M Calculus II - Chapter 10 Sequences and Series. Rob Malo

Transcription

1 M Calculus II - Chapter 10 Sequeces ad Series Rob Malo Jue 20, 2016

2 Cotets 10 Sequeces ad Series Sequeces Itroductio to Series Geometric Series Fractal Dimesio Power Series Ratio ad Root Tests Covergece of Positive Series Coditioal ad Absolute Covergece Ituitio

3 10 Sequeces ad Series 10.1 Sequeces Dichotomy Paradox, Zeo BC: To travel a distace of 1, first oe must travel 1/2, the half of what remais, i.e. 1/4, the half of what remais, i.e. 1/8, etc. Sice the sequece is ifiite, the distace caot be traveled. Remark. The steps are terms i the sequece. { 1 2, 1 4, 1 } 8, Sequeces of values of this type is the topic of this first sectio. Remark. The sum of the steps forms a ifiite series, the topic of Sectio 10.2 ad the rest of Chapter = 1 2 = 1 We will eed to be careful, but it turs out that we ca ideed walk across a room! Defiitio A sequece is a fuctio with domai N = {1, 2, 3,...}, the Natural Numbers. Examples ad Notatio:

4 4 CHAPTER 10. SEQUENCES AND SERIES Defiitio We say the sequece {a } coverges to L ad write lim a = L or a L as if for every ɛ > 0, there exists M such that a L < ɛ whe > M. If the limit does ot exist, we say the sequece diverges. If L =, we say the sequece diverges to ifiity. Examples:

5 10.1. SEQUENCES 5 Theorem If lim x f (x) = L, the lim f () = L. Remark. The implicatio does ot work the other directio, i.e. lim f () = L = lim f (x) = L, for example: x { } l Example Show coverges. Remark. Do ot apply L Hopital s Rule to terms of a sequece. Sequeces are ot differetiable fuctios, ot eve cotiuous.

6 6 CHAPTER 10. SEQUENCES AND SERIES May previous results regardig limits apply i the sequece case as well. For coveiece they are summarized here. Theorem If lim a = A ad lim b = B the lim (a + b ) = A + B, lim (a b ) = AB, ( ) a lim = A provided B = 0, ad B b lim (ca ) = ca for ay costat c. Theorem (Squeeze Theorem). If a b c for M, lim a = L, ad lim c = L, the lim b = L. Example Sice a a a, by the Squeeze Theorem, if lim a = 0, the lim a = 0. { 3 } Example Use the Squeeze Theorem to show coverges.!

7 10.1. SEQUENCES 7 Oe more result from earlier is useful for us. Theorem If f is cotiuous ad a L as, the f (a ) f (L) as. ( ) Example Fid lim l Defiitio Bouded. {a } is bouded above if there exist M such that a M for all. {a } is bouded below if there exist N such that a N for all. {a } is bouded if it is bouded above ad below. Examples:

8 8 CHAPTER 10. SEQUENCES AND SERIES Defiitio Mootoe. {a } is icreasig if a a +1 for all. {a } is decreasig if a a +1 for all. {a } is mootoe if either of the above hold. Examples: The two previous defiitios are useful by themselves, but combied they give us the oe Big Gu of the theory of sequeces. Theorem (Mootoe Covergece Theorem). If {a } is bouded ad mootoe, the it coverges.

9 10.1. SEQUENCES 9 We fiish our survey of sequeces with five limits, may of which will prove useful for the remaider of this chapter. Theorem Let p > lim p = 0 2. lim p = 1 3. lim = 1 4. If r < 1, lim r = 0. ( 5. lim 1 + c ) = e c Remark. Parts 2 ad 3 imply that if P() is ay polyomial i, the P() 1 as. This will be particularly useful i a upcomig sectio. Selected proofs.

