Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it s just a facy way of sayig the domai cosists of a set of cosecutive itegers startig with some iteger but ever edig. I most cases, the domai will be either the set of positive itegers or the set of o-egative itegers. Try to recogize that the etire defiitio is just a facy, but precise, way of sayig a sequece is a list of umbers. We ca have sequeces of objects other tha real umbers, but i this course we will restrict ourselves to sequeces of real umbers ad will from ow o just refer to sequeces. Notatio We geerally use the otatio {a } to deote a sequece, just as we ofte use the otatio f(x) to deote a fuctio. is the idepedet variable, but whe studyig sequeces we refer to it as the idex. We ll ofte defie a sequece by givig a formula for a, just as we ofte defie a ordiary fuctio f(x) by givig a formula. Example: a =. This sequece ca also be described by, 2, 3, 4,.... Example: b = 2. This sequece ca also be described by, 4, 9, 6, 25,.... Covergece of a Sequece We ofte wat to kow whether the terms of a sequece {a } approach some limit as. This is aalogous to a ordiary limit at ifiity, so we defie a limit of a sequece by appropriately modifyig the defiitio of a ordiary limit at ifiity. Recall: Defiitio 2 (Limit at Ifiity). lim x f(x) = L if for every ɛ > 0 there is some real umber N such that f(x) L < ɛ wheever x > N. We get a defiitio of a limit of a sequece by replacig f(x) by a ad replacig x by, obtaiig: Defiitio 3 (Limit of a Sequece). lim a = L if for every ɛ > 0 there is some real umber N such that a L < ɛ wheever > N. Covergece of a Sequece
Note: We may write lim a, but it is acceptible to simply write lim a sice there is o reasoable iterpretatio other tha for. If a sequece has a limit, we say it coverges; otherwise, we say it diverges. Properties of Limits of Sequeces Limits of sequeces share may properties with ordiary limits. Each of the followig properties may be prove essetially the same way the aalogous properties are prove for ordiary limits. (Each of these properties depeds o the limit o the right side existig.) lim(a ± b ) = lim a ± lim b lim ka = k lim a lim k = k lim a b = lim a lim b If lim b 0, lim a b = lim a lim b Sequeces Through Ordiary Fuctios The similarity of the defiitios of limits of sequeces ad limits at ifiity yield the followig corollary: Theorem. Cosider a sequece {a } ad a ordiary fuctio f. If a = f() ad lim x f(x) = L, the lim a = L. Proof. Suppose the hypotheses are satisfied ad let ɛ > 0. Sice lim x f(x) = L, if follows there must be some N R such that f(x) L < ɛ wheever x > N. Sice a = f(), it follows that a L < ɛ wheever > N. Applyig the Aalogy This theorem implies each of the followig limits, which ca also be prove idepedetly. lim = 0 lim = lim( + /) = e lim l = 0 lim 2 = 0 A similarly flavored limit which eeds to be prove separately is lim 2! = 0. Usig L Hôpital s Rule 2
L Hôpital s Rule caot be used directly to fid limits of sequeces, but it ca be used idirectly. We ca ofte fid lim a by fidig a fuctio f(x) such that a = f() ad the usig L Hôpital s Rule to fid lim x f(x). Example We wat to fid lim l 2 + L Hôpital s Rule to fid lim f(x) = lim x x so lim l 2 + = 0. x l x. We let f(x) =. We ca the use x 2 + x l x x + (l x) x 2 + = lim x x 2x = lim x + l x 2x /x = lim x 2 = 0, Mootoic Sequeces Sometimes it is possible ad eve ecessary to determie whether a sequece coverges without havig to fid what it coverges to. This is ofte the case with mootoic sequeces. Defiitio 4 (Icreasig). A sequece {a } is icreasig if a k a k+ for all k i its domai. Defiitio 5 (Strictly Icreasig). A sequece {a } is strictly icreasig if a k < a k+ for all k i its domai. Defiitio 6 (Decreasig). A sequece {a } is decreasig if a k a k+ for all k i its domai. Defiitio 7 (Strictly Decreasig). A sequece {a } is strictly decreasig if a k > a k+ for all k i its domai. Mootoic Sequeces Defiitio 8 (Mootoicity). If a sequece is either icreasig or decreasig, it is said to be mootoic. Defiitio 9 (Boudedess). A sequece {a } is said to be bouded if there is a umber B R such that a B for all i the domai of the sequece. B is referred to as a boud. Theorem 2 (Mootoe Covergece Theorem). A mootoic sequece coverges if ad oly if it is bouded. 3
