Math 201 (Introduction to Analysis) Fall

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1 Mth (Itrodutio to Alysis) Fll 4-5 Istrutor: Dr. Ki Y. Li Offie: Room 47 Offie Phoe: e-mil ddress: Offie Hours: Tu. :pm - 4:pm (or y ppoitmets) Prerequisite: A-level Mth or Oe Vrile Clulus Wesite for Leture Notes d Prtie Eerises: myli/ug.html Grde Sheme: Homewors (4%), Tutoril Presettios (6%), Midterm (4%), Fil Emitio (5%) All reords of grdes will e put o the site s soo s they re ville. This ourse is essetilly grded y urve with oe eeptio, mely studets who hieve 4% or less of the overll grde will fil the ourse. Studets who te the sme midterm will e grouped uder the sme urve. Studets should me opies of homewors efore sumittig the origils. I se homewors re ot reeived, studets will e required to resumit opies withi short period of time (my e less th dy). For tutoril presettios, studets will form groups (of to studets) d preset solutios to ssiged prolems i the tutoril sessios. All memers of group must tted the sessios to ssist i srig possile questios from presettios. Mrs will e deduted for filure to preset solutios or for see i supportig his/her group. Course Desriptio: This is the first of two required ourses o lysis for Mth mjors. It is to e follod y Mth (Rel Alysis). This ourse will fous o the proofs of si theorems of lysis, s ppered i oe vrile lulus. Alog the wy to estlish the proofs, my ew oepts will e itrodued. These ilude outility, sequees d series of umers, of futios, supremum/ifimum, Cuhy oditio, Riem itegrls d improper itegrls. Uderstdig them d their properties re importt for the developmet of the preset d further ourses. Tetoos: Keeth A. Ross, Elemetry Alysis : The Theory of Clulus, Spriger-Verlg, 98. Roert Brtle d Dold Sherert, Itrodutio to Rel Alysis, rd ed., Wiley,. Ross is for studets who re tig this ourse for the first time. Studets who re repetig my use either Ross or Brtle d Sherert. Referees:. J. A. Fridy, Itrodutory Alysis d ed., Ademi Press,.. Mfred Stoll, Itrodutio to Rel Alysis, Addiso Wesley, Wlter Rudi, Priiples of Mthemtil Alysis, rd ed., MGrw-Hill, Tom Apostol, Mthemtil Alysis, d ed., Addiso-Wesley, 974. *5. Chiese Solutio Mul to Tom Apostol s Mthemtil Alysis, d ed. Remiders: Studets re highly eourged to ome to offie hours for osulttio. This is diffiult ourse for my, ut ot ll studets. Although re leture otes, studets should tted ll letures d tutorils s letures otes re oly rief reords of mterils overed i lss, whih my oti typogrphil errors. Of ourse, questios from studets d srs from istrutors or other digressios will ot e reorded. Studets re dvised to te your ow otes. All mterils preseted i letures d tutorils s ll s proper lss odut re the studets resposiility. The istrutor reserves the right to me y hges to the ourse throughout the semester. The oly wy to sueed i this ourse is to do the wor.