10 10 CHAPTER 10. SEQUENCES AND SERIES 10.2 Itroductio to Series We retur to the Dichotomy Paradox. It is clear that I ca walk across the room, so we should be able to do somethig like this = 1 2 = 1 Of course, we will eed to formalize this. We start with the stadard sigma otatio for fiite sums For example, N a = a 1 + a a N. N We will defie the Nth partial sum to be S N = a. =? Remark. The startig idex is ot particularly importat. Although may examples will start at = 1 or = 0, there is o reaso it caot start at = 47. Additioally, there is othig special about the idex. We ofte use i, j, k, or m as well. For example, the 5th partial sum for the followig series are S 5 = 5 k = k=2 S 5 = 5 k=1 1 k 2 = S 5 = 5 k=4 1 k 3 1 =

11 10.2. INTRODUCTION TO SERIES 11 We would like to defie a ifiite series, i.e. a = a 1 + a 2 + a 3 + The stadard 172 idea holds, approximate with partial sums ad take the limit. Defiitio Let S N = If lim S = S, we say If lim S =, we say N a be the sequece of Nth partial sums. a coverges to S ad write a diverges to ifiity ad write If lim S does ot exist, we say a diverges. a = S. a =. Oe type of series that we ca approach usig the defiitio directly are telescopig series. Example Fid the sum of the series 2 ( + 2).

12 12 CHAPTER 10. SEQUENCES AND SERIES There is oe additioal type of series that we ca use the defiitio directly for, they are the topic of the followig sectio. For ow, we tur our attetio to oe issue of theoretical importace ad fially oe fudametal example. Theorem If a k coverges, a k 0 as k. Remark. Covergece or divergece of a series depeds o the behavior of the tail of the series, i.e. we ca throw away the first te, or hudred, or eve millio terms ad ot chage the covergece of the series. If the idex is immaterial to the topic at had, as i the theorem above, we will ofte supress the otatio for coveiece. Proof. The above theorem is typically used i its cotrapositive form. Theorem (Test for Divergece). If lim a = 0, the a diverges. Remark. We are sayig that the terms goig to zero is a ecessary coditio for the covergece of a series. However, as the ext example shows, it is ot a sufficiet coditio.

13 10.2. INTRODUCTION TO SERIES 13 Example The Harmoic Series diverges, i.e. 1 = =. Proof.

14 14 CHAPTER 10. SEQUENCES AND SERIES Mostly Empty Page

15 10.3. GEOMETRIC SERIES Geometric Series A importat, perhaps the most importat, type of series is the geometric series. We have already see oe example, our walk across the room. 1 2 = = 1 Before we dive ito the geeral theory, we should look closely at this example. As before, we cosider the Nth partial sum S N.

16 16 CHAPTER 10. SEQUENCES AND SERIES Theorem (Geometric Series). For c = 0, diverges otherwise. cr = =0 c 1 r for r < 1 ad Proof. Remark. The more geeral result is cr = first term 1 r for r < 1.

17 10.3. GEOMETRIC SERIES 17 Examples: =0 3 2 = = =3 3 2 = = = = 1 + ( 2) =0 3 2 We will eed additioal machiery to deal with this last example.

18 18 CHAPTER 10. SEQUENCES AND SERIES Sice fiite sums ad limits are both liear, so are series. Theorem (Liearity of Series). Assume the followig series are coverget, the ca = c a, ad (a + b ) = a + b. We ca ow retur to the example from the previous page ad a similar example. 1 + ( 2) = ( 4) =0 5 Remark. The assumptio that all the series coverged i the theorem is ecessary. For example, Sice either of the last series coverge. 0 = 0 = (1 1) = 1 1

19 10.3. GEOMETRIC SERIES 19 I our discussio of geometric series, the commo ratio r was costat. What happes if we let r vary? Example Fid the values of x for which the followig series coverges ad fid what it coverges to. =0 x Example Fid the values of x for which the followig series coverges ad fid what it coverges to. 2( 1) x 2 =0 4 Remark. The two series o this page are represetatios of fuctios. They are examples of series we will refer to to as power series, the topic sectio Homework From sectio 10.2 i the text, # 23, 25, 27, 29, 33, 39, 43, 47, 49, 57

