CALCULUS II. Sequences and Series. Paul Dawkins

Transcription

1 CALCULUS II Sequeces ad Series Paul Dawkis

2 Table of Cotets Preface... ii Sequeces ad Series... 3 Itroductio... 3 Sequeces... 5 More o Sequeces...5 Series The Basics... Series Covergece/Divergece...7 Series Special Series...36 Itegral Test...44 Compariso Test / Limit Compariso Test...53 Alteratig Series Test...6 Absolute Covergece...68 Ratio Test...7 Root Test...79 Strategy for Series...8 Estimatig the Value of a Series...85 Power Series...96 Power Series ad Fuctios...4 Taylor Series... Applicatios of Series... Biomial Series Paul Dawkis i

3 Preface Here are my olie otes for my Calculus II course that I teach here at Lamar Uiversity. Despite the fact that these are my class otes, they should be accessible to ayoe watig to lear Calculus II or eedig a refresher i some of the topics from the class. These otes do assume that the reader has a good workig kowledge of Calculus I topics icludig limits, derivatives ad basic itegratio ad itegratio by substitutio. Calculus II teds to be a very difficult course for may studets. There are may reasos for this. The first reaso is that this course does require that you have a very good workig kowledge of Calculus I. The Calculus I portio of may of the problems teds to be skipped ad left to the studet to verify or fill i the details. If you do t have good Calculus I skills, ad you are costatly gettig stuck o the Calculus I portio of the problem, you will fid this course very difficult to complete. The secod, ad probably larger, reaso may studets have difficulty with Calculus II is that you will be asked to truly thik i this class. That is ot meat to isult ayoe; it is simply a ackowledgmet that you ca t just memorize a buch of formulas ad epect to pass the course as you ca do i may math classes. There are formulas i this class that you will eed to kow, but they ted to be fairly geeral. You will eed to uderstad them, how they work, ad more importatly whether they ca be used or ot. As a eample, the first topic we will look at is Itegratio by Parts. The itegratio by parts formula is very easy to remember. However, just because you ve got it memorized does t mea that you ca use it. You ll eed to be able to look at a itegral ad realize that itegratio by parts ca be used (which is t always obvious) ad the decide which portios of the itegral correspod to the parts i the formula (agai, ot always obvious). Fially, may of the problems i this course will have multiple solutio techiques ad so you ll eed to be able to idetify all the possible techiques ad the decide which will be the easiest techique to use. So, with all that out of the way let me also get a couple of warigs out of the way to my studets who may be here to get a copy of what happeed o a day that you missed.. Because I wated to make this a fairly complete set of otes for ayoe watig to lear calculus I have icluded some material that I do ot usually have time to cover i class ad because this chages from semester to semester it is ot oted here. You will eed to fid oe of your fellow class mates to see if there is somethig i these otes that was t covered i class.. I geeral I try to work problems i class that are differet from my otes. However, with Calculus II may of the problems are difficult to make up o the spur of the momet ad so i this class my class work will follow these otes fairly close as far as worked problems go. With that beig said I will, o occasio, work problems off the top of my head whe I ca to provide more eamples tha just those i my otes. Also, I ofte 7 Paul Dawkis ii

4 do t have time i class to work all of the problems i the otes ad so you will fid that some sectios cotai problems that were t worked i class due to time restrictios. 3. Sometimes questios i class will lead dow paths that are ot covered here. I try to aticipate as may of the questios as possible i writig these up, but the reality is that I ca t aticipate all the questios. Sometimes a very good questio gets asked i class that leads to isights that I ve ot icluded here. You should always talk to someoe who was i class o the day you missed ad compare these otes to their otes ad see what the differeces are. 4. This is somewhat related to the previous three items, but is importat eough to merit its ow item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Usig these otes as a substitute for class is liable to get you i trouble. As already oted ot everythig i these otes is covered i class ad ofte material or isights ot i these otes is covered i class. Sequeces ad Series Itroductio I this chapter we ll be takig a look at sequeces ad (ifiite) series. Actually, this chapter will deal almost eclusively with series. However, we also eed to uderstad some of the basics of sequeces i order to properly deal with series. We will therefore, sped a little time o sequeces as well. Series is oe of those topics that may studets do t fid all that useful. To be hoest, may studets will ever see series outside of their calculus class. However, series do play a importat role i the field of ordiary differetial equatios ad without series large portios of the field of partial differetial equatios would ot be possible. I other words, series is a importat topic eve if you wo t ever see ay of the applicatios. Most of the applicatios are beyod the scope of most Calculus courses ad ted to occur i classes that may studets do t take. So, as you go through this material keep i mid that these do have applicatios eve if we wo t really be coverig may of them i this class. Here is a list of topics i this chapter. Sequeces We will start the chapter off with a brief discussio of sequeces. This sectio will focus o the basic termiology ad covergece of sequeces More o Sequeces Here we will take a quick look about mootoic ad bouded sequeces. Series The Basics I this sectio we will discuss some of the basics of ifiite series. Series Covergece/Divergece Most of this chapter will be about the covergece/divergece of a series so we will give the basic ideas ad defiitios i this sectio. 7 Paul Dawkis iii

5 Series Special Series We will look at the Geometric Series, Telescopig Series, ad Harmoic Series i this sectio. Itegral Test Usig the Itegral Test to determie if a series coverges or diverges. Compariso Test/Limit Compariso Test Usig the Compariso Test ad Limit Compariso Tests to determie if a series coverges or diverges. Alteratig Series Test Usig the Alteratig Series Test to determie if a series coverges or diverges. Absolute Covergece A brief discussio o absolute covergece ad how it differs from covergece. Ratio Test Usig the Ratio Test to determie if a series coverges or diverges. Root Test Usig the Root Test to determie if a series coverges or diverges. Strategy for Series A set of geeral guidelies to use whe decidig which test to use. Estimatig the Value of a Series Here we will look at estimatig the value of a ifiite series. Power Series A itroductio to power series ad some of the basic cocepts. Power Series ad Fuctios I this sectio we will start lookig at how to fid a power series represetatio of a fuctio. Taylor Series Here we will discuss how to fid the Taylor/Maclauri Series for a fuctio. Applicatios of Series I this sectio we will take a quick look at a couple of applicatios of series. Biomial Series A brief look at biomial series. 7 Paul Dawkis 4

6 Sequeces Let s start off this sectio with a discussio of just what a sequece is. A sequece is othig more tha a list of umbers writte i a specific order. The list may or may ot have a ifiite umber of terms i them although we will be dealig eclusively with ifiite sequeces i this class. Geeral sequece terms are deoted as follows, a first term a secod term th a term a + ( ) st + term Because we will be dealig with ifiite sequeces each term i the sequece will be followed by aother term as oted above. I the otatio above we eed to be very careful with the subscripts. The subscript of + deotes the et term i the sequece ad NOT oe plus the th term! I other words, a a + + so be very careful whe writig subscripts to make sure that the + does t migrate out of the subscript! This is a easy mistake to make whe you first start dealig with this kid of thig. There is a variety of ways of deotig a sequece. Each of the followig are equivalet ways of deotig a sequece. {,,,,, } { } { } a a a a a a + = I the secod ad third otatios above a is usually give by a formula. A couple of otes are ow i order about these otatios. First, ote the differece betwee the secod ad third otatios above. If the startig poit is ot importat or is implied i some way by the problem it is ofte ot writte dow as we did i the third otatio. Net, we used a startig poit of = i the third otatio oly so we could write oe dow. There is absolutely o reaso to believe that a sequece will start at =. A sequece will start where ever it eeds to start. Let s take a look at a couple of sequeces. 7 Paul Dawkis 5

7 Eample Write dow the first few terms of each of the followig sequeces. + (a) (b) = + ( ) b = (c) { } = [Solutio] [Solutio] th, where b digit of π = [Solutio] Solutio + (a) = To get the first few sequece terms here all we eed to do is plug i values of ito the formula give ad we ll get the sequece terms =,,,,, = = = = 3 = 4 = 5 Note the iclusio of the at the ed! This is a importat piece of otatio as it is the oly thig that tells us that the sequece cotiues o ad does t termiate at the last term. [Retur to Problems] + (b) = This oe is similar to the first oe. The mai differece is that this sequece does t start at =. +,,,,, = = Note that the terms i this sequece alterate i sigs. Sequeces of this kid are sometimes called alteratig sequeces. [Retur to Problems] b = (c) { } th, where b = digit of π This sequece is differet from the first two i the sese that it does t have a specific formula for each term. However, it does tell us what each term should be. Each term should be the th digit of π. So we kow that π = The sequece is the, { 3,,4,,5,9,,6,5,3,5, } [Retur to Problems] 7 Paul Dawkis 6