2 f f f f f ] while f f f f ] Ojetives of the Course The ojetives of the ourse re to ler lysis d ler proofs. Questios: Wht is lysis? How is it differet from other rhes of mthemtis? Alysis is the rh of mthemtis tht studies it d oepts derived from it, suh s otiuity, differetitio d itegrtio, while geometry dels with figures, lger dels with equtios d iequlities ivolvig dditio, sutrtio, multiplitio, divisio d umer theory studies itegers. Whe try to solve prolems ivolvig rel or omple umers, suh s fidig roots of polyomils or solvig differetil equtios, my ot get the right srs the first time. Hover, get pproimtios d the its of these pproimtios, hope, will give us the right srs. At lest, ow solutios eist eve though my ot e le to write them epliitly. Prolems ivolvig itegers e muh hrder sie itegers re disrete, i.e. re miimum diste seprtig distit itegers so tht oe ot fid itegers ritrrily lose to other iteger. Try to see if is rel solutio to the equtio The try to see if is iteger solutio. Wht re the differees i the wy you solve these two prolems? Questio: Why should ler proofs? A sttemet is true ot euse your teher tells you it is true. A teher me mistes! There re fmous mthemtiis who mde ojetures tht re disovered to e wrog yers lter. How e erti the fts lered re true? How judge whe more th oe proposed solutios re give, whih is orret? si Suppose wt to fid Sie the umertor is ete d the si umertor hs it. si The deomitor lso hs it. So, y l Hopitl s rule, si os Hover, the ew si os umertor does ot hve it euse os hs o it s the ew deomitor hs it. So the it of the origil prolem does ot eist. Is this resoig orret? No. Where is the miste? Sometimes epli fts y emples or pitures. For iste, the sttemet tht every odd degree polyomil with rel oeffiiets must hve t lest oe root is ofte eplied y some emples or some pitures. I our lifetime, oly do fiitely my emples d drw fiitely my pitures. Should elieve somethig is true y seeig few pitures or emples? Drw the grphs of few otiuous futios o [ Do you thi every otiuous futio o [ ] is differetile i t lest oe poit o? Or do you thi eists otiuous futio o [ ot differetile t y poit of? Cosider the futio f 4 Note f re prime umers. Should you elieve tht f is prime umer for every positive iteger? Wht is the first tht f is ot prime? I order to hve ofidet, you hve to e le to judge the fts you lered re solutely orret. Almost orret is ot good eough i mthemtis.

3 y Chpter. Logi To reso orretly, hve to follow some rules. These rules of resoig re wht lled logi. We will oly eed few of these rules, mily to del with tig opposite of sttemets d to hdle oditiol sttemets. We will use the symol (or ) to deote the word ot. Also, will use the symol to deote for ll, for y, for every. Similrly, the symol will deote is (t lest oe), eists, re (some) d usully follod y suh tht. The symols d re lled qutifiers. Negtio. Below will loo t rules of egtio (i.e. tig opposite). They re eeded whe do idiret proofs (or proofs y otrditio). For y epressio p hve p p Emples. () epressio : p d q opposite epressio : or rule : p d q () epressio : or opposite epressio : d rule : p or q p or q p d q () sttemet : For every qutified sttemet : hs squre root. (True) hs squre root opposite sttemet : There eists suh tht does ot hve squre root. (Flse) qutified opposite sttemet : hs squre root (4) sttemet : For every qutified sttemet : is y suh tht y (True) y y opposite sttemet : There eists suh tht for every y qutified opposite sttemet : y y (Flse) From emples () d (4), see tht the rule for egtig sttemets with qutifiers is first swith every to d every to, egte the remiig prt of the sttemet. If- Sttemets. If- sttemets our frequetly i mthemtis. We will eed to ow some equivlet wys of epressig if- sttemet to do proofs. The sttemet if p q my lso e stted s p implies q, p oly if q, p is suffiiet for q, q is eessry for p d is ommoly deoted y p q. For emple, the sttemet if d y 4 y 5 my lso e stted s d y 4 re suffiiet for y 5 or y 5 is eessry for d y 4. Emple. (5) sttemet : If (True) opposite sttemet : d (Flse) rule : p q p d q Remr. Note p q p q p d q p or q p or q