20 20 CHAPTER 10. SEQUENCES AND SERIES K 0 K 1 K 2 K 3 K 4. K Figure 10.1: Middle third Cator set 10.4 Fractal Dimesio I this brief diversio we cosider the questio, how big is a fractal? We begi by discussig four elemetary examples that will give us a framework for the upcomig discussio. The examples are classic costructios that date from the late ieteeth ad early twetieth cetury. Oe of the simplest fractals to costruct is the middle third Cator set, amed for the Germa mathematicia Georg Cator, see Figure Let K 0 be the uit iterval [0, 1]. Remove the ope middle third (1/3, 2/3) to obtai K 1 = [0, 1/3] [2/3, 1]. Removig the ope middle third of each remaiig iterval gives K 2 = [0, 1/9] [2/9, 1/3] [2/3, 7/9] [8/9, 1]. Cotiuig i this maer, we costruct the middle third Cator set K := lim i K i Alteratively we could view the costructio at each step ot as removig itervals, but rather as replacig each previous iterval with two itervals of oe third the legth. The secod example we cosider is the vo Koch curve, ofte called a sowflake curve, amed after the Swedish mathematicia Helge vo Koch. While the Cator set removed a iterval at each iteratio, the vo Koch curve replaces each middle third with two segmets of the same legth. I essece, replacig a segmet with the two other sides of a equilateral triagle. Alteratively, we

21 10.4. FRACTAL DIMENSION 21 V 0 V 1 V 2... V Figure 10.2: vo Koch curve G 0 G 1 G 2... G Figure 10.3: Sierpiński gasket ca view the costructio as replacig each iterval with four itervals of oe third the legth, see Figure We will soo see that the Cator set has zero legth. Curiously, the sowflake curve has ifiite legth. Noetheless, the curve has zero area i the plae. It will be useful to fid a measure of the size of this curve, ad similar objects, that is a more useful measuremet tha ifiite legth or zero area. Oe such measure of size is the dimesio, which we will discuss soo. However, two additioal examples are worth familiarizig ourselves with before we dive i. The fial two examples are similar to the costructio of the Cator set ad both due to the Polish mathematicia Wacław Sierpiński. The Sierpiński gasket is costructed from a equilateral triagle. At each iteratio we divide the triagle, or triagles, ito four cogruet subtriagles ad remove the cetral subtriagle, see Figure A Sierpiński carpet is costructed from a square. At each iteratio we divide the square, or squares, ito ie cogruet subsquares ad remove the cetral subsquare, see Figure We will soo see that the area of the gasket ad the carpet are both zero. Agai we see that area is a poor measure of these objects. A more useful aalytic tool

22 22 CHAPTER 10. SEQUENCES AND SERIES S 0 S 1 S 2... S Figure 10.4: Sierpiński carpet for these types of fractals is their dimesio. Before we cosider the cocept of dimesio, we should verify the claims made earlier. Exercise Show the legth of the Cator set is 0. This is homework. Exercise Show the leght of the vo Koch curve is ifiite. Solutio. It is clear the lie segmet V 0 has legth 1. At the ext iteratio we replace the segmet by four pieces oe third the legth, so the legth of V 1 is (4/3). The legth grows by this factor at each step, so V 2 has legth (4/3) 2, V 3 has legth (4/3) 3, ad i geeral V has legth (4/3). Sice (4/3) as we see the legth of the vo Koch curve V is ifiite. This is a example of a diverget geometric sequece, see Theorem Exercise Show the area of the Sierpiński gasket is 0. This is homework. Exercise Show the area of the Sierpiński carpet is 0. Solutio. The area ca be computed by subtractig the removed squares from the total area. Covietly, the area of removed squares form a geometric series. ( ) 1 Area = = 1 1/9 1 8/9 = 1 1 = 0

23 10.4. FRACTAL DIMENSION 23 For these types of mathematical toys, ad may real world objects that have self-simliar structures at various scales, a useful measure is the dimesio. There are may cocepts of dimesio, we will discuss a very basic versio. What is dimesio? Roughly, umber of pieces = (1/( size of pieces)) dimesio

24 24 CHAPTER 10. SEQUENCES AND SERIES Solvig for the dimesio i the previous gives. dim X = l (umber of pieces) l ( size of pieces) Or formally, if X is self similar shape made of N copies of itself, each scaled by a similarity with cotractio factor r the we defie the similarity dimesio as dim S X = l N l r The examples we have see have the followig similarity dimesios. Cator set K, dim S K = l 2 l vo Koch curve V, dim S V = l 4 l Sierpiński gasketg, dim S G = l 3 l Sierpiński carpet S, dim S S = l 8 l Although similarity dimesio is very easy to compute for our examples, it is ot very flexible. For more complicated objects, mathematical or real world, other more rigorous methods are eeded. But, we are already beyod the scope of the class, so we should retur to the topic at had.