8 I the first two parts of the previous eample ote that we were really treatig the formulas as fuctios that ca oly have itegers plugged ito them. Or, + + ( ) f = g = This is a importat idea i the study of sequeces (ad series). Treatig the sequece terms as fuctio evaluatios will allow us to do may thigs with sequeces that could t do otherwise. Before delvig further ito this idea however we eed to get a couple more ideas out of the way. First we wat to thik about graphig a sequece. To graph the sequece { a } we plot the poits ( a, ) as rages over all possible values o a graph. For istace, let s graph the + sequece =. The first few poits o the graph are, ,,,, 3,, 4,, 5,, The graph, for the first 3 terms of the sequece, is the, This graph leads us to a importat idea about sequeces. Notice that as icreases the sequece terms i our sequece, i this case, get closer ad closer to zero. We the say that zero is the limit (or sometimes the limitig value) of the sequece ad write, + lim a = lim = This otatio should look familiar to you. It is the same otatio we used whe we talked about the limit of a fuctio. I fact, if you recall, we said earlier that we could thik of sequeces as fuctios i some way ad so this otatio should t be too surprisig. Usig the ideas that we developed for limits of fuctios we ca write dow the followig workig defiitio for limits of sequeces. 7 Paul Dawkis 7

9 Workig Defiitio of Limit. We say that lim a = L if we ca make a as close to L as we wat for all sufficietly large. I other words, the value of the a s approach L as approaches ifiity.. We say that lim a = if we ca make a as large as we wat for all sufficietly large. Agai, i other words, the value of the a s get larger ad larger without boud as approaches ifiity. 3. We say that lim a = if we ca make a as large ad egative as we wat for all sufficietly large. Agai, i other words, the value of the a s are egative ad get larger ad larger without boud as approaches ifiity. The workig defiitios of the various sequece limits are ice i that they help us to visualize what the limit actually is. Just like with limits of fuctios however, there is also a precise defiitio for each of these limits. Let s give those before proceedig Precise Defiitio of Limit. We say that lim a = L if for every umber ε > there is a iteger N such that. We say that lim a a L < ε wheever > N 3. We say that lim a = if for every umber M > there is a iteger N such that a > M wheever > N = if for every umber M < there is a iteger N such that a < M wheever > N We wo t be usig the precise defiitio ofte, but it will show up occasioally. Note that both defiitios tell us that i order for a limit to eist ad have a fiite value all the sequece terms must be gettig closer ad closer to that fiite value as icreases. Now that we have the defiitios of the limit of sequeces out of the way we have a bit of termiology that we eed to look at. If lim a eists ad is fiite we say that the sequece is coverget. If lim a does t eist or is ifiite we say the sequece diverges. Note that sometimes we will say the sequece diverges to if lim a = ad if lim a = we will sometimes say that the sequece diverges to. 7 Paul Dawkis 8

10 Get used to the terms coverget ad diverget as we ll be seeig them quite a bit throughout this chapter. So just how do we fid the limits of sequeces? Most limits of most sequeces ca be foud usig oe of the followig theorems. Theorem Give the sequece { a } if we have a fuctio f ( ) such that f = a ad lim the lim a = L f = L This theorem is basically tellig us that we take the limits of sequeces much like we take the limit of fuctios. I fact, i most cases we ll ot eve really use this theorem by eplicitly writig dow a fuctio. We will more ofte just treat the limit as if it were a limit of a fuctio ad take the limit as we always did back i Calculus I whe we were takig the limits of fuctios. So, ow that we kow that takig the limit of a sequece is early idetical to takig the limit of a fuctio we also kow that all the properties from the limits of fuctios will also hold. Properties lim a ± b = lim a ± lim b.. lim ca = c lim a 3. lim ( ab) = ( lim a)( lim b) 4. a lim a lim =, provided lim b b lim b p 5. lim a = lim a provided a p These properties ca be proved usig Theorem above ad the fuctio limit properties we saw i Calculus I or we ca prove them directly usig the precise defiitio of a limit usig early idetical proofs of the fuctio limit properties. Net, just as we had a Squeeze Theorem for fuctio limits we also have oe for sequeces ad it is pretty much idetical to the fuctio limit versio. Squeeze Theorem for Sequeces a c b for all > N for some N ad lim a = lim b = L the lim c If = L. 7 Paul Dawkis 9

11 Note that i this theorem the for all > N for some N is really just tellig us that we eed to have a c b for all sufficietly large, but if it is t true for the first few that wo t ivalidate the theorem. As we ll see ot all sequeces ca be writte as fuctios that we ca actually take the limit of. This will be especially true for sequeces that alterate i sigs. While we ca always write these sequece terms as a fuctio we simply do t kow how to take the limit of a fuctio like that. The followig theorem will help with some of these sequeces. Theorem If lim a = the lim a =. Note that i order for this theorem to hold the limit MUST be zero ad it wo t work for a sequece whose limit is ot zero. This theorem is easy eough to prove so let s do that. Proof of Theorem The mai thig to this proof is to ote that, a a a The ote that, ( a ) lim = lim a = We the have ( a ) lim = lim a = ad so by the Squeeze Theorem we must also have, lim a = The et theorem is a useful theorem givig the covergece/divergece ad value (for whe it s coverget) of a sequece that arises o occasio. Theorem 3 The sequece { } r = coverges if < r ad diverges for all other value of r. Also, lim r if < r < = if r = Here is a quick (well ot so quick, but defiitely simple) partial proof of this theorem. Partial Proof of Theorem 3 We ll do this by a series of cases although the last case will ot be completely prove. Case : r > We kow from Calculus I that lim r = if r > ad so by Theorem above we also kow 7 Paul Dawkis

12 that lim r = ad so the sequece diverges if r >. Case : r = I this case we have, lim r = lim = lim = So, the sequece coverges for r = ad i this case its limit is. Case 3 : < r < We kow from Calculus I that lim r = if < r < ad so by Theorem above we also kow that lim r = ad so the sequece coverges if < r < ad i this case its limit is zero. Case 4 : r = I this case we have, lim r = lim = lim = So, the sequece coverges for r = ad i this case its limit is zero. Case 5 : < r < First let s ote that if < r < the < r < the by Case 3 above we have, lim r = lim r = Theorem above ow tells us that we must also have, lim r = ad so if < r < the sequece coverges ad has a limit of. Case 6 : r = I this case the sequece is, { r } { } {,,,,,,,, } = = = = = lim ad hopefully it is clear that does t eist. Recall that i order of this limit to eist the terms must be approachig a sigle value as icreases. I this case however the terms just alterate betwee ad - ad so the limit does ot eist. So, the sequece diverges for r =. Case 7 : r < I this case we re ot goig to go through a complete proof. Let s just see what happes if we let r = for istace. If we do that the sequece becomes, { r } { } {,, 4, 8,6, 3, } = = = = = So, if r = we get a sequece of terms whose values alterate i sig ad get larger ad larger 7 Paul Dawkis

13 ad so lim does t eist. It does ot settle dow to a sigle value as icreases or do the terms ALL approach ifiity. So, the sequece diverges for r =. We could do somethig similar for ay value of r such that r < ad so the sequece diverges for r <. Let s take a look at a couple of eamples of limits of sequeces. Eample Determie if the followig sequeces coverge or diverge. If the sequece coverges determie its limit. 3 (a) [Solutio] + 5 = e (b) [Solutio] (c) ( ) (d) = = { } = [Solutio] [Solutio] Solutio 3 (a) + 5 = I this case all we eed to do is recall the method that was developed i Calculus I to deal with the limits of ratioal fuctios. See the Limits At Ifiity, Part I sectio of my Calculus I otes for a review of this if you eed to. To do a limit i this form all we eed to do is factor from the umerator ad deomiator the largest power of, cacel ad the take the limit lim = lim = lim = So the sequece coverges ad its limit is 3 5. [Retur to Problems] 7 Paul Dawkis

