Chapter 6 Infinite Series

Transcription

1 Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat similar, the topic of ifiite series. A ifiite series is a sum cotaiig a ifiite umber of terms. For example, are each examples of ifiite series. Sice the umber of operatios eeded to compute such a sum is ifiite, it is ot clear whether or ot a fiite value ca be associated with the sum. It is eve less clear how the value of the sum ca be foud whe it is kow to be fiite. We will show how it ca be decided if the series ca be associated with a sum ad, i special cases, the value of the sum ca be determied. Ifiite Series Let a deote a sequece of real umbers, ad defie a ew sequece, S N by The N S N a, N, 2, S a, S 2 a a 2, S 3 a a 2 a 3, etc. We refer to the sequece, a, as the sequece of terms for the ifiite series S a. We refer to the related sequece, S N, as the sequece of partial sums for the ifiite series. If the sequece of partial sums is a coverget sequece, the the ifiite series is said to be a coverget ifiite series. If the sequece of partial sums is ot coverget the the ifiite series is ot coverget. Note that the sequece of partial sums may fail to coverge due to the presece of more tha a sigle limit poit or because the sequece of partial sums is ot bouded. I the latter case, we say the ifiite series diverges to ifiity ad i the case of a sequece of partial sums havig more tha a sigle limit poit, we say the ifiite series simply fails to coverge. I either case, we ca say that the ifiite series diverges. (a) Cosider the followig series, kow as the harmoic series, 2 3 Note that for ay positive iteger, N, The N S N 2 3 N 2N S 2N 2 N N 2N.

2 S 2N S N N 2N 2N 2N N 2N 2. Sice S 2N S N /2 for every N, the sequece of partial sums is ot a Cauchy sequece. It follows that the sequece of partial sums is ot coverget ad the ifiite series is diverget. (b) Cosider the series, The method used i (a) fails to lead to a coclusio. We will show later that this series is coverget. (c) Cosider the geometric series, a r a arar 2 0 This geometric series has first term equal to a ad ratio term equal to r. Note that The ad if r, we fid S N a arar 2 ar N ad rs N arar 2 ar N ar N. S N rs N S N r a ar N, S N a arn r a r if r if r Evidetly, if r, the sequece of partial sums is coverget to a, hece the series is r also coverget ad this is the value of the sum. If r, the the series is just the sum a a a, which is diverget. The geometric series is oe of the very few ifiite series for which it is possible to express S N i terms of N ad to the use this expressio to determie the covergece of the series. Tests for Covergece The covergece or divergece of a ifiite series for which o explicit formula for S N is available must be decided by idirect tests for covergece. Theorem 6. (th term test) If S N coverges to a limit S, the a must coverge to zero. Proof: Note that a S S. The if lim S S, it is easy to see that lim a lim S lim S S S 0. Note that theorem 6. asserts that if a does ot coverge to zero the the sequece of partial sums does ot coverge ad the associated ifiite series is diverget. The theorem. 2

3 does ot say that if a teds to zero the the series coverges. The ext result is a simple cosequece of the arithmetic with limits theorem from the chapter o sequeces. Theorem 6.2 If a, b are coverget ifiite series with sums A, B respectively, the a b is also coverget to the suma B. Positive Term Series A ifiite series for which the sequece of terms cotais oly positive umbers is called a positive term series. The ext two results are true for series whose terms are all positive. Theorem 6.3 (Compariso test) If a ad b are two sequeces of positive umbers such that i for some iteger N, a b for all N the the covergece of the series b implies the covergece of a ad the divergece of the series a implies the divergece of b. If the coditio i) is replaced by ii lim a L, b the the two series are either both coverget or both diverget. The test i) is called the direct compariso test while ii) is referred to as the limit compariso test. Clearly these results are related to the squeeze theorem for sequeces. The ext result is based o the similarity betwee improper itegrals ad ifiite series. Theorem 6.4 (Itegral test) Suppose ax is a cotiuous ad mootoically decreasig fuctio that is positive o D,. The the ifiite series a, where a a, ad the improper itegral axdx, either both coverge or both diverge. Proof: Let P deote the partitio, 2, 3, for D. The, by drawig a sketch it is easy to see that a sa, P axdx Sa, P 2 from which the result follows. Example (a) Cosider the series of the previous example, part (b) a a ad ote that a a defies the fuctio ax, o the domai, D,. The 2 x ax is positive, cotiuous ad mootoe decreasig o D so the itegral test applies. 3