4 ut For the sttemet if p q p q re two relted sttemets: the overse of the sttemet is if q, p (q p) d the otrpositive of the sttemet is if q p ( q p). Emples. (6) sttemet : If 9 (True) overse : If 9 (Flse, s my e.) otrpositive : If 9 (True) (7) sttemet : 6 (True) overse : 6 (True) otrpositive : 6 (True) (8) sttemet : If (Flse, s my e.) overse : If (True) otrpositive : If (Flse, s my e.) Remrs. Emples (6) d (7) shod tht the overse of if- sttemet is ot the sme s the sttemet or the opposite of the sttemet i geerl. Emples (6), (7) d (8) shod tht if- sttemet d its otrpositive re either oth true or oth flse. I ft, this is lwys the se euse y the remr o the lst pge, q p q or p q or p p or q p q So if- sttemet d its otrpositive sttemet re equivlet. Filly, itrodue the termiology p if d oly if q to me if p q d if q p. The sttemet p if d oly if q is the sme s p is eessry d suffiiet for q. We revite p if d oly if q y p q. So p q mes p q d q p. The phrse if d oly if is ofte revited s iff. Cutio! Note d For emple, every studet is ssiged umer is the sme s studet, umer suh tht the studet is ssiged the umer. This sttemet implies differet studets my e ssiged possily differet umers. Hover, if swith the order of the qutifiers, the sttemet eomes umer suh tht studet, the studet is ssiged the umer. This sttemet implies is umer d every studet is ssiged tht sme umer! 4

5 7] 4] B B P y 4 4 C A [ y z ] y 5] ] A the A m Chpter. Sets To red d write mthemtil epressios urtely d oisely, will itrodue the lguge of sets. A set is olletio of ojets (usully umers, ordered pirs, futios, et.) If ojet is i set S, sy is elemet (or memer) of S d write S If is ot elemet of S, write S A set hvig fiitely my elemets is lled fiite set, otherwise it is lled ifiite set. The empty set is the set hvig o ojets d is deoted y A set my e show y listigits elemets elosed i res (eg. is set otiigthe ojets the positive iteger, the iteger, the empty set ) or y desriptio elosed i res (eg. the rtiol umers m : m rel umers : is rel umer d the omple umers iy : ) I desriig sets, the usul ovetio is to put the form of the ojets o the left side of the olo d to stte the oditios o the ojets o the right side of the olo. It is lso ommo to use vertil r i ple of olo i set desriptios. Emples. (i) The losed itervl with edpoits (ii) The set of squre umers is is [ : : d (iii) The set of ll positive rel umers is : d (If wt to emphsize this is suset of my stress is rel i the form of the ojets d write : If umers re lwys te to me rel umers, my write simply : ) (iv) The set of poits (or ordered pirs) o the lie m with equtio y m is : For sets A sy A is suset of B (or B otis A) iff every elemet of A is lso elemet of B I tht se, write A B (For the se of the empty set, hve S for every set S ) Two sets A d B re equl if d oly if they hve the sme elemets (i.e. A B mes A B d B A ) So A B if d oly if ( A B). If A B d A B sy A is proper suset of B d write A B (For emple, if A A B is true, ut B C is flse. I ft, B C Repeted elemets re outed oly oe time so tht C hs elemets, ot 4 elemets.) For set S ollet ll its susets. This is lled the por set of S d is deoted y P S or S For emples, P d P For set with elemets, its por set will hve elemet. This is the reso for the ltertive ottio S for the por set of S Por set is oe opertio of set. There re few other ommo opertios of sets. Defiitios. For sets A A A (i) their uio is A A A : A or A or or A (ii) their itersetio is A A A : A d A d d A (iii) their Crtesi produt is A A A : A d A d d A (iv) the omplemet of A i A is A A : A d A Emples. (i) (ii) [ (iii) [ : (iv) z : [ [4 Remrs. (i) For the se of the empty set, hve A A A A [ 6] : is rtiol d is irrtiol A A d A 5

