower series n =1 na n x

n 1 . According to (9.16) and Theorem 9.8.1, the term-by-term di ff erentiated series has the same interval of convergence as the original.Its partial sums are the derivatives of the partial sums of the original series andTheorem 9.8.2 guarantees they converge uniformly on any compact subset of I .Corollary 9.7.3 shows

ddx

n =0an xn =

n =0

ddx

an xn =

n =1na n xn 1 , xI.

This process can be continued inductively to obtain the same results for allhigher order derivatives. We have proved the following theorem.



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9.8. POWER SERIES 9-19

Theorem 9.8.5. If f (x) =

n =0 an (x c)n is a power series with nontrivial interval of convergence, I , then f is di ff erentiable to all orders on I with

f (m ) (x) =

n = m

n!(n m)!

an (x c)n m . (9.17)

Moreover, the di ff erentiated series has I as its interval of convergence.

9.8.3 Taylor Series

Suppose f (x) = n =0 an xn has I = ( R, R ) as its interval of convergence forsome R > 0. According to Theorem 9.8.5,

f (m ) (0) = m!

(m m)!am =

am = f (m ) (0)

m! ,

m

.

Therefore,

f (x) =

n =0

f (n ) (0)n!

xn , xI.

This is a remarkable result! It shows that the values of f on I are completelydetermined by its values on any neighborhood of 0. This is summarized in thefollowing theorem.

Theorem 9.8.6. If a power series f (x) = n =0 an (x c)n has nontrivial

interval of convergence I , then

f (x) =

n =0f

(n )

(c)n! (x c)n , xI. (9.18)

The series (9.18) is called the Taylor series 3 for f centered at c. The Taylorseries can be formally dened for any function that has derivatives of all ordersat c, but, as Example 7.4.2 shows, there is no guarantee it will converge to thefunction anywhere except at c.

9.8.4 The Endpoints of the Interval of Convergence

We have seen that at the endpoints of its interval of convergence a power seriesmay diverge or even absolutely converge. A natural question when it doesconverge is the following: What is the relationship between the value at theendpoint and the values inside the interval of convergence?

Theorem 9.8.7 (Abel) . If f (x) = n =0 an (x c)n has a nite radius of

convergence R > 0 and f (c + R) exists, then c + RC (f ).3 When c = 0, it is often called the Maclaurin series for f .



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Proof. It can be assumed c = 0 and R = 1. There is no loss of generality witheither of these assumptions because otherwise just replace f (x) with f ((x +c)/R ). Set s = f (1), s1 = 0 and sn = nk=0 ak for n

. For |x | < 1,n

k=0

ak xk =n

k=0

(sk sk1)xk

=n

k=0

sk xk n

k=1

sk1xk

= sn xn +n 1

k=0

sk xk xn 1

k =0

sk xk

= sn xn + (1 x)n 1

k=0

sk xk

When n , since sn is bounded,

f (x) = (1 x)

k =0

sk xk . (9.19)

Since (1 x) n =0 xn = 1, (9.19) implies

|f (x) s | = (1 x)

k =0

(sk s)xk . (9.20)

Let > 0. Choose N N such that |sN s | < / 2, (0, 1) so

N

k=0

|sn s | < / 2.

and 1 < x < 1. With these choices, (9.20) becomes

|f (x) s | (1 x)N

k =0

(sk s)xk + (1 x)

k = N +1

(sk s)xk

< N

k=0

|sk

s | +

2

(1

x)

k= N +1

xk <

2

+

2

=

It has been shown that lim x1 f (x) = f (1), so 1C (f ).

Abels theorem opens up a more general idea for the summation of series.Suppose, as in the proof of the theorem, f (x) = n =0 an x

n with interval of



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9.8. POWER SERIES 9-21

convergence ( 1, 1). If limx1 f (x) exists, then this limit is called the Abel sum of the coeffi cients of the series. In this case, the following notation is used

limx1

f (x) = A

n =0an .

From Theorem 9.8.7, it is clear that when n =0 an exists, then n =0 an =A n =0 an . The converse of this statement may not be true.Example 9.8.5. Notice that

n =0(x)n =

11 + x

,

has interval of convergence ( 1, 1) and diverges when x = 1. (It is the sum1 1 + 1 1 + .) But,

A

n =0(1)n = limx1

11 + x

= 12

.

Finally, here is an example showing the power of these techniques.Example 9.8.6. The series

n =0(1)n x2n =

11 + x2

has (1, 1) as its interval of convergence and (0 , 1] as its domain. If 0 |x | < 1,then Corollary 9.6.2 justies

arctan( x) = x

0

dt1 + t2

= x

0

n =0(1)n t2n dt =

n =0

(1)n2n + 1

x2n +1 .

This series for the arctangent converges by the alternating series test whenx = 1, so Theorem 9.8.7 implies

n =0

(1)n2n + 1

= limx1

arctan( x) = arctan(1) =

4.



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