4 Sice, x 2 lim L L x 2 lim L x x xl lim L L it follows from theorem 6.4 that the series is coverget. (b) More geerally, cosider The a a becomes I this case, x p a p 2 p 3 p ax x p, o D,, ad lim L L x p lim xp L p x xl p lim L Lp. lim L L p if p Div if p ad theorem 6.4 leads to the result that the series coverges if p. (c) We ca apply the itegral test to the harmoic series a 2 3 I this case a a becomes ax x, o D,, ad L x limlx xl x lim ll. L L x lim L Sice the improper itegral is diverget, the ifiite series is diverget as well. This has already bee proved by aother argumet. Absolute ad Coditioal Covergece Not every ifiite series is a positive term series. If a is a ifiite series cotaiig both positive ad egative terms, the we say that the series is absolutely coverget if the associated positive terms series a is coverget. Every series which is absolutely coverget ca be show to be coverget, but the coverse is false. A series which is coverget but ot absolutely coverget is said to be coditioally coverget. A series that is coditioally coverget depeds o the cacellatio of positive ad egative terms for its covergece while a absolutely coverget series coverges simply because the terms decrease so rapidly that there is o eed to rely o cacellatio. Example The series cotais both positive ad egative terms. Note that 4

5 ad sice we have if The the series is coverget by theorem 6.3. O the other had, the series is ot absolutely coverget sice we have already show that the series whose terms are a, is diverget. Therefore the origial series is coditioally coverget but ot absolutely coverget. Theorem 6.5 If a ifiite series is absolutely coverget the it is coverget. The previous example illustrates that the coverse of this theorem is false. If a is a sequece of positive umbers the the series a a a 2 a 3 a 4 is called a alteratig series. The covergece of a alteratig series is easily determied. Theorem 6.6 (Alteratig Series Test) Suppose the positive term sequece a satisfies a a lim a 0 for all The the alteratig series a coverges to a fiite sum S, ad for each N 0 we have S S N a N. Proof: Note that for all N, S 2N2 S 2N a 2N a 22 S 2N S 2N a 2N a 2 Sice a a for all, it follows that the eve order partial sums S 2 are a mootoe icreasig sequece while the odd order partial sums are a mootoe decreasig sequece. I additio, for ay m ad, such that 2 2m, S 2 S 2m a 2 a 2 a 2m3 a 2m2 a 2m 0. O the other had, for ay m ad, such that 2 2m, S 2m S 2 a 2m a 2m a 22 a 2 a

6 I either case, S 2m S 2, which implies that the mootoe icreasig sequece of eve order sums is bouded above be every odd order sum, while the mootoe decreasig sequece of odd order sums is bouded below by every eve order sum. The the mootoe sequece theorem asserts that both sequeces must coverge. To see that they coverge to the same limit, ote that lim S 2 lim S 2 lim S 2 S 2 lim a 2 0. If we deote the limit of the sequece of partial sums by S, the 0 S 2 S S 2 S 2 a 2 which implies that for all, 0 S S 2 S 2 S 2 a 2 S S a. i.e., we ca approximate S by S with a error less tha a. Suppose a is a ifiite series, whose terms eed ot be positive, but for which lim a a exists ad equals r. This meas that this series behaves ultimately like a geometric series havig ratio term r ad we ca show that if r, the series is coverget, (although we are ot etitled to coclude that the sum equals. Similarly, if r, we r ca coclude that the series is diverget. If r the we ca oly coclude that the series does ot behave like a geometric series ad some other approach must be foud to determie the covergece or divergece. We ca state this result i a theorem. Theorem 6.7 (Ratio test) Suppose the sequece a satisfies i lim a a r. The the ifiite series a is absolutely coverget if r ad is diverget if r. If r, the test fails. The coditio i) implies that the series behaves ultimately like a geometric series with ratio term equal to r. The same result is true if the coditio i) is replaced by the coditio, ii lim a / r. Example (a) Cosider the alteratig series 2 a 2. We are free to apply either theorem 6.6 or 6.7. Note that, ad sice a a , 6