25 10.5. POWER SERIES Power Series A importat applicatio of series is that of a power series, i.e. a way to represet a fuctio by a series. I essece, turig a fuctio like e x ito somethig computable. Although it is useful to thik of e x as the fuctio that is its ow derivative ad that has slope 1 whe x = 0, that is hard to use from a computatioal poit of view. We saw two examples of power series at the ed of sectio The most importat beig F(x) = 1 1 x = x = 1 + x + x 2 + for x < 1. (10.1) =0 The goal of this chapter is to be able to fid series represetatios for fuctios. I essece, fidig ifiite polyomial represetatios of fuctios. Defiitio A power series with ceter c is a series of the form F(x) = a (x c) = a 0 + a 1 (x c) + a 2 (x c) 2 + =0 I the power series (10.1) above, a = 1 for all ad the ceter is c = 0. Oe thig to ote is that the series oly coverges for x < 1. This type of restrictio o x is typical. Theorem (Radius of Covergece). Every power series F(x) = a (x c) = a 0 + a 1 (x c) + a 2 (x c) 2 + =0 has a radius of covergece R with R = 0, R > 0, or R =. If R = 0, the series oly coverges at x = c. If R > 0, the series coverges for x c < R. If R =, the series coverges for all x R. We will be iterested i determiig the radius of covergece for power series soo, but we start with geeratig some power series based off of (10.1).

26 26 CHAPTER 10. SEQUENCES AND SERIES Games we ca play with power series. 1. Multiplicatio. Fid a power series represetatio for f (x) = 2x 1 x. 2. Substitutio. Fid a power series represetatio for g(x) = x 2.

27 10.5. POWER SERIES Shiftig the ceter. Fid a power series represetatio cetered at x = 2 for h(x) = 1 1 x. Oe reaso power series are so useful is that they are very easy to differetiate ad itegrate. Theorem Assume F(x) = a (x c) = a 0 + a 1 (x c) + a 2 (x c) 2 + =0 has radius of covergece R. The F is differetiable ad itegrable o (c R, c + R) with F (x) = a (x c) 1 = a 1 + 2a 2 (x c) + 3a 3 (x c) 2 + ad, for ay costat A F(x) dx = A + =0 a + 1 (x c)+1. Furthermore, all three series have the same radius of covergece.

28 28 CHAPTER 10. SEQUENCES AND SERIES Example Fid a power series represetatio for l(1 + x).

29 10.5. POWER SERIES 29 Example Fid a power series represetatio for arcta x. Remark. Although the radius of covergece does ot chage whe we itegrate, it is possible that the iterval of covergece does. We will discuss the covergece at edpoits soo, but for ow we simply ote that the represetatio of arcta x is valid for x [ 1, 1] which gives the followig curious result. arcta 1 = π 4 =

30 30 CHAPTER 10. SEQUENCES AND SERIES Example Fid a power series represetatio for e x. We start by a quick diversio ito differetial equatios, i particular it is major result i the subject that there is a uique solutio to a Iital Value Problem (IVP)of the followig type y = y, y(0) = 1 i.e. there is a uique fuctio that is its ow derivative with slope oe whe x = 1. The fuctio e x clearly satifies the IVP ad so is the uique solutio. We will put that kowledge o hold for a miute ad search for a power series solutio to the IVP. Usig power series to solve differetial equatios is a fudametal tool i the subject. Which leads us to the recursive relatioship a = a 1. (10.2)

31 10.5. POWER SERIES 31 We ca choose a 0 arbitrarily, ad use (10.2) to fid the other coefficiets i terms of a 0. We coclude that F(x) = a 0 =0 We also wat it to satisfy the iital data, i.e. F(0) = 1. Sice F(0) = a 0 we coclude that a 0 = 1 ad e x x = =0!. (10.3) x!. We are left with oe major cocer; for what values of x does this series coverge? That questio will be the topic of the followig sectio.