14 (b) e = We will eed to be careful with this oe. We will eed to use L Hospital s Rule o this sequece. The problem is that L Hospital s Rule oly works o fuctios ad ot o sequeces. Normally this would be a problem, but we ve got Theorem from above to help us out. Let s defie f ( ) = e ad ote that, f = e Theorem says that all we eed to do is take the limit of the fuctio. e e e lim = lim = lim = So, the sequece i this part diverges (to ). More ofte tha ot we just do L Hospital s Rule o the sequece terms without first covertig to s sice the work will be idetical regardless of whether we use or. However, we really should remember that techically we ca t do the derivatives while dealig with sequece terms. [Retur to Problems] (c) ( ) = We will also eed to be careful with this sequece. We might be tempted to just say that the limit of the sequece terms is zero (ad we d be correct). However, techically we ca t take the limit of sequeces whose terms alterate i sig, because we do t kow how to do limits of fuctios that ehibit that same behavior. Also, we wat to be very careful to ot rely too much o ituitio with these problems. As we will see i the et sectio, ad i later sectios, our ituitio ca lead us astray i these problem if we are t careful. So, let s work this oe by the book. We will eed to use Theorem o this problem. To this we ll first eed to compute, ( ) lim = lim = Therefore, sice the limit of the sequece terms with absolute value bars o them goes to zero we kow by Theorem that, ( ) lim = which also meas that the sequece coverges to a value of zero. [Retur to Problems] 7 Paul Dawkis 3

15 { } = (d) ( ) For this theorem ote that all we eed to do is realize that this is the sequece i Theorem 3 above usig r =. So, by Theorem 3 this sequece diverges. [Retur to Problems] We ow eed to give a warig about misusig Theorem. Theorem oly works if the limit is zero. If the limit of the absolute value of the sequece terms is ot zero the the theorem will ot hold. The last part of the previous eample is a good eample of this (ad i fact this warig is the whole reaso that part is there). Notice that lim = lim = ad yet, lim does t eve eist let aloe equal. So, be careful usig this Theorem. You must always remember that it oly works if the limit is zero. Before movig oto the et sectio we eed to give oe more theorem that we ll eed for a proof dow the road. Theorem 4 For the sequece { } lim a = L. a if both lim a = L ad lim a + = L the { a } is coverget ad Proof of Theorem 4 Let ε >. The sice lim a Likewise, because lim a = L there is a N > such that if > N we kow that, + a L < ε = L there is a N > such that if > N we kow that, a + L < ε Now, let N = ma{ N, N + } ad let > N. The either a = ak for some k > N or a = a k + for some k > N ad so i either case we have that, a L < ε Therefore, lim a = L ad so { a } is coverget. 7 Paul Dawkis 4

16 More o Sequeces I the previous sectio we itroduced the cocept of a sequece ad talked about limits of sequeces ad the idea of covergece ad divergece for a sequece. I this sectio we wat to take a quick look at some ideas ivolvig sequeces. Let s start off with some termiology ad defiitios. Give ay sequece { a } we have the followig.. We call the sequece icreasig if a < a + for every.. We call the sequece decreasig if a > a + for every. 3. If { a } is a icreasig sequece or { } a is a decreasig sequece we call it mootoic. 4. If there eists a umber m such that m a for every we say the sequece is bouded below. The umber m is sometimes called a lower boud for the sequece. 5. If there eists a umber M such that a M for every we say the sequece is bouded above. The umber M is sometimes called a upper boud for the sequece. 6. If the sequece is both bouded below ad bouded above we call the sequece bouded. Note that i order for a sequece to be icreasig or decreasig it must be icreasig/decreasig for every. I other words, a sequece that icreases for three terms ad the decreases for the rest of the terms is NOT a decreasig sequece! Also ote that a mootoic sequece must always icrease or it must always decrease. Before movig o we should make a quick poit about the bouds for a sequece that is bouded above ad/or below. We ll make the poit about lower bouds, but we could just as easily make it about upper bouds. A sequece is bouded below if we ca fid ay umber m such that m a for every. Note however that if we fid oe umber m to use for a lower boud the ay umber smaller tha m will also be a lower boud. Also, just because we fid oe lower boud that does t mea there wo t be a better lower boud for the sequece tha the oe we foud. I other words, there are a ifiite umber of lower bouds for a sequece that is bouded below, some will be better tha others. I my class all that I m after will be a lower boud. I do t ecessarily eed the best lower boud, just a umber that will be a lower boud for the sequece. Let s take a look at a couple of eamples. 7 Paul Dawkis 5

17 Eample Determie if the followig sequeces are mootoic ad/or bouded. Solutio (a) { } [Solutio] = (a) { } + { } (b) (c) = [Solutio] = = 5 [Solutio] This sequece is a decreasig sequece (ad hece mootoic) because, > ( + ) for every. Also, sice the sequece terms will be either zero or egative this sequece is bouded above. We ca use ay positive umber or zero as the boud, M, however, it s stadard to choose the smallest possible boud if we ca ad it s a ice umber. So, we ll choose M = sice, for every This sequece is ot bouded below however sice we ca always get below ay potetial boud by takig large eough. Therefore, while the sequece is bouded above it is ot bouded. As a side ote we ca also ote that this sequece diverges (to if we wat to be specific). [Retur to Problems] + { } (b) ( ) = The sequece terms i this sequece alterate betwee ad - ad so the sequece is either a icreasig sequece or a decreasig sequece. Sice the sequece is either a icreasig or decreasig sequece it is ot a mootoic sequece. The sequece is bouded however sice it is bouded above by ad bouded below by -. Agai, we ca ote that this sequece is also diverget. [Retur to Problems] (c) = 5 This sequece is a decreasig sequece (ad hece mootoic) sice, > + The terms i this sequece are all positive ad so it is bouded below by zero. Also, sice the 7 Paul Dawkis 6

18 sequece is a decreasig sequece the first sequece term will be the largest ad so we ca see that the sequece will also be bouded above by 5. Therefore, this sequece is bouded. We ca also take a quick limit ad ote that this sequece coverges ad its limit is zero. [Retur to Problems] Now, let s work a couple more eamples that are desiged to make sure that we do t get too used to relyig o our ituitio with these problems. As we oted i the previous sectio our ituitio ca ofte lead us astray with some of the cocepts we ll be lookig at i this chapter. Eample Determie if the followig sequeces are mootoic ad/or bouded. (a) [Solutio] + Solutio (a) + (b) = 4 = 3 + = [Solutio] We ll start with the bouded part of this eample first ad the come back ad deal with the icreasig/decreasig questio sice that is where studets ofte make mistakes with this type of sequece. First, is positive ad so the sequece terms are all positive. The sequece is therefore bouded below by zero. Likewise each sequece term is the quotiet of a umber divided by a larger umber ad so is guarateed to be less tha oe. The sequece is the bouded above by oe. So, this sequece is bouded. Now let s thik about the mootoic questio. First, studets will ofte make the mistake of assumig that because the deomiator is larger the quotiet must be decreasig. This will ot always be the case ad i this case we would be wrog. This sequece is icreasig as we ll see. To determie the icreasig/decreasig ature of this sequece we will eed to resort to Calculus I techiques. First cosider the followig fuctio ad its derivative. f ( ) = f ( ) = + + We ca see that the first derivative is always positive ad so from Calculus I we kow that the fuctio must the be a icreasig fuctio. So, how does this help us? Notice that, f = = a + Therefore because < + ad f is icreasig we ca also say that, 7 Paul Dawkis 7

19 + a = = f < f + = = a a < a I other words, the sequece must be icreasig. Note that ow that we kow the sequece is a icreasig sequece we ca get a better lower boud for the sequece. Sice the sequece is icreasig the first term i the sequece must be the smallest term ad so sice we are startig at = we could also use a lower boud of for this sequece. It is importat to remember that ay umber that is always less tha or equal to all the sequece terms ca be a lower boud. Some are better tha others however. A quick limit will also tell us that this sequece coverges with a limit of. Before movig o to the et part there is a atural questio that may studets will have at this poit. Why did we use Calculus to determie the icreasig/decreasig ature of the sequece whe we could have just plugged i a couple of s ad quickly determied the same thig? The aswer to this questio is the et part of this eample! [Retur to Problems] (b) = This is a messy lookig sequece, but it eeds to be i order to make the poit of this part. First, otice that, as with the previous part, the sequece terms are all positive ad will all be less tha oe (sice the umerator is guarateed to be less tha the deomiator) ad so the sequece is bouded. Now, let s move o to the icreasig/decreasig questio. As with the last problem, may studets will look at the epoets i the umerator ad deomiator ad determie based o that that sequece terms must decrease. This however, is t a decreasig sequece. Let s take a look at the first few terms to see this. a =.9999 a = a3 =.5678 a4 = a5 =.756 a6 = a7 =.766 a8 = a9 =.44 a = = The first terms of this sequece are all icreasig ad so clearly the sequece ca t be a 7 Paul Dawkis 8