7 we have lim 2, lim a a 2. It follows from theorem 6.7 that the series is absolutely coverget. We ca also use the N 2 alteratig series test which tells us that the series is coverget with S S N. 2 N (b) If we apply theorem 6.7 to the series a we fid a p p a p as. The the ratio test fails to tell us aythig about the covergece or divergece for this series. As we have see i a previous example the itegral test implies the series coverges if p. p Exercises Test the followig series for covergece e l 2 2 l l 7

8 / l Power Series For a give sequece of terms a, ad a fixed real umber c, defie Fx a x c 0 The Fx is a fuctio whose domai, D, is the set of poits x, where the ifiite series coverges. We refer to this as a power series i x, expaded about the poit x c. I order to determie D, it is ofte possible to use the ratio test. (a) Cosider the fuctio F defied by the followig power series F x!x c. 0 Here we have a a!! x c x c x c 0 if x c if x c The D cosists of just the sigle poit, x c. (b) Cosider the fuctio F 2 defied by I this case, a a F 2 x 0 2 x c 2 x c 2 x c x c L 2 ad L if x c 2. Therefore F 2 has as its domai, D 2 x c 2.. This is a ope iterval about x c of radius 2 or total width 4. c Let F 3 be defied by Here we have F 3 x x!. 0 8

9 a a!! x x x 0 for all x. Sice L 0 for ay value of x, this series is coverget for all values of x; i.e., D 3 x. We observe from these examples that a power series is coverget o a set of the form D x c R where /R L lim a a. The umber R /L is called the radius of covergece of the series ad, as we see, R ca be 0, ifiity or aythig i betwee. Note that formally, we have ad F x a x c 0 c x F 0 a x c, ad, by the ratio test, these series have the same domai of covergece as does the series for Fx. Of course these are just formal remarks at this poit, with o proof that these represetatios are valid. However, if we let N F N x a x c, 0 the for each iteger N, F N x is a polyomial i x of degree N ad it therefore cotiuous with cotiuous derivatives of all orders. The we ca ask, i what sese is it true that F N coverges to F as N teds to ifiity? does it follow from this that F N coverges to F as N teds to ifiity? does it also follow that c x FN coverges to c x F as N teds to ifiity? if F N does coverge to F what properties of F N are carried over to F? Aswerig these questios will occupy the rest of the time i this course. The followig examples show that the properties of the limit fuctio are ot ecessarily the same as the properties of the fuctios i the sequece. (a) Let ad f x x 2 x 2 for x R,, 2,... Fx 0 f x 0 x 2 x 2. Sice f 0 0 for each, we have F0 0. For x 0, the series for F is just a geometric series with r x 2, ad it is easy to show that 9

10 The Fx 0 Fx x 2 x 2 x 2 for x if x 0 x 2 if x 2 0 so a coverget series of cotiuous fuctios eed ot have a cotiuous sum. (b) Let g x si 2 x/ for 0 x,, 2,.... The it is easy to show that g x coverges to zero as teds to ifiity for every x i 0,; i. e, g x coverges to gx 0. It is also clear that g x cos 2 x. ad this sequece diverges at every x where cosx is differet from zero. Evidetly differetiatig a coverget sequece of differetiable fuctios eed ot lead to a coverget sequece much less a sequece which coverges to the derivative of the limit of the origial sequece. (c) Let 2 2 x if 0 x 2 h x x if x 2 The 0 if x 0 h xdx for every but h x 0 as teds to ifiity. To see this ote that for ay c, 0 c, h c 0 if c; i.e., if c.. Thus the sequece of fuctios coverges to the limit hx 0 o 0, but the itegrals of the h s does ot coverge to the itegral of hx Sequeces of Fuctios I order to aswer the questios we have about power series, it will be ecessary to first cosider a apparetly simpler topic, covergece of sequeces of fuctios. Let f x :, 2,... deote a family of fuctios o a commo domai, D a, b. For each x 0 i D, cosider the sequece of real umbers, f x 0 ad let fx 0 deote the limit of this sequece as teds to ifiity. Sice the value of this limit will likely deped o x 0, we deote the limit by fx 0. The whe we write lim f x fx what we mea is that for each x 0 i D, the sequece of real umbers, f x 0 coverges to a limit, deoted by fx 0. More precisely, we mea that: Defiitio lim f x fx if ad oly if for each x 0 i D,ad for each 0, there exists a N 0 such that f x 0 fx 0 for all N. Example (a)cosider 0