6 ] let let whih 4 ] 4] defie 5] let ut 5 whih y (ii) The otios of uio, itersetio d Crtesi produt my e eteded to ifiitely my sets similrly. The uio is the set of ojets i t lest oe of the sets. The itersetio is the set of ojets i every oe of the sets. The Crtesi produt is the set of ordered tuples suh tht the i-th oordite must elog to the i-th set. (iii) The set A A A my e writte s ottio A A A my e revited s the uio of ll the sets A s for ll Crtesi produt. A If for every positive iteger S is deoted y A or S Emples. (i) [ [ [ [4 [ (ii) [ 4 [ ] (iii) For every (iv) For eh m d m m A A A A A If for every S is set A the is set A A Similr revitios eist for itersetio d : eh is or for m e the lie with equtio y m o the ple, m m y : y We shll sy tht sets re disjoit iff their itersetio is the empty set. Also, sy they re mutully disjoit iff the itersetio of every pir of them is the empty set. A reltio o set E is y suset of E E The followig is importt oept tht is eeded i lmost ll rhes of mthemtis. It is tool to divide (or prtitio) the set of ojets lie to study ito mutully disjoit susets. Defiitio. A equivlee reltio R o set E is suset R of E () (refleive property) for every E R () (symmetri property) if y R y R () (trsitive property) if y y z R z R E suh tht We write y if y R For eh E [] y : y This is lled the equivlee lss otiig Note tht every [] y () so tht [] E If y [] [y] euse y () d (), E z [] z z y z [y] If y [] [y] euse ssumig z [] [y] will led to z d z y imply y otrditio. So every pir of equivlee lsses re either the sme or disjoit. Therefore, R prtitios the set E ito mutully disjoit equivlee lsses. Emples. () (Geometry) For trigles T d T defie T T if d oly if T is similr to T This is equivlee reltio o the set of ll trigles s the three properties ove re stisfied. For trigle T [T ] is the set of ll trigles similr to T () (Arithmeti) For itegers m d m if d oly if m is eve. Agi, properties (), (), () esily e verified. So this is lso equivlee reltio o There re etly two equivlee lsses, mely [] 4 (eve itegers) d [] 5 (odd itegers). Two itegers i the sme equivlee lss is sid to e of the sme prity. () Some people thi tht properties () d () imply property () y usig (), lettig z i () to olude R This is flse s show y the outeremple tht E d R stisfies properties () d (), ut ot property (). R fils property () euse E R s is ot i y ordered pir i R 6

7 4 9 B 6 g the whih f the odom rd A futio (or mp or mppig) f from set A to set B (deoted y f : A B) is method of ssigig to every A etly oe B This is deoted y f d is lled the vlue of f t Thus, futio must e ll-defied i the sese tht if f f The set A is lled the domi of f (deoted y dom f ) d the set B is lled the odomi of f (deoted y odom f ). We sy f is B-vlued futio (eg. if B sy f is rel-vlued futio.) Whe the odomi B is ot emphsized, my simply sy f is futio o A The imge or rge of f (deoted y f A or im f or r f ) is the set f : A (To emphsize this is suset of B lso write it s f B : A ) The set G : A is lled the grph of f Two futios re equl if d oly if they hve the sme grphs. I prtiulr, the domis of equl futios re the sme set. Emples. The futio f : give y f hs dom f f Also, r f This is differet from the futio g : give y g euse dom g dom f Also, futio my hve more th oe prts i its defiitio, eg. the solute vlue futio h : defied y h if if Be reful i defiig futios. The followig is d: let d i The rule is ot ll-defied euse ut i i Defiitios. (i) The idetity futio o set S is I S : S S give y I S for ll S (ii) Let f : A B : B C e futios d f A B The ompositio of g y f is the futio g f : A C defied y g f g f for ll A (iii) Let f : A B e futio d C A The futio f C : C B defied y f C is lled the restritio of f to C (iv) A futio f : A B is surjetive (or oto) iff f A (v) A futio f : A B is ijetive (or oe-to-oe) iff f (vi) A futio f : A (vii) For ijetive futio f : A B f y f y B f y implies y f for every C B is ijetio (or oe-to-oe orrespodee) iff it is ijetive d surjetive. iverse futio of f is the futio f : f A A defied y Remrs. A futio f : A B is surjetive mes f A B is the sme s syig every B is f for t lest oe A I this sese, the vlues of f do ot omit ythig i B We will loosely sy f does ot omit y elemet of B for oveiee. Hover, my possily e more th oe A tht re ssiged the sme B Hee, the rge of f my repet some elemets of B If A d B re fiite sets, f surjetive implies the umer of elemets i A is greter th or equl to the umer of elemets i B Net, futio f : A B is ijetive mes, i the otrpositive sese, tht y implies f f y whih my loosely sy f does ot repet y elemet of B Hover, f my omit elemets of B s my possily e elemets i B tht re ot i the rge of f So if A d B re fiite sets, f ijetive implies the umer of elemets i A is less th or equl to the umer of elemets i B Therefore, ijetio from A to B is futio whose vlues do ot omit or repet y elemet of B If A d B re fiite sets, f ijetive implies the umer of elemets i A d B re the sme. Remrs (Eerises). () Let f : A B e futio. We hve f is ijetio if d oly if is futio g : B A suh tht g f I A d f g I B (I ft, for f ijetive, hve g f is ijetive.) () If f : A B d h : B C re ijetios, h f : A C is ijetio. () Let A e susets of d f : A B e futio. If for every B horizotl lie y itersets the grph of f etly oe, f is ijetio. To del with the umer of elemets i set, itrodue the followig oept. For sets S d S will defie S S d sy they hve the sme rdility (or the sme rdil umer) if d oly if eists ijetio from S to S This is esily heed to e equivlee reltio o the olletio of ll sets. For set S the equivlee lss [S] is ofte lled the rdil umer of S d is deoted y rd S or S This is wy to ssig symol for the umer of elemets i set. It is ommo to deote, for positive iteger rd (red leph-ught) d rd (ofte lled the rdility of the otiuum). 7

8 whih Chpter. Coutility Ofte ompre two sets to see if they re differet. I se oth re ifiite sets, the oept of outle sets my help to distiguish these ifiite sets. Defiitios. A set S is outly ifiite iff eists ijetio f : S (i.e. d S hve the sme rdil umer ) A set is outle iff it is fiite or outly ifiite set. A set is uoutle iff it is ot outle. Remrs. Suppose f : S is ijetio. The f is ijetive mes f f f re ll distit d f is surjetive mes f f f S So f S is oe-to-oe orrespodee ete d S Therefore, the elemets of S e listed i orderly wy (s f f f ) without repetitio or omissio. Coversely, if the elemets of S e listed s s s without repetitio or omissio, f : S defied y f s will e ijetio s o repetitio implies ijetivity d o omissio implies surjetivity. Bijetio Theorem. Let g: S T e ijetio. S is outle if d oly if T is outle. (Resos. The theorem is true euse S outle implies is ijetive futio f : S implies h g f : T is ijetive, i.e. T is outle. For the overse, h is ijetive implies f g h is ijetive.) Remrs. Similrly, tig otrpositive, S is uoutle if d oly if T is uoutle. Bsi Emples. () is outly ifiite (euse the idetity futio I is ijetio). () is outly ifiite euse the followig futio is ijetio (oe-to-oe orrespodee): f The futio f : is give y f y g m () m if is eve m if m m if m Just he g f I d f g I : m is outly ifiite. (Digol Coutig Sheme) Usig the digrm o the right, defie f : y f, f, f, f 4, f 5, f 6,, f is ijetive euse o ordered pir is repeted. Also, f is surjetive euse m f m f m m if is odd d its iverse futio g : is give 4 4 (4) The ope itervl : d is uoutle. Also, is uoutle. f f f f Suppose is outly ifiite d f : is ijetio s show o the left. Cosider the umer whose deiml represettio is 4, where if if. The d f for ll euse. So f ot e surjetive, otrditio. Net is uoutle euse t provides ijetio from oto To determie the outility of more omplited sets, will eed the theorems elow. 8