20 decreasig sequece. Recall that a sequece ca oly be decreasig if ALL the terms are decreasig. Now, we ca t make aother commo mistake ad assume that because the first few terms icrease the whole sequece must also icrease. If we did that we would also be mistake as this is also ot a icreasig sequece. This sequece is either decreasig or icreasig. The oly sure way to see this is to do the Calculus I approach to icreasig/decreasig fuctios. I this case we ll eed the followig fuctio ad its derivative f ( ) = f 4 ( ) = This fuctio will have the followig three critical poits, 4 4 =, = , = Why critical poits? Remember these are the oly places where the fuctio may chage sig! Our sequece starts at = ad so we ca igore the third oe sice it lies outside the values of that we re cosiderig. By pluggig i some test values of we ca quickly determie that the 4 derivative is positive for < < ad so the fuctio is icreasig i this rage. Likewise, we ca see that the derivative is egative for > ad so the fuctio will be decreasig i this rage. So, our sequece will be icreasig for 3 ad decreasig for 3. Therefore the fuctio is ot mootoic. Fially, ote that this sequece will also coverge ad has a limit of zero. [Retur to Problems] So, as the last eample has show we eed to be careful i makig assumptios about sequeces. Our ituitio will ofte ot be sufficiet to get the correct aswer ad we ca NEVER make assumptios about a sequece based o the value of the first few terms. As the last part has show there are sequeces which will icrease or decrease for a few terms ad the chage directio after that. Note as well that we said first few terms here, but it is completely possible for a sequece to decrease for the first, terms ad the start icreasig for the remaiig terms. I other words, there is o magical value of for which all we have to do is check up to that poit ad the we ll kow what the whole sequece will do. The oly time that we ll be able to avoid usig Calculus I techiques to determie the icreasig/decreasig ature of a sequece is i sequeces like part (c) of Eample. I this case icreasig oly chaged (i fact icreased) the deomiator ad so we were able to determie the behavior of the sequece based o that. 7 Paul Dawkis 9

21 I Eample however, icreasig icreased both the deomiator ad the umerator. I cases like this there is o way to determie which icrease will wi out ad cause the sequece terms to icrease or decrease ad so we eed to resort to Calculus I techiques to aswer the questio. We ll close out this sectio with a ice theorem that we ll use i some of the proofs later i this chapter. Theorem If { a } is bouded ad mootoic the { } a is coverget. Be careful to ot misuse this theorem. It does ot say that if a sequece is ot bouded ad/or ot mootoic that it is diverget. Eample b is a good case i poit. The sequece i that eample was ot mootoic but it does coverge. Note as well that we ca make several variats of this theorem. If { a } is bouded above ad icreasig the it coverges ad likewise if { a } is bouded below ad decreasig the it coverges. 7 Paul Dawkis

22 Series The Basics I this sectio we will itroduce the topic that we will be discussig for the rest of this chapter. That topic is ifiite series. So just what is a ifiite series? Well, let s start with a sequece { a} = (ote the = is for coveiece, it ca be aythig) ad defie the followig, s = a s = a+ a s3 = a+ a + a3 s = a + a + a + a s = a + a + a + a + + a = a 3 4 i i= s are called partial sums ad otice that they will form a sequece, { } The. Also recall = that the Σ is used to represet this summatio ad called a variety of ames. The most commo ames are : series otatio, summatio otatio, ad sigma otatio. s You should have see this otatio, at least briefly, back whe you saw the defiitio of a defiite itegral i Calculus I. If you eed a quick refresher o summatio otatio see the review of summatio otatio i my Calculus I otes. s. = Now back to series. We wat to take a look at the limit of the sequece of partial sums, { } Notatioally we ll defie, We will call i= a i where our origial sequece, { } lim s = lim a = a i i i = i = a ifiite series ad ote that the series starts at i = because that is a =, started. Had our origial sequece started at the our ifiite series would also have started at. The ifiite series will start at the same value that the sequece of terms (as opposed to the sequece of partial sums) starts. s = If the sequece of partial sums, { } ifiite series, i= series is also called diverget. a i, is coverget ad its limit is fiite the we also call the coverget ad if the sequece of partial sums is diveret the the ifiite Note that sometimes it is coveiet to write the ifiite series as, ai = a+ a + a3+ + a + i= 7 Paul Dawkis

23 We do have to be careful with this however. This implies that a ifiite series is just a ifiite sum of terms ad as we ll see i the et sectio this is ot really true. I the et sectio we re goig to be discussig i greater detail the value of a ifiite series, provided it has oe of course as well as the ideas of covergece ad divergece. This sectio is goig to be devoted mostly to otatioal issues as well as makig sure we ca do some basic maipulatios with ifiite series so we are ready for them whe we eed to be able to deal with them i later sectios. First, we should ote that i most of this chapter we will refer to ifiite series as simply series. If we ever eed to work with both ifiite ad fiite series we ll be more careful with termiology, but i most sectios we ll be dealig eclusively with ifiite series ad so we ll just call them series. Now, i i= a i the i is called the ide of summatio or just ide for short ad ote that the letter we use to represet the ide does ot matter. So for eample the followig series are all the same. The oly differece is the letter we ve used for the ide = = i + k + + i= k= = etc. It is importat to agai ote that the ide will start at whatever value the sequece of series terms starts at ad this ca literally be aythig. So far we ve used = ad = but the ide could have started aywhere. I fact, we will usually use a to represet a ifiite series i which the startig poit for the ide is ot importat. Whe we drop the iitial value of the ide we ll also drop the ifiity from the top so do t forget that it is still techically there. We will be droppig the iitial value of the ide i quite a few facts ad theorems that we ll be seeig throughout this chapter. I these facts/theorems the startig poit of the series will ot affect the result ad so to simplify the otatio ad to avoid givig the impressio that the startig poit is importat we will drop the ide from the otatio. Do ot forget however, that there is a startig poit ad that this will be a ifiite series. Note however, that if we do put a iitial value of the ide o a series i a fact/theorem it is there because it really does eed to be there. Now that some of the otatioal issues are out of the way we eed to start thikig about various ways that we ca maipulate series. We ll start this off with basic arithmetic with ifiite series as we ll eed to be able to do that o occasio. We have the followig properties. 7 Paul Dawkis

24 Properties If a ad b are both coverget series the, 6. ca, where c is ay umber, is also coverget ad = ca c a 7. a ± b = k = k is also coverget ad, a ± b = ( a ± b). = k = k = k The first property is simply tellig us that we ca always factor a multiplicative costat out of a ifiite series ad agai recall that if we do t put i a iitial value of the ide that the series ca start at ay value. Also recall that i these cases we wo t put a ifiity at the top either. The secod property says that if we add/subtract series all we really eed to do is add/subtract the series terms. Note as well that i order to add/subtract series we eed to make sure that both have the same iitial value of the ide ad the ew series will also start at this value. Before we move o to a differet topic let s discuss multiplicatio of series briefly. We ll start both series at = for a later formula ad the ote that, a b ( ab ) = = = To covice yourself that this is t true cosider the followig product of two fiite sums. ( + )( 3 5+ ) = Yeah, it was just the multiplicatio of two polyomials. Each is a fiite sum ad so it makes the poit. I doig the multiplicatio we did t just multiply the costat terms, the the terms, etc. Istead we had to distribute the through the secod polyomial, the distribute the through the secod polyomial ad fially combie like terms. Multiplyig ifiite series (eve though we said we ca t thik of a ifiite series as a ifiite sum) eeds to be doe i the same maer. With multiplicatio we re really askig us to do the followig, a b = ( a + a + a + a3 + )( b + b + b + b3 + ) = = To do this multiplicatio we would have to distribute the a through the secod term, distribute the a through, etc the combie like terms. This is pretty much impossible sice both series have a ifiite set of terms i them, however the followig formula ca be used to determie the product of two series. a b = c where c= ab i i = = = i= 7 Paul Dawkis 3