11 f x x 2 x 2 o D, The we have f 0 0, ad for 0 x 0 The f x 0 f x 0 x 0 x 0 x x x 0 x 0 if x 0 Nx 0,. This shows that for each x 0 i,, ad for each 0, there exists N 0, such that f x 0 0 for all N; i.e., f x coverges to the limit zero. (b)cosider g x We have g 0 0 ad g 2 The for x 0 If 0 x 0, the g x 0 g x 0 x x 2 o D 0,. g x 0. I additio, for 0 x 0 x 0 x 0 x 0 2 x 0. x x 0 x 0 x 0 if log log x 0. x 0 x 2 0 for log x 0 log x 0 This shows that for each x 0 i 0,, ad for each 0, there exists Nx 0 0, such that g x 0 0 for all Nx 0 ; i.e., g x coverges to the limit gx /2 if x 0 otherwise (c)cosider h x x x o D 0, I this case we have h 0 0 ad h 2 for 0 x 0, h x 0 x 0 x 0 x 0 ad for x 0 The h x 0 x 0 x 0 x 0. I additio,

12 h x 0 0 whe 0 x 0 h x 0. whe x 0 h 2 hece lim h x 0 if 0 x /2 if x if x (d)cosider p x if x 0 if x o D, By the defiitio of p, for ay x 0, p x 0 if x 0 ; i.e., for ay x 0, ad ay 0 The for all x, lim p x. (e) Cosider the sequece, The for all, p x 0 0 if x 0. q x x 2 o D 0,. q 0 0 ad q x 2 for 0 x It follows that lim q x 0; i.e., for all x 0,, q x 0 2 if log log2 Note that N here depeds oly o ad ot o x. N. Poitwise ad Uiform Covergece I these examples, we were able to show that f x fx for Nx,. I each example but the last, the iteger N depeded ot just o but o the poit x i D. These examples satisfy the coditio i the defiitio of covergece of f x to fx. We say for each of these previous examples that the sequeces coverge poitwise to their limit fuctio. For the last example sequece q x, for each 0, there exists N 0 such that sup xd q x qx for all N. It is importat to otice that i this example, N depeds o but it does ot deped o x D. I such cases, we say the sequece q x coverges uiformly to the limit fuctio, q 0. Example (a)cosider 2

13 f x si x o D 0, The f /2 for every ad, for x /2, f x si x r with r si x. The for each x 0 /2, r 0 si x 0, f x 0 r 0 if log N, r logr 0 0 ad lim f x if x /2 0 if x /2 Sice N, r 0 depeds o ad o x 0 (via r 0 ), we coclude that the sequece coverges poitwise to its limit. (b)cosider I this case, ad g x g x si x o D 0, si x o D g x if N. Sice N does ot deped o x, we coclude that g coverges uiformly to zero o D. I geeral, to determie whether a give sequece of fuctios coverges uiformly to a limit, it is ecessary to proceed as i the most recet example to see whether N ca be chose idepedetly of x. For ifiite series, there is a simple test which idicates uiform covergece. Theorem 6.8 Let f x :, 2,... deote a family of fuctios o a commo domai, D a, b, ad suppose there exists a sequece of positive costatsm that satisfies the followig two coditios: i) for each, f x M for all x D, ad ii) M is coverget. The f x coverges uiformly o D. Proof To prove that f x coverges it will be sufficiet to prove that the sequece of N partial sums, S N x f x, is a Cauchy sequece, uiformly i x. That is, it must be prove that for every 0, there exists a iteger N 0 (idepedet of x ) such that S N x S M x wheever, M, N N. To show this write Sice M N S N x S M x M N f x M N f x M. is coverget, for every 0, there exists a iteger N 0 such that N M, ad the result follows. 3