The Mootoe Covergece Theorem becomes very importat i determiig the covergece of ifiite series. The Completeess Axiom The proof of the Mootoe Covergece Theorem depeds o: The Completeess Axiom: If a oempty set has a lower boud, it has a greatest lower boud; if a oempty set has a upper boud, it has a least upper boud. The terms lower boud, greatest lower boud, upper boud ad least upper boud mea precisely what they soud like. Exercise: Write dow precise defiitios. We will give a proof of the Mootoe Covergece Theorem for a icreasig sequece. A similar proof ca be created for a decreasig sequece. Proof of the Mootoe Covergece Theorem Proof. If a sequece is icreasig ad has a limit, it is clearly bouded below by its first term ad bouded above by its limit ad thus must be bouded, so we ll just show that a sequece which is icreasig ad bouded must have a limit. So suppose {a } is icreasig ad bouded. It must have a upper boud ad thus, by the Completeess Axiom, must have a least upper boud B. Let ɛ > 0. There must be some elemet a N of the sequece such that B ɛ < a N B. a N B sice B is a upper boud. If B ɛ < a N was false for all elemets of the sequece, B ɛ would be a upper boud smaller tha B, so B would ot be the least upper boud. It follows that if > N, B ɛ < a N < a B, so a B < ɛ ad it follows from the defiitio of a limit that lim a = B. Ifiite Series Defiitio 0 (Ifiite Series). A expressio a + a 2 + a 3 + = a k is called a ifiite series. The terms of a series form a sequece, but i a series we attempt to add them together rather tha simply list them. We do t actually have to start with k = ; we could start with ay iteger value although we will almost always start with either k = or k = 0. Covergece of Ifiite Series We wat to assig some meaig to a sum for a ifiite series. It s aturally to add the terms oe-by-oe, effectively gettig a sum for part of the series. This is called a partial sum. 4
Defiitio (Partial Sum). S = a k is called the th partial sum of the series a k. If the sequece {S } of a series coverges to some umber S, we say the series coverges to S ad write a k = S. We call S the sum of the series. If the series does t coverge, we say it diverges. The Series 0.33333... With the defiitio of a series, we are able to give a meaig to a o-termiatig decimal such as 0.33333... by viewig it as 0.3 + 0.03 + 0.003 + 0.0003 + = 3 0 + 3 0 + 3 2 0 + = 3 3 0. k Usig the defiitio of covergece ad a little algebra, we ca show this series coverges to 3 as follows. The th partial sum S = The Series 0.33333... 3 0 = 3 k 0 + 3 0 + 3 2 0 + + 3 3 Multiplyig both sides by 0, we get 0S = 3 0 + + 3 2 0. 2 5 0. 3 0 = 3+ 3 k 0 + Subtractig, we get 0S S = 3 3 0, so 9S = 3 3 0 S = 3 /3 0. Clearly, lim S = 3, so the series 3 0 k coverges to 3. Geometric Series A similar aalysis may be applied to ay geometric series. ad Defiitio 2 (Geometric Series). A geometric series is a series which may be writte i the form ark = a + ar + ar 2 + ar 3 +.... I other words, a k = ar k. The first term is geerally referred to as a ad r is called the commo ratio. We ca obtai a compact formula for the partial sums as follows: Geometric Series Lettig S = a+ar+ar 2 +ar 3 +... ar, we ca multiply both sides by the commo ratio r to get rs = ar + ar 2 + ar 3 +... ar + ar. Subtractig, we get S rs = a ar ( r)s = a( r ) S = a( r ) r if r.
Geometric Series S = a( r ) if r. r If r <, it is clear that r 0 as, so S a r. If r >, the r as, so {S } clearly diverges. If r =, the S oscillates back ad forth betwee 0 ad 2a, so {S } clearly diverges. If r =, the S = a + a + a + + a = a, so {S } clearly diverges. We may summarize this iformatio by otig the geometric series a ark coverges to if r < but diverges if r. r Note o a Alterate Derivatio We could have foud S differetly by otig the factorizatio r = ( r)( + r + r 2 +... r ), which is a special case of the geeral factorizatio formula a b = (a b)(a +a 2 b+a 3 b 2 +... ab 2 + b ). It immediately follows that + r + r 2 +... r = r r. Positive Term Series Defiitio 3 (Positive Term Series). A series a k is called a Positive Term Series is a k 0 for all k. 6 Theorem 3. A positive term series coverges if ad oly if its sequece of partial sums is bouded. + Proof. Lookig at the sequece of partial sums, S + = a k = a k + a + = S + a + S, sice a + 0. Thus {S } is mootoic ad, by the Mootoe Covergece Theorem, coverges is ad oly if it s bouded. Note ad Notatio This ca be used to show a series coverges but its more importat purpose is to eable us to prove the Compariso Test for Series. Notatio: Whe dealig with positive term series, we may write a k < whe the series coverges ad a k = whe the series diverges. This is aalogous to the otatio used for covergece of improper itegrals with positive itegrads.