25 We also ca t say a lot about the covergece of the product. Eve if both of the origial series are coverget it is possible for the product to be diverget. The reality is that multiplicatio of series is a somewhat difficult process ad i geeral is avoided if possible. We will take a brief look at it towards the ed of the chapter whe we ve got more work uder our belt ad we ru across a situatio where it might actually be what we wat to do. Util the, do t worry about multiplyig series. The et topic that we eed to discuss i this sectio is that of ide shift. To be hoest this is ot a topic that we ll see all that ofte i this course. I fact, we ll use it oce i the et sectio ad the ot use it agai i all likelihood. Despite the fact that we wo t use it much i this course does t mea however that it is t used ofte i other classes where you might ru across series. So, we will cover it briefly here so that you ca say you ve see it. The basic idea behid ide shifts is to start a series at a differet value for whatever the reaso (ad yes, there are legitimate reasos for doig that). Cosider the followig series, = + 5 Suppose that for some reaso we wated to start this series at =, but we did t wat to chage the value of the series. This meas that we ca t just chage the = to = as this would add i two ew terms to the series ad thus chage its value. Performig a ide shift is a fairly simple process to do. We ll start by defiig a ew ide, say i, as follows, i = Now, whe =, we will get i =. Notice as well that if = the i = =, so oly the lower limit will chage here. Net, we ca solve this for to get, = i+ We ca ow completely rewrite the series i terms of the ide i istead of the ide simply by pluggig i our equatio for i terms of i. + 5 ( i + ) + 5 i+ 7 = = i+ i+ = i= i= To fiish the problem out we ll recall that the letter we used for the ide does t matter ad so we ll chage the fial i back ito a to get, = + = = To covice yourselves that these really are the same summatio let s write out the first couple of terms for each of them, 7 Paul Dawkis 4

26 = = = = So, sure eough the two series do have eactly the same terms. There is actually a easier way to do a ide shift. The method give above is the techically correct way of doig a ide shift. However, otice i the above eample we decreased the iitial value of the ide by ad all the s i the series terms icreased by as well. This will always work i this maer. If we decrease the iitial value of the ide by a set amout the all the other s i the series term will icrease by the same amout. Likewise, if we icrease the iitial value of the ide by a set amout, the all the s i the series term will decrease by the same amout. Let s do a couple of eamples usig this shorthad method for doig ide shifts. Eample Perform the followig ide shifts. (a) Write ar as a series that starts at =. = (b) Write as a series that starts at = = Solutio (a) I this case we eed to decrease the iitial value by ad so the s (okay the sigle ) i the term must icrease by as well. ( + ) ar = ar = ar = = = (b) For this problem we wat to icrease the iitial value by ad so all the s i the series term must decrease by. ( ) ( ) = = + ( ) = = 3 = 3 The fial topic i this sectio is agai a topic that we ll ot be seeig all that ofte i this class, although we will be seeig it more ofte tha the ide shifts. This fial topic is really more about alterate ways to write series whe the situatio requires it. Let s start with the followig series ad ote that the = startig poit is oly for coveiece sice we eed to start the series somewhere. = a = a + a + a + a + a Paul Dawkis 5

27 Notice that if we igore the first term the remaiig terms will also be a series that will start at = istead of = So, we ca rewrite the origial series as follows, a = a + a = = I this eample we say that we ve stripped out the first term. We could have stripped out more terms if we wated to. I the followig series we ve stripped out the first two terms ad the first four terms respectively. a = a + a + a = = 3 a = a + a + a + a + a 3 4 = = 5 Beig able to strip out terms will, o occasio, simplify our work or allow us to reuse a prior result so it s a importat idea to remember. Notice that i the secod eample above we could have also deoted the four terms that we stripped out as a fiite series as follows, 4 a = a + a + a + a + a = a + a 3 4 = = 5 = = 5 This is a coveiet otatio whe we are strippig out a large umber of terms or if we eed to strip out a udetermied umber of terms. I geeral, we ca write a series as follows, N a = a + a = = = N+ We ll leave this sectio with a importat warig about termiology. Do t get sequeces ad series cofused! A sequece is a list of umbers writte i a specific order while a ifiite series is a limit of a sequece of fiite series ad hece, if it eists will be a sigle value. So, oce agai, a sequece is a list of umbers while a series is a sigle umber, provided it makes sese to eve compute the series. Studets will ofte cofuse the two ad try to use facts pertaiig to oe o the other. However, sice they are differet beasts this just wo t work. There will be problems where we are usig both sequeces ad series so we ll always have to remember that they are differet. 7 Paul Dawkis 6

28 Series Covergece/Divergece I the previous sectio we spet some time gettig familiar with series ad we briefly defied covergece ad divergece. Before worryig about covergece ad divergece of a series we wated to make sure that we ve started to get comfortable with the otatio ivolved i series ad some of the various maipulatios of series that we will, o occasio, eed to be able to do. As oted i the previous sectio most of what we were doig there wo t be doe much i this chapter. So, it is ow time to start talkig about the covergece ad divergece of a series as this will be a topic that we ll be dealig with to oe etet or aother i almost all of the remaiig sectios of this chapter. So, let s recap just what a ifiite series is ad what it meas for a series to be coverget or diverget. We ll start with a sequece { a} = ad agai ote that we re startig the sequece at = oly for the sake of coveiece ad it ca, i fact, be aythig. Net we defie the partial sums of the series as, s = a s = a+ a s3 = a+ a + a3 s = a + a + a + a ad these form a ew sequece, { } s = a + a + a + a + + a = a 3 4 i i= s. = A ifiite series, or just series here sice almost every series that we ll be lookig at will be a ifiite series, is the the limit of the partial sums. Or, i= a = lim s i If the sequece of partial sums is a coverget sequece (i.e. its limit eists ad is fiite) the the series is also called coverget ad i this case if lim s = s the, i= a i = s. Likewise, if the sequece of partial sums is a diverget sequece (i.e. its limit does t eist or is plus or mius ifiity) the the series is also called diverget. Let s take a look at some series ad see if we ca determie if they are coverget or diverget ad see if we ca determie the value of ay coverget series we fid. Eample Determie if the followig series is coverget or diverget. If it coverges 7 Paul Dawkis 7

29 determie its value. Solutio To determie if the series is coverget we first eed to get our hads o a formula for the geeral term i the sequece of partial sums. s = = i i= This is a kow series ad its value ca be show to be, + s = i = i= Do t worry if you did t kow this formula (I d be surprised if ayoe kew it ) as you wo t be required to kow it i my course. So, to determie if the series is coverget we will first eed to see if the sequece of partial sums, ( + ) = is coverget or diverget. That s ot terribly difficult i this case. The limit of the sequece terms is, ( + ) lim = Therefore, the sequece of partial sums diverges to ad so the series also diverges. So, as we saw i this eample we had to kow a fairly obscure formula i order to determie the covergece of this series. I geeral fidig a formula for the geeral term i the sequece of partial sums is a very difficult process. I fact after the et sectio we ll ot be doig much with the partial sums of series due to the etreme difficulty faced i fidig the geeral formula. This also meas that we ll ot be doig much work with the value of series sice i order to get the value we ll also eed to kow the geeral formula for the partial sums. We will cotiue with a few more eamples however, sice this is techically how we determie covergece ad the value of a series. Also, the remaiig eamples we ll be lookig at i this sectio will lead us to a very importat fact about the covergece of series. So, let s take a look at a couple more eamples. Eample Determie if the followig series coverges or diverges. If it coverges determie its sum. 7 Paul Dawkis 8

30 = Solutio This is actually oe of the few series i which we are able to determie a formula for the geeral term i the sequece of partial fractios. However, i this sectio we are more iterested i the geeral idea of covergece ad divergece ad so we ll put off discussig the process for fidig the formula util the et sectio. The geeral formula for the partial sums is, 3 s = = i= i 4 ( + ) ad i this case we have, 3 3 lim s = lim = 4 ( ) + 4 The sequece of partial sums coverges ad so the series coverges also ad its value is, 3 = 4 = Eample 3 Determie if the followig series coverges or diverges. If it coverges determie its sum. = ( ) Solutio I this case we really do t eed a geeral formula for the partial sums to determie the covergece of this series. Let s just write dow the first few partial sums. s = s = = s = + = s3 = + = etc. So, it looks like the sequece of partial sums is, { s } = {,,,,,,,,, } = ad this sequece diverges sice lim s does t eist. Therefore, the series also diverges. Eample 4 Determie if the followig series coverges or diverges. If it coverges determie its sum. 7 Paul Dawkis 9