14 Corollary Suppose the power series a x c 0 has radius of covergece R, 0 R. The the series coverges uiformly o the set x c R 0 for every R 0 R. Proof Note that o the set x c R 0, we have f x a x c satisfyig f x a R 0. It is furthermore evidet by applyig the ratio test to the series a R 0 the series is coverget; ie. we compute, 0 that a lim R 0 a R 0 R 0 lim a a R 0 L R 0 /R, ad sice the limit of the ratio of cosecutive terms is less tha oe, the series coverges. Here we recall from the ratio test that the radius of covergece R is the reciprocal of the limit, lim a a L. Cosequeces of Uiform Covergece For ifiite series of fuctios we ca use theorem 6.8 to determie if the series coverges uiformly. O the other had, we have o such simple approach to determie whether the covergece of a sequece f x of fuctios o a commo domai D coverges uiformly or just poitwise. I geeral, to determie whether a give sequece of fuctios coverges uiformly to a limit, it is ecessary to show that for each give 0, the correspodig N ca be chose idepedetly of x i D. As to why it is importat to kow whether or ot a sequece coverges uiformly, cosider the followig theorem. Theorem 6.9 Let f x :, 2,... deote a family of cotiuous fuctios o a commo domai, D a, b, ad suppose the sequece f coverges uiformly o D to the limit f. The f is ecessarily cotiuous o D. Corollary Let f x :, 2,... deote a family of cotiuous fuctios o a commo domai, D a, b, ad suppose the ifiite series f coverges uiformly o D to the sum, Fx. The Fx is ecessarily cotiuous o D. With these results i had, we ca ow aswer each of the questios that were posed i sectio 2. Theorem 6.0 Let f x :, 2,... deote a family of cotiuous fuctios o a commo domai, D a, b, ad suppose the sequece f coverges uiformly o D to the limit f. The Corollary b b b lim f a lim a f f a Let f x :, 2,... deote a family of cotiuous fuctios o a commo domai, D a, b, ad suppose the ifiite series f coverges uiformly o D to 4

15 the sum, Fx. The b b a f a f a b F Theorem 6. Let f x :, 2,... deote a family of cotiuously differetiable fuctios o a commo domai, D a, b, ad suppose that: i) the sequece f coverges uiformly o D to the limit f, ad ii) the sequece of derivatives, f coverges uiformly o D to the limit g. The f is differetiable ad f x gx Corollary Let f x :, 2,... deote a family of cotiuously differetiable fuctios o a commo domai, D a, b, ad suppose that: i) the ifiite series f coverges uiformly o D to the sum, Fx, ad ii) the differetiated ifiite series f coverges uiformly o D to the sum, Gx. The F G; i.e., F x d dx f x f x Gx. Proof of theorem 6.8- Let f x :, 2,... deote a family of cotiuous fuctios o a commo domai, D a, b, ad suppose the sequece f coverges uiformly o D to the limit f. Let x, y deote arbitrary poits i a, b ad let 0 be give. Now write fx fy fx f x f x f y f y fy fx f x f x f y f y fy. Sice f coverges uiformly o D to the limit f, there exists a iteger N 0 such that fx f x 3, ad fy f y 3 for all N Note that we have used the uiform cotiuity here to esure that the same N works for both x ad y. Now fix a N ad use the fact that every f is cotiuous o D to coclude that there exists a 0 such that f x f y 3 wheever x y. The for ay 0 there is a 0 such that fx fy wheever x y. Exercises Fid the values of x for which the followig power series coverge:. x 2. x 5

16 x 2 l 2x2 x 2 2 x2 2 3 l 2x2! l 2x4 3 6