32 32 CHAPTER 10. SEQUENCES AND SERIES Homework. Exercise True / False 1. T / F : If a (x 3) coverges for x = 5 it coverges for x = T / F : If a (x + 3) coverges for x = 5 it coverges for x = T / F : There exists a power series that oly coverges for x > 0. Exercise If a power series coverges for 4 < x < 2, what is the ceter ad radius of covergece? Exercise Use (10.1) to fid a power series represetatio for the followig fuctios. Specify where each series coverges. Uless otherwise specified, ceter the series at x = f (x) = x 2. f (x) = x 3 3. f (x) = x 1 x 4. f (x) = 3x2 2 + x 4 5. f (x) = 1 1 x 6. f (x) = x cetered at x = 4 cetered at x = 2 7. f (x) = 1 (1 + x) f (x) = (1 + x) 3 Exercise Use (10.3) to itegrate e x2 dx.

33 10.6. RATIO AND ROOT TESTS Ratio ad Root Tests I this sectio we develop two tests useful for determiig the covergece or divergece of series with a particular emphasis o power series. Both are geeralizatios of the geometric series from sectio Theorem (The Ratio Test). Let ρ = lim 1. if ρ < 1, a coverges. 2. if ρ > 1, a diverges. a +1 a 3. if ρ = 1 or the limit does ot exist, the test is icoclusive. Sketch of Proof. Theorem (The Root Test). Let L = lim a 1. if L < 1, a coverges. 2. if L > 1, a diverges. 3. if L = 1 or the limit does ot exist, the test is icoclusive. Sketch of Proof.

34 34 CHAPTER 10. SEQUENCES AND SERIES Iitial Examples. Show the followig series coverge. ( ) + 1 = =0!

35 10.6. RATIO AND ROOT TESTS 35 Example I the previous sectio we showed that e x = x =0!. Show that the series coverges for all x, i.e. the radius of covergece is ifiite. Both tests are useful for determiig the radius of covergece of power series. For kow power series, there is o eed to reivet the wheel, but to determie the radius of covergece of a ukow series either of the tests is our first step. There are two importat poits to ote. First, both tests are icoclusive at the edpoits. Neither will give us ay isight ito the iterval of covergece. To address that issue we will have to develop more subtle tests; that will be the topic of the ext few sectios. Secod, although the Root Test seems trickier, we have some ice results that streamlie the process; see Theorem I particular...

36 36 CHAPTER 10. SEQUENCES AND SERIES Power Series Examples. Fid the radius of covergece for the followig. 2 (x 2) 2 9 ( ) 2 (x + 1) + 2

37 10.6. RATIO AND ROOT TESTS 37 Two Additioal Examples. Fid the radius of covergece for the followig. =0 (2x) 2 + 1!x =0

38 38 CHAPTER 10. SEQUENCES AND SERIES Homework I sectio 10.5 do #5, 11, 17, 19, 23, 29, 37, 41 Exercise Fid the radius of covergece for the followig. Express your solutio i the form x c < R, where possible (x 3) 2 l (x + 4) 3 2 (x 1) =0 + 2 =0 =0 2 x (2)!!x ( 2) (x + 2) =0 3 ( 1) x 2 =0 (2)!

39 10.7. CONVERGENCE OF POSITIVE SERIES Covergece of Positive Series I this sectio we do a brief survey of methods for testig covergece at edpoits of itervals of covergece. Theorem (Itegral Test). Let a = f (), where f is positive ad decreasig for x 1. f (x) dx coverges if ad oly if a coverges. 1 Sketch of Proof.

40 40 CHAPTER 10. SEQUENCES AND SERIES Example Covergece of p-series. 1 p coverges if ad oly if p > 1.

41 10.7. CONVERGENCE OF POSITIVE SERIES 41 Theorem (Direct Compariso Test). Assume 0 a b for M. 1. If 2. If Examples. Show b coverges, the a coverges. a diverges, the b diverges diverges. Show coverges.