31 3 = Solutio Here is the geeral formula for the partial sums for this series. 3 s = = i i= 3 3 Agai, do ot worry about kowig this formula. This is ot somethig that you ll ever be asked to kow i my class. I this case the limit of the sequece of partial sums is, 3 3 lim s = lim = 3 The sequece of partial sums is coverget ad so the series will also be coverget. The value of the series is, 3 = 3 = As we already oted, do ot get ecited about determiig the geeral formula for the sequece of partial sums. There is oly goig to be oe type of series where you will eed to determie this formula ad the process i that case is t too bad. I fact, you already kow how to do most of the work i the process as you ll see i the et sectio. So, we ve determied the covergece of four series ow. Two of the series coverged ad two diverged. Let s go back ad eamie the series terms for each of these. For each of the series let s take the limit as goes to ifiity of the series terms (ot the partial sums!!). lim = ( ) this series diverged lim = this series coverged lim does't eist this series diverged lim = this series coverged 3 Notice that for the two series that coverged the series term itself was zero i the limit. This will always be true for coverget series ad leads to the followig theorem. Theorem If a coverges the lim a =. Proof First let s suppose that the series starts at =. If it does t the we ca modify thigs 7 Paul Dawkis 3

32 as appropriate below. The the partial sums are, s = a = a + a + a + a + + a s = a = a + a + a + a + + a + a i 3 4 i 3 4 i= i= Net, we ca use these two partial sums to write, a = s s Now because we kow that is also = coverget ad that lim s = sfor some fiite value s. However, sice as we also have lim s = s. a is coverget we also kow that the sequece { } s We ow have, lim a = lim s s = lim s lim s = s s = Be careful to ot misuse this theorem! This theorem gives us a requiremet for covergece but ot a guaratee of covergece. I other words, the coverse is NOT true. If lim a = the series may actually diverge! Cosider the followig two series. = = I both cases the series terms are zero i the limit as goes to ifiity, yet oly the secod series coverges. The first series diverges. It will be a couple of sectios before we ca prove this, so at this poit please believe this ad kow that you ll be able to prove the covergece of these two series i a couple of sectios. Agai, as oted above, all this theorem does is give us a requiremet for a series to coverge. I order for a series to coverge the series terms must go to zero i the limit. If the series terms do ot go to zero i the limit the there is o way the series ca coverge sice this would violate the theorem. This leads us to the first of may tests for the covergece/divergece of a series that we ll be seeig i this chapter. Divergece Test If lim a the a will diverge. Agai, do NOT misuse this test. This test oly says that a series is guarateed to diverge if the series terms do t go to zero i the limit. If the series terms do happe to go to zero the series may or may ot coverge! Agai, recall the followig two series, 7 Paul Dawkis 3

33 = = diverges coverges Oe of the more commo mistakes that studets make whe they first get ito series is to assume that if lim a = the a will coverge. There is just o way to guaratee this so be careful! Let s take a quick look at a eample of how this test ca be used. Eample 5 Determie if the followig series is coverget or diverget = + Solutio With almost every series we ll be lookig at i this chapter the first thig that we should do is take a look at the series terms ad see if they go to zero or ot. If it s clear that the terms do t go to zero use the Divergece Test ad be doe with the problem. That s what we ll do here lim = The limit of the series terms is t zero ad so by the Divergece Test the series diverges. The divergece test is the first test of may tests that we will be lookig at over the course of the et several sectios. You will eed to keep track of all these tests, the coditios uder which they ca be used ad their coclusios all i oe place so you ca quickly refer back to them as you eed to. Net we should briefly revisit arithmetic of series ad covergece/divergece. As we saw i the previous sectio if a ad b are both coverget series the so are ca ad = k ( a ± b ). Furthermore, these series will have the followig sums or values. = ( ± ) = ± ca c a a b a b = k = k = k We ll see a eample of this i the et sectio after we get a few more eamples uder our belt. At this poit just remember that a sum of coverget series is coverget ad multiplyig a coverget series by a umber will ot chage its covergece. We eed to be a little careful with these facts whe it comes to diverget series. I the first case if a is diverget the ca will also be diverget (provided c is t zero of course) sice multiplyig a series that is ifiite i value or does t have a value by a fiite value (i.e. c) wo t 7 Paul Dawkis 3

34 chage the fact that the series has a ifiite or o value. However, it is possible to have both a ad b be diverget series ad yet have ( a ± b ) be a coverget series. = k Now, sice the mai topic of this sectio is the covergece of a series we should metio a stroger type of covergece. A series a is said to coverge absolutely if a also coverges. Absolute covergece is stroger tha covergece i the sese that a series that is absolutely coverget will also be coverget, but a series that is coverget may or may ot be absolutely coverget. I fact if a coverges ad a diverges the series a is called coditioally coverget. At this poit we do t really have the tools at had to properly ivestigate this topic i detail or do we have the tools i had to determie if a series is absolutely coverget or ot. So we ll ot say aythig more about this subject for a while. Whe we fially have the tools i had to discuss this topic i more detail we will revisit it. Util the do t worry about it. The idea is metioed here oly because we were already discussig covergece i this sectio ad it ties ito the last topic that we wat to discuss i this sectio. I the previous sectio after we d itroduced the idea of a ifiite series we commeted o the fact that we should t thik of a ifiite series as a ifiite sum despite the fact that the otatio we use for ifiite series seems to imply that it is a ifiite sum. It s ow time to briefly discuss this. First, we eed to itroduce the idea of a rearragemet. A rearragemet of a series is eactly what it might soud like, it is the same series with the terms rearraged ito a differet order. For eample, cosider the followig the ifiite series. = A rearragemet of this series is, = a = a + a + a + a + a + a + a a = a + a + a + a + a + a + a The issue we eed to discuss here is that for some series each of these arragemets of terms ca have differet values despite the fact that they are usig eactly the same terms. Here is a eample of this. It ca be show that, + ( ) = = l () = Sice this series coverges we kow that if we multiply it by a costat c its value will also be multiplied by c. So, let s multiply this by to get, 7 Paul Dawkis 33

35 = l () Now, let s add i a zero betwee each term as follows = l (3) Note that this wo t chage the value of the series because the partial sums for this series will be the partial sums for the () ecept that each term will be repeated. Repeatig terms i a series will ot affect its limit however ad so both () ad (3) will be the same. We kow that if two series coverge we ca add them by addig term by term ad so add () ad (3) to get, = l (4) Now, otice that the terms of (4) are simply the terms of () rearraged so that each egative term comes after two positive terms. The values however are defiitely differet despite the fact that the terms are the same. Note as well that this is ot oe of those tricks that you see occasioally where you get a cotradictory result because of a hard to spot math/logic error. This is a very real result ad we ve ot made ay logic mistakes/errors. Here is a ice set of facts that gover this idea of whe a rearragemet will lead to a differet value of a series. Facts Give the series a,. If a is absolutely coverget ad its value is s the ay rearragemet of a will also have a value of s.. If a is coditioally coverget ad r is ay real umber the there is a rearragemet of a whose value will be r. Agai, we do ot have the tools i had yet to determie if a series is absolutely coverget ad so do t worry about this at this poit. This is here just to make sure that you uderstad that we have to be very careful i thikig of a ifiite series as a ifiite sum. There are times whe we ca (i.e. the series is absolutely coverget) ad there are times whe we ca t (i.e. the series is coditioally coverget). As a fial ote, the fact above tells us that the series, = + 7 Paul Dawkis 34

36 must be coditioally coverget sice two rearragemets gave two separate values of this series. Evetually it will be very simple to show that this series coditioally coverget. 7 Paul Dawkis 35

37 Series Special Series I this sectio we are goig to take a brief look at three special series. Actually, special may ot be the correct term. All three have bee amed which makes them special i some way, however the mai reaso that we re goig to look at two of them i this sectio is that they are the oly types of series that we ll be lookig at for which we will be able to get actual values for the series. The third type is diverget ad so wo t have a value to worry about. I geeral, determiig the value of a series is very difficult ad outside of these two kids of series that we ll look at i this sectio we will ot be determiig the value of series i this chapter. So, let s get started. Geometric Series A geometric series is ay series that ca be writte i the form, = ar or, with a ide shift the geometric series will ofte be writte as, = These are idetical series ad will have idetical values, provided they coverge of course. If we start with the first form it ca be show that the partial sums are, a( r ) a ar s = = r r r The series will coverge provided the partial sums form a coverget sequece, so let s take the limit of the partial sums. a ar lim s = lim r r a ar = lim lim r r a a = lim r r r Now, from Theorem 3 from the Sequeces sectio we kow that the limit above will eist ad be fiite provided < r. However, ote that we ca t let r = sice this will give divisio by zero. Therefore, this will eist ad be fiite provided < r < ad i this case the limit is zero ad so we get, a lim s = r ar 7 Paul Dawkis 36