Proof. Let S = Example: k Coverges 2 k. Sice, for k 2, 2 k if k x k, it 2 x2 k follows that k k = 2 k k dx 2 k x dx. 2 Thus 0 S = k = + k 2 k + 2 k=2 k=2 k x dx = + 2 x dx = + ] = + ] ( ) = 2 / 2. 2 x Sice the sequece of partial sums is bouded, the series coverges. Estimatig the Error Estimatig k by leaves a error 2 k2 k. 2 k=+ Usig the same type of reasoig used to show the series coverges shows this sum is o greater tha dx, which ca be evaluated x2 as follows: dx = lim x2 t t dx = lim x2 t x = lim ] t t ] t ] = lim t ] = t We thus see estimatig the series by the th partial sum leaves a error o larger tha, which ca be made as small as desired by makig large eough. The Compariso Test Recall: Theorem 4 (Compariso Test for Improper Itegrals). Let 0 f(x) g(x) for x a. () If g(x) dx <, the f(x) dx <. a a (2) If f(x) dx =, the g(x) dx =. a a The Compariso Test 7
The Compariso Test for Improper Itegrals has a atural aalogue for Positive Term Series: Theorem 5 (Compariso Test for Positive Term Series). Let 0 a b for sufficietly large. () If b <, the a <. (2) If = a =, the = = b =. = The Compariso Test for Positive Term Series is used aalogously to the way the Compariso Test for Improper Itegrals is used. Proof of the Compariso Test We prove a special case of the first case as follows: Proof. Suppose 0 a k b k for k ad show the sequece of partial sums S = 8 b k <. It suffices to a k is bouded. Sice is a positive term series, its sequece of partial sums has a boud B. Clearly, S = a k b k B. Notes About the Proof () The proof assumed 0 a k b k for k. Sice a chage i ay fiite umber of terms does t affect covergece, the coclusio must hold as log as 0 a k b k for k sufficetly large. (2) The secod case is the cotrapositive of the first, so it does ot have to be prove separately. Usig the Compariso Test I order to use the Compariso Test, oe eeds a kowledge of stadard series with whose covergece oe is familiar. There are provided by Geometric ad P-Series. Geometric series have already bee aalyzed. b k P-Series are aalogous to the improper itegrals used with the P-Test for Covergece of Improper Itegrals. The P-Test for Series ca be prove usig the Itegral Test. The Itegral Test
Theorem 6 (Itegral Test). Let f(x) 0, itegrable for x large eough, mootoic ad lim x f(x) = 0 ad suppose a k = f(k). It follows that a k < if ad oly if f(x) dx <. Proof of the Itegral Test Proof. Suppose the itegral coverges. For coveiece, we will assume α =. The proof ca easily be modified if the itegral is defied for some other α, but the argumet is made most clearly without that complicatio. Sice f(x) is clearly decreasig, a k f(x) for k x k, so a k k f(x) dx ad S k = a k a + f(x) dx. Sice the improper itegral coverges, the itegral o the right is bouded. Thus the sequece of partial sums is bouded ad the series must coverge. If the itegral diverges, we may use the observatio S + f(x) dx to show the sequece of partial sums is ot bouded ad the series must diverge. Error Estimatio The proof of the Itegral Test provides a clue about the error ivolved if oe uses a partial sum to estimate the sum of a ifiite series. If oe estimates the sum of a series by its th partial sum s = α a k, the error will equal the sum a k k=+ a k 9 of the terms ot icluded i the partial sum. If the series is a positive term series ad a k = f(k) for a decreasig fuctio f(x), the aalysis used i provig the Itegral Test leads to the coclusio that this error is bouded by f(x) dx. Suppose we estimate the sum the error. Example k 2 The error will be bouded by lim t ] t = lim t ] x t 00 00 by s 00 ad wat a boud o x dx = lim 2 t ] = 00. 00 t Determiig a Number of Terms to Use 00 x dx = 2