42 42 CHAPTER 10. SEQUENCES AND SERIES Theorem (Limit Compariso Test). Let a, b 0. Assume the followig limit exists, or is. a L = lim b 1. If 0 < L <, the a coverges if ad oly if b coverges. 2. If L = 0, the a coverges if b coverges. 3. If L =, the b coverges if a coverges. The limits of 0 ad are rarely used. Typically we try to compare with a series that is of a similar size ad hece have a limit that is positive ad fiite, i.e. 0 < L <. Geerally it is easier to write a argumet for the Direct Compariso Test, but i cases where the iequalities go the wrog directio we make use of the Limit Compariso Test. Two examples. ( ) 1 Show si coverges.

43 10.7. CONVERGENCE OF POSITIVE SERIES 43 Show coverges. (x 2) Example Usig the Ratio Test we ca show that 2 coverges for x 2 < 2 ad diverges for x 2 > 2. Does it coverge whe x 2 = 2? Whe x = 4 we have Whe x = 0 we have Our curret tools do ot address this situatio, but the followig sectio will. Exercise From sectio 10.3 i the text do # 79 usig the Itegral Test, 19, 21, 25, 31, 39, 41, 43, 47.

44 44 CHAPTER 10. SEQUENCES AND SERIES 10.8 Coditioal ad Absolute Covergece Iitially we ote that everythig we discussed regardig positive series a, i.e. a > 0, i the previous sectio applies to egative series b, i.e. b < 0, after we factor out the egative (series are liear). For series with both sigs, we will eed additioal tools. Defiitio a coverges absolutely if a coverges. Remark. A coverget positive series is absolutely coverget by defiitio. Theorem If a coverges absolutely, the a coverges. Proof. Examples. ( 1) 1 2 = cos Remark. To use either of the compariso tests we must have oegative terms, so ofte testig for absolute covergece is ecessary.

45 10.8. CONDITIONAL AND ABSOLUTE CONVERGENCE 45 Example ( 1) 1 = is the Alteratig Harmoic Series. Sice the Harmoic Series diverges, the Alteratig Harmoic Series does ot coverge absolutely. But, does it coverge? This was the questio we left off with i Example Defiitio a coverges coditioally if it coverges, but ot absolutely. Defiitio For a > 0, a series of the form ( 1) a or ( 1) +1 a is a alteratig series. Theorem [Alteratig Series Test] If a > 0, a decreasig, ad a 0 as, the S = estimates for N 1: ( 1) +1 a coverges. Additioally, we have the followig 1. 0 < S < a 1, 2. S 2N < S < S 2N+1, ad 3. S S N < a N+1. Sketch of Proof. To fiish Example above,

46 46 CHAPTER 10. SEQUENCES AND SERIES Example Fid the iterval of covergece for the power series ( 1) (x + 2) =0 ( + 1)3. As before, we use either the Ratio or Root test to fid the radius of covergece. Now we check the edpoits idividually. For x = 5 we have For x = 1 we have

47 10.8. CONDITIONAL AND ABSOLUTE CONVERGENCE 47 Example I Example we foud a power series represetatio for arcta x by itegratig the power series for 1/(1 + x 2 ), i particular arcta x = ( 1) x 2+1 = = x x3 3 + x5 5 x7 7 + (10.4) which was valid for x < 1. Theorem assured us that the radius of covergece remaied 1, but it does ot tell us aythig about the covergece or divergece at the edpoits of the iterval. We test the two edpoits ow. At x = 1 we have At x = 1 we have We coclude that (10.4) is valid for 1 x 1.

48 48 CHAPTER 10. SEQUENCES AND SERIES Example I Example we foud a power series represetatio for l (1 + x) by itegratig the power series for 1/(1 + x), i particular l (1 + x) = ( 1) x +1 =0 + 1 = x x2 2 + x3 3 x4 4 + (10.5) which was valid for x < 1. As i the previous example, we would like to check the edpoits for covergece. At x = 1 we have At x = 1 we have We coclude that (10.5) is valid for 1 < x 1. It would be a problem if the series coverged at x = 1 sice we would be computig the value of l 0.

49 10.8. CONDITIONAL AND ABSOLUTE CONVERGENCE 49 Example Show coverges coditioally. This argumet requires two parts. (10.6) First, we must show (10.6) does ot coverge absolutely. Secod, we must show (10.6) does coverge.