38 Therefore, a geometric series will coverge if < r <, which is usually writte r <, its value is, a = = r ar ar = = Note that i usig this formula we ll eed to make sure that we are i the correct form. I other words, if the series starts at = the the epoet o the r must be. Likewise if the series starts at = the the epoet o the r must be. Eample Determie if the followig series coverge or diverge. If they coverge give the value of the series. + + (a) 9 4 [Solutio] Solutio (a) = (b) 9 4 = 3 4 [Solutio] = This series does t really look like a geometric series. However, otice that both parts of the series term are umbers raised to a power. This meas that it ca be put ito the form of a geometric series. We will just eed to decide which form is the correct form. Sice the series starts at = we will wat the epoets o the umbers to be. It will be fairly easy to get this ito the correct form. Let s first rewrite thigs slightly. Oe of the s i the epoet has a egative i frot of it ad that ca t be there i the geometric form. So, let s first get rid of that ( ) = 9 4 = 9 = = = Now let s get the correct epoet o each of the umbers. This ca be doe usig simple epoet properties = = = = = Now, rewrite the term a little = 6( 9) = 44 = = 9 = 9 4 So, this is a geometric series with a = 44 ad r = 9 <. Therefore, sice r < we kow the series will coverge ad its value will be, 7 Paul Dawkis 37

39 = = = = [Retur to Problems] (b) = ( 4) 3 5 Agai, this does t look like a geometric series, but it ca be put ito the correct form. I this case the series starts at = so we ll eed the epoets to be o the terms. Note that this meas we re goig to eed to rewrite the epoet o the umerator a little 3 ( 4) = = 5 = = = = = 64 So, we ve got it ito the correct form ad we ca see that a = 5 ad r = 5. Also ote that r ad so this series diverges. [Retur to Problems] Back i the Series Basics sectio we talked about strippig out terms from a series, but did t really provide ay eamples of how this idea could be used i practice. We ca ow do some eamples. Eample Use the results from the previous eample to determie the value of the followig series. + + (a) 9 4 [Solutio] Solutio (a) = (b) 9 4 = [Solutio] = I this case we could just ackowledge that this is a geometric series that starts at = ad so we could put it ito the correct form ad be doe with it. However, this does provide us with a ice eample of how to use the idea of strippig out terms to our advatage. Let s otice that if we strip out the first term from this series we arrive at, = = = = = From the previous eample we kow the value of the ew series that arises here ad so the value of the series i this eample is, 7 Paul Dawkis 38

40 = = + = 5 5 [Retur to Problems] (b) = I this case we ca t strip out terms from the give series to arrive at the series used i the previous eample. However, we ca start with the series used i the previous eample ad strip terms out of it to get the series i this eample. So, let s do that. We will strip out the first two terms from the series we looked at i the previous eample = = = = 3 = 3 We ca ow use the value of the series from the previous eample to get the value of this series = = 8 = = 3 = 5 5 [Retur to Problems] Notice that we did t discuss the covergece of either of the series i the above eample. Here s why. Cosider the followig series writte i two separate ways (i.e. we stripped out a couple of terms from it). Let s suppose that we kow = 3 a a = a + a + a + a = = 3 is a coverget series. This meas that it s got a fiite value ad addig three fiite terms oto this will ot chage that fact. So the value of fiite ad so is coverget. Likewise, suppose that = a this value we will remai fiite ad arrive at the value of so this series will also be coverget. = a is also is coverget. I this case if we subtract three fiite values from = 3 a. This is ow a fiite value ad I other words, if we have two series ad they differ oly by the presece, or absece, of a fiite umber of fiite terms they will either both be coverget or they will both be diverget. The differece of a few terms oe way or the other will ot chage the covergece of a series. This is a importat idea ad we will use it several times i the followig sectios to simplify some of the tests that we ll be lookig at. 7 Paul Dawkis 39

41 Telescopig Series It s ow time to look at the secod of the three series i this sectio. I this portio we are goig to look at a series that is called a telescopig series. The ame i this case comes from what happes with the partial sums ad is best show i a eample. Eample 3 Determie if the followig series coverges or diverges. If it coverges fid its value. = + 3+ Solutio We first eed the partial sums for this series. s = i + 3i+ Now, let s otice that we ca use partial fractios o the series term to get, = = i + 3i+ i+ i+ i+ i+ i= I ll leave the details of the partial fractios to you. By ow you should be fairly adept at this sice we spet a fair amout of time doig partial fractios back i the Itegratio Techiques chapter. If you eed a refresher you should go back ad review that sectio. So, what does this do for us? Well, let s start writig out the terms of the geeral partial sum for this series usig the partial fractio form. s = i= i+ i+ = = + Notice that every term ecept the first ad last term caceled out. This is the origi of the ame telescopig series. This also meas that we ca determie the covergece of this series by takig the limit of the partial sums. lim s = lim = + The sequece of partial sums is coverget ad so the series is coverget ad has a value of = = + 3+ I telescopig series be careful to ot assume that successive terms will be the oes that cacel. Cosider the followig eample. 7 Paul Dawkis 4

42 Eample 4 Determie if the followig series coverges or diverges. If it coverges fid its value. = Solutio As with the last eample we ll leave the partial fractios details to you to verify. The partial sums are, s = i i i 3 = = + + i= i+ i+ 3 = = I this case istead of successive terms cacelig a term will cacel with a term that is farther dow the list. The ed result this time is two iitial ad two fial terms are left. Notice as well that i order to help with the work a little we factored the out of the series. The limit of the partial sums is, 5 5 lim s = lim = So, this series is coverget (because the partial sums form a coverget sequece) ad its value is, 5 = = Note that it s ot always obvious if a series is telescopig or ot util you try to get the partial sums ad the see if they are i fact telescopig. There is o test that will tell us that we ve got a telescopig series right off the bat. Also ote that just because you ca do partial fractios o a series term does ot mea that the series will be a telescopig series. The followig series, for eample, is ot a telescopig series despite the fact that we ca partial fractio the series terms. 3+ = = = I order for a series to be a telescopig series we must get terms to cacel ad all of these terms are positive ad so oe will cacel. Net, we eed to go back ad address a issue that was first raised i the previous sectio. I that sectio we stated that the sum or differece of coverget series was also coverget ad that the presece of a multiplicative costat would ot affect the covergece of a series. Now that we have a few more series i had let s work a quick eample showig that. 7 Paul Dawkis 4

43 Eample 5 Determie the value of the followig series = Solutio To get the value of this series all we eed to do is rewrite it ad the use the previous results = = = = + + = = = 5 96 = = 5 We did t discuss the covergece of this series because it was the sum of two coverget series ad that guarateed that the origial series would also be coverget. Harmoic Series This is the third ad fial series that we re goig to look at i this sectio. Here is the harmoic series. = The harmoic series is diverget ad we ll eed to wait util the et sectio to show that. This series is here because it s got a ame ad so I wated to put it here with the other two amed series that we looked at i this sectio. We re also goig to use the harmoic series to illustrate a couple of ideas about diverget series that we ve already discussed for coverget series. We ll do that with the followig eample. Eample 6 Show that each of the followig series are diverget. 5 (a) = (b) = 4 Solutio 5 (a) = To see that this series is diverget all we eed to do is use the fact that we ca factor a costat out of a series as follows, 5 = 5 = = 7 Paul Dawkis 4

44 Now, is diverget ad so five times this will still ot be a fiite umber ad so the series = has to be diverget. I other words, if we multiply a diverget series by a costat it will still be diverget. (b) = 4 I this case we ll start with the harmoic series ad strip out the first three terms. = = = = 4 = 4 = 6 I this case we are subtractig a fiite umber from a diverget series. This subtractio will ot chage the divergece of the series. We will either have ifiity mius a fiite umber, which is still ifiity, or a series with o value mius a fiite umber, which will still have o value. Therefore, this series is diverget. Just like with coverget series, addig/subtractig a fiite umber from a diverget series is ot goig to chage the divergece of the series. So, some geeral rules about the covergece/divergece of a series are ow i order. Multiplyig a series by a costat will ot chage the covergece/divergece of the series ad addig or subtractig a costat from a series will ot chage the covergece/divergece of the series. These are ice ideas to keep i mid. 7 Paul Dawkis 43