We ca also figure out how may terms are eeded to estimate a sum to withi a predetermied tolerace ɛ. Usig the same otatio as before, this ca be guarateed if () f(x) dx ɛ. We ca look at () as a iequality i ad solve for. This may be easier said tha doe. Example Suppose we wat to estimate k to withi 3 0 8. We eed to fid such that x dx 3 0 8. t Itegratig: dx = lim x3 t x dx = lim 3 t ] t 2x 2 = lim t ] ] = 2t 2 2 2 2. 2 So we eed 2 2 0 8, which may be solved as follows: 0 8 2 2 5 0 7 2 5 07 Sice 5 0 7 707.07, we eed to add 7072 terms to estimate the sum to withi 0 8. Stadard Series P-Test for Series { k p < if p > = if p. Geometric Series { ar k coverges if r < diverges if r. k=0 Absolute Covergece Defiitio 4 (Absolute Covergece). a k is said to be absolutely coverget if a k is coverget. Theorem 7. A series which is absolutely coverget is coverget. 0
Proof of Absolute Covergece Theorem Proof. Suppose { a k is absolutely coverget. Let a + k = a k if a k 0 0 if a k < 0. { Let a k = a k if a k 0 0 if a k > 0. The terms of the positive term series a k+ ad a k are both bouded by the terms of the coverget series a k. It follows immediately that a k = (a+ k a k ) = a+ k a k also coverges. Defiitio 5 (Coditioal Covergece). A coverget series which is ot absolutely coverget is said to be coditioally coverget. Testig for Absolute Covergece All the tests devised for positive term series automatically double as tests for absolute covergece. We will study oe more test for covergece, the Ratio Test. Ratio Test The Ratio Test is useful for series which behave almost like geometric series but for which it ca be difficult to use the Compariso Test. It is ot very useful for series that ordiarily would be compared to P-series. The ratio test is usually stated as a test for absolute covergece, but ca also be thought of as a test for covergece of positive term series. We state both versios below ad use whichever versio is more coveiet. Ratio Tests Theorem 8 (Ratio Test for Positive Term Series). Cosider a positive term series a a k+ k ad let r = lim k. a k If r <, the the series coverges. If r >, the the series diverges. If r = or the limit does t exist, the ratio test is icoclusive. Theorem 9 (Ratio Test for Absolute Covergece). Cosider a series a a k+ k ad let r = lim k a k. If r <, the the series coverges absolutely. If r >, the the series diverges. If r = or the limit does t exist, the ratio test is icoclusive.
Proof of the Ratio Test We prove the ratio test for positive term series. Proof. If r >, the terms of the series evetually keep gettig larger, so the series clearly must diverge. Thus, we eed oly cosider the case r <. Choose some R R such that r < R <. There must be some N Z such that a k+ < R wheever k N. We thus have a k a N+ < a N R, a N+2 < a N+ R < a N R 2, a N+3 < a N+2 R < a N R 3, ad so o. Sice a N + a N R + a N R 2 + a N R 3 +... is a geometric series which commo ratio 0 < R <, it must coverge. By the Compariso Test, the origial series must coverge as well. Strategy For Testig Covergece It s importat to have a strategy to determie whether a series a k coverges. The followig is oe reasoable strategy. Begi by makig sure the idividual terms coverge to 0, sice if the terms do t approach 0 the we kow the series must diverge ad there s o reaso to check further. Next, check whether the series is oe of the stadard series, such as a P-Series or a Geometric Series ar k. If so, we k p ca immediately determie whether it coverges. Otherwise, we cotiue. We start by testig for absolute covergece. Strategy For Testig Covergece Fid a reasoable series to compare it to. Oe way is to look at the differet terms ad factors i the umerator ad deomiator, pickig out the largest (usig the geeral criteria powers of logs << powers << expoetials << factorials), ad replacig aythig smaller tha the largest type by, as appropriate, 0 (for terms) or (for factors). If we are lucky, we ca use the resultig series alog with the compariso test to determie whether our origial series is absolutely coverget. If we re ot lucky, we have to try somethig else. If the series seems to almost be geometric, the Ratio Test is likely to work. As a last resort, we ca try the Itegral Test. If the series is ot absolutely coverget, we may be able to show it coverges coditioally either by direct examiatio or by usig the Alteratig Series Test. Strategy for Aalyzig Improper Itegrals 2
Essetially the same strategy may be used to aalyze covergece of improper itegrals. 3