50 50 CHAPTER 10. SEQUENCES AND SERIES Coditioal Covergece is Curious, or, a brief bit a mathematical weirdess. I the previous example we saw that l 2 = We cosider rearragig the terms as follows This turs out to be the case with all coditioally coverget series, if we rearrage the terms we ca chage the sum, or eve make the series diverge. Theorem (Riema Rearragemet Theorem). If a is a coditioally coverget series, the for ay M R there exists a rearragemet of {a } ito {b }, i.e. a oe-to-oe oto mappig, such that b = M. Furthermore, there is a differet rearragemet of {a } ito {c } such that c =. Theorem If a is absolutely coverget, the ay rearragemet also coverges absolutely ad to the same value.

51 10.8. CONDITIONAL AND ABSOLUTE CONVERGENCE 51 Remark. Now that we have the appropriate laguage it should be oted that the coclusio i both the Root ad Ratio test is that the series i questio coverges absolutely whe the appropriate limit is less tha oe. Similarly, power series coverge absolutely o their radius of covergece. Sice absolute covergece implies covergece, the results as writte are true, but ot as strog as they are properly. Homework Exercise True or False. 1. T / F : If a coverges coditioally, the a coverges. 2. T / F : If a coverges absolutely, the a coverges. 3. T / F : If a coverges, the a coverges. 4. T / F : If a 0 as, the a coverges. 5. T / F : If a 0 as, the a coverges. Exercise Determie if the followig coverge absolutely, coverge coditioally, or diverge. Proper justificatio will be expected o quizzes ad exams, ow is a good time to practice. 1. si cos ( 1) Exercise Fid a error estimate for Hit: See Theorem ad Example π 4 ( ) <. Exercise From sectio 10.6 i the text do # 11, 13, 19, 23 (Note: for 1, l < ), 29

52 52 CHAPTER 10. SEQUENCES AND SERIES 10.9 Ituitio I this sectio we cocetrate o tryig to build ad ituitio ad uderstadig of the behavior of series. Although we should be able to craft proper argumets to verify all of the claims we will make, it is importat to be able to idetify a coverget series. Of equal importace, beig able to idetify the correct method of showig covergece or divergece. Methods we have discussed. 1. Test for Divergece. Examples. 2. Geometric Series. Examples.

53 10.9. INTUITION Ratio Test. Examples. 4. Root Test. Examples. Questio: For what types of series are the Root ad Ratio Test icoclusive?

54 54 CHAPTER 10. SEQUENCES AND SERIES 5. Itegral Test. Examples. 6. Compariso Tests. Examples.

55 10.9. INTUITION 55 Exercise For each of the followig series determie if they coverge or diverge ad the choose a test that ca be used to show that. ( ) 2 Coverges / Diverges by the Root Test / Divergece Test. ( ) 1 Coverges / Diverges by the Root Test / Divergece Test l Coverges / Diverges Coverges / Diverges by the Ratio Test / Itegral Test. by the LCT / Ratio Test. 5. ( 1) Coverges / Diverges by the Itegral Test / AST. 6. ( ) Coverges / Diverges by the Divergece Test / AST. Exercise For the followig series, specify what series you would compare each to (either direct or limit compariso) ad based o your compariso, decide if it coverges or diverges. No formal justificatio is eeded = compare to so it coverges / diverges 2. =2 3 2 compare to so it coverges / diverges 3. = compare to so it coverges / diverges =2 2 + compare to so it coverges / diverges 5. =2 2 3 compare to so it coverges / diverges

56 56 CHAPTER 10. SEQUENCES AND SERIES Exercise Choose all series below that ca be show to coverge usig the Divergece Test. (a) 4 l (b) 2 (c) 1 e ca be show to diverge usig the Divergece Test. (a) 1 (b) 1 (c) 1 l 3. ca be show to coverge usig either Compariso Test. (a) 1 l (b) ( 1) (c) l 4. ca be show to coverge usig the Alteratig Series Test. (a) ( 1) (b) cos(π) (c) ( 1) ca be show to diverge usig the Alteratig Series Test. (a) cos (b) ( 1) 6. ca be show to coverge usig the Ratio or Root Test. (a) (b) ca be show to diverge usig the Ratio or Root Test. (c) ( 1) + 1 (c) 2! (a) (b) (c)!