45 Itegral Test The last topic that we discussed i the previous sectio was the harmoic series. I that discussio we stated that the harmoic series was a diverget series. It is ow time to prove that statemet. This proof will also get us started o the way to our et test for covergece that we ll be lookig at. So, we will be tryig to prove that the harmoic series, = diverges. We ll start this off by lookig at a apparetly urelated problem. Let s start off by askig what the area uder f ( ) = o the iterval [, ). From the sectio o Improper Itegrals we kow that this is, d = ad so we called this itegral diverget (yes, that s the same term we re usig here with series.). So, just how does that help us to prove that the harmoic series diverges? Well, recall that we ca always estimate the area by breakig up the iterval ito segmets ad the sketchig i rectagles ad usig the sum of the area all of the rectagles as a estimate of the actual area. Let s do that for this problem as well ad see what we get. We will break up the iterval ito subitervals of width ad we ll take the fuctio value at the left edpoit as the height of the rectagle. The image below shows the first few rectagles for this area. So, the area uder the curve is approimately, 7 Paul Dawkis 44

46 A = = = Now ote a couple of thigs about this approimatio. First, each of the rectagles overestimates the actual area ad secodly the formula for the area is eactly the harmoic series! Puttig these two facts together gives the followig, A > d = Notice that this tells us that we must have, > = = = = Sice we ca t really be larger tha ifiity the harmoic series must also be ifiite i value. I other words, the harmoic series is i fact diverget. So, we ve maaged to relate a series to a improper itegral that we could compute ad it turs out that the improper itegral ad the series have eactly the same covergece. Let s see if this will also be true for a series that coverges. Whe discussig the Divergece Test we made the claim that = coverges. Let s see if we ca do somethig similar to the above process to prove this. We will try to relate this to the area uder f ( ) = is o the iterval [ ) Improper Itegral sectio we kow that, ad so this itegral coverges. d =,. Agai, from the We will oce agai try to estimate the area uder this curve. We will do this i a almost idetical maer as the previous part with the eceptio that istead of usig the left ed poits for the height of our rectagles we will use the right ed poits. Here is a sketch of this case, 7 Paul Dawkis 45

47 I this case the area estimatio is, A = This time, ulike the first case, the area will be a uderestimatio of the actual area ad the estimatio is ot quite the series that we are workig with. Notice however that the oly differece is that we re missig the first term. This meas we ca do the followig, = < + d = + = = Area Estimatio Or, puttig all this together we see that, = < With the harmoic series this was all that we eeded to say that the series was diverget. With this series however, this is t quite eough. For istace < ad if the series did have a value of the it would be diverget (whe we wat coverget). So, let s do a little more work. First, let s otice that all the series terms are positive (that s importat) ad that the partial sums are, s = i Because the terms are all positive we kow that the partial sums must be a icreasig sequece. I other words, + s = < s = + i i i= i= i= 7 Paul Dawkis 46

48 I s + we are addig a sigle positive term oto s ad so must get larger. Therefore, the partial sums form a icreasig (ad hece mootoic) sequece. Also ote that, sice the terms are all positive, we ca say, s = < < s < i= i = ad so the sequece of partial sums is a bouded sequece. I the secod sectio o Sequeces we gave a theorem that stated that a bouded ad mootoic sequece was guarateed to be coverget. This meas that the sequece of partial sums is a coverget sequece. So, who cares right? Well recall that this meas that the series must the also be coverget! So, oce agai we were able to relate a series to a improper itegral (that we could compute) ad the series ad the itegral had the same covergece. We wet through a fair amout of work i both of these eamples to determie the covergece of the two series. Luckily for us we do t eed to do all this work every time. The ideas i these two eamples ca be summarized i the followig test. Itegral Test Suppose that f ( ) is a cotiuous, positive ad decreasig fuctio o the iterval [, ) f that = a the, d is coverget so is a.. If f. If f k k = k d is diverget so is a. = k A formal proof of this test ca be foud at the ed of this sectio. k ad There are a couple of thigs to ote about the itegral test. First, the lower limit o the improper itegral must be the same value that starts the series. Secod, the fuctio does ot actually eed to be decreasig ad positive everywhere i the iterval. All that s really required is that evetually the fuctio is decreasig ad positive. I other words, it is okay if the fuctio (ad hece series terms) icreases or is egative for a while, but evetually the fuctio (series terms) must decrease ad be positive for all terms. To see why this is true let s suppose that the series terms icrease ad or are egative i the rage k N ad the decrease ad are positive for N +. I this case the series ca be writte as, N a = a + a = k = k = N+ Now, the first series is othig more tha a fiite sum (o matter how large N is) of fiite terms ad so will be fiite. So the origial series will be coverget/diverget oly if the secod ifiite series o the right is coverget/diverget ad the test ca be doe o the secod series as it satisfies the coditios of the test. 7 Paul Dawkis 47

49 A similar argumet ca be made usig the improper itegral as well. The requiremet i the test that the fuctio/series be decreasig ad positive everywhere i the rage is required for the proof. I practice however, we oly eed to make sure that the fuctio/series is evetually a decreasig ad positive fuctio/series. Also ote that whe computig the itegral i the test we do t actually eed to strip out the icreasig/egative portio sice the presece of a small rage o which the fuctio is icreasig/egative will ot chage the itegral from coverget to diverget or from diverget to coverget. There is oe more very importat poit that must be made about this test. This test does NOT give the value of a series. It will oly give the covergece/divergece of the series. That s it. No value. We ca use the above series as a perfect eample of this. All that the test gave us was that, < = So, we got a upper boud o the value of the series, but ot a actual value for the series. I fact, from this poit o we will ot be askig for the value of a series we will oly be askig whether a series coverges or diverges. I a later sectio we look at estimatig values of series, but eve i that sectio still wo t actually be gettig values of series. Just for the sake of completeess the value of this series is kow. π = = < 6 Let s work a couple of eamples. = Eample Determie if the followig series is coverget or diverget. = l Solutio I this case the fuctio we ll use is, f ( ) = l This fuctio is clearly positive ad if we make larger the deomiator will get larger ad so the fuctio is also decreasig. Therefore, all we eed to do is determie the covergece of the followig itegral. t d = lim d u = l l t l t ( ( ) ) ( ( t) ) = lim l l t = lim l l l l t = The itegral is diverget ad so the series is also diverget by the Itegral Test. 7 Paul Dawkis 48

50 Eample Determie if the followig series is coverget or diverget. e = Solutio The fuctio that we ll use i this eample is, f ( ) = e This fuctio is always positive o the iterval that we re lookig at. Now we eed to check that the fuctio is decreasig. It is ot clear that this fuctio will always be decreasig o the iterval give. We ca use our Calculus I kowledge to help us however. The derivative of this fuctio is, f = e This fuctio has two critical poits (which will tell us where the derivative chages sig) at = ±. Sice we are startig at = we ca igore the egative critical poit. Pickig a couple of test poits we ca see that the fuctio is icreasig o the iterval, ad it is decreasig o, ). Therefore, evetually the fuctio will be decreasig ad that s all that s required for us to use the Itegral Test. t e lim t e d = d u = = lim t e t = lim e = t The itegral is coverget ad so the series must also be coverget by the Itegral Test. We ca use the Itegral Test to get the followig fact/test for some series. Fact ( The p series Test) If k > the coverges if p > ad diverges if p = k p. Sometimes the series i this fact are called p-series ad so this fact is sometimes called the p- series test. This fact follows directly from the Itegral Test ad a similar fact we saw i the Improper Itegral sectio. This fact says that the itegral, d p k coverges if p > ad diverges if p. t 7 Paul Dawkis 49

51 Usig the p-series test makes it very easy to determie the covergece of some series. Eample 3 Determie if the followig series are coverget or diverget. (a) 7 = 4 (b) = Solutio (a) I this case p = 7> ad so by this fact the series is coverget. (b) For this series p = ad so the series is diverget by the fact. The last thig that we ll do i this sectio is give a quick proof of the Itegral Test. We ve essetially doe the proof already at the begiig of the sectio whe we were itroducig the Itegral Test, but let s go through it formally for a geeral fuctio. Proof of Itegral Test First, for the sake of the proof we ll be workig with the series = a. The origial test statemet was for a series that started at a geeral = k ad while the proof ca be doe for that it will be easier if we assume that the series starts at =. Aother way of dealig with the = k is we could do a ide shift ad start the series at = ad the do the Itegral Test. Either way provig the test for = will be sufficiet. Let s start off ad estimate the area uder the curve o the iterval [, ] ad we ll uderestimate the area by takig rectagles of width oe ad whose height is the right edpoit. This gives the followig figure. Now, ote that, f ( ) = a f ( 3) = a3 f = a 7 Paul Dawkis 5