9 y r where the is 5 B r Coutle Suset Theorem. Let A uoutle, B is uoutle.) Coutle Uio Theorem. If A is outle for every B If B is outle, A is outle. (Tig otrpositive, if A is (sy f : S is ijetio) d A s is outle for every s S outle uio of outle sets is outle.) A is outle. I geerl, if S is outle s S A s A f is outle. (Briefly, Produt Theorem. If A, B re outle, A B : A B is outle. I ft, if A A re outle, A A A is outle (y mthemtil idutio). A (Seth of Resos. For the outle suset theorem, if B is outle, list the elemets of B d to out the elemets of A, sip over those elemets of B tht re ot i A For the outle uio theorem, if list the elemets of A i the first row, the elemets of A i the seod row, out ll the elemets y usig the digol outig sheme. As for the produt theorem, imitte the emple of d lso use the digol outig sheme.) Emples. (5) S where S m : m For every futio f : S give y f m m m is ijetio (with f m), so S is outle y the ijetio theorem. Therefore, is outle y the outle uio theorem. (The susets of lie re lso outle.) (6) is uoutle. (I ft, if A is uoutle d B is outle, A B is uoutle s A B outle implies A B A B A outle y the outle uio theorem, whih is otrditio). (7) iy : otis d is uoutle, so y the outle suset theorem, is uoutle. (8) Show tht the set A r m : m is uoutle, ut the set B r m : m is outle. Solutio. Tig m see tht A Sie is uoutle, A is uoutle. Net will oserve tht B B m where B m r m : r r m for eh m Sie m is outle d r m hs elemet for every r theorem. Filly, sie is outle d B m is outle for every m uio theorem. (9) Show tht the set L of ll lies with equtio y m m r B m is outle y the outle uio is outle y the outle outle. Solutio. Note tht for eh pir m of rtiol umers, is uique lie y m i the set L So the futio f : L defied y lettig f m e the lie y m (with f sedig the lie to m ) is ijetio. Sie is outle y the produt theorem, so the set L is outle y the ijetio theorem. () Show tht if A for every A A A is uoutle. (I prtiulr, this shows tht the produt theorem is ot true for ifitely my outle sets.) Solutio. Assume A A A : eh i or is outle d f : A A A is ijetio. Followig emple (4), hge the -th oordite of f (from to or from to ) to produe elemet of A A A ot equl to y f whih is otrditio. So it must e uoutle. () Show tht the por set P of ll susets of is uoutle. Solutio. As i emple (), let A for every Defie g : P A A A y g S where if m S m if m (For emple, g ) Note S g hs the iverse futio g m : m Hee g is ijetio. Sie A A A is uoutle, so P is uoutle y the ijetio theorem. 9

10 it whih the every () Show tht the set S of ll polyomils with iteger oeffiiets is outle. Solutio. Let S For set of S of ll polyomils of degree with iteger oeffiiets is outle euse the futio f : S defied y f is ijetio d is outle y the produt theorem. So, S S S is outle y the outle uio theorem. () Show tht eists rel umer, whih is ot root of y oostt polyomil with iteger oeffiiets. Solutio. For eh polyomil f with iteger oeffiiets, let R f deotes the set of roots of f The R f hs t most deg f elemets, hee R f is outle. Let S e the set of ll oostt polyomils with iteger oeffiiets, whih is the suset S of S i the lst emple. The f S R f is the set of ll roots of oostt polyomils with iteger oeffiiets. It is outle y the outle uio theorem. Sie is uoutle, R f is uoutle y the ft i emple (6). So eist uoutly my rel umers, whih re f S ot roots of y oostt polyomil with iteger oeffiiets. Remrs. Ay umer whih is root of oostt polyomil with iteger oeffiiets is lled lgeri umer. A umer whih is ot root of y oostt polyomil with iteger oeffiiets is lled trsedetl umer. Are y lgeri umers? Are y trsedetl umers? If so, re fiitely my or outly my suh umers? Sie every rtiol umer is the root of the polyomil rtiol umer is lgeri. There re irrtiol umers lie re lgeri euse they re the roots of Usig the idetity os 4 os os the irrtiol umer os is esily see to e lgeri s it is root of 8 6 Emple () d the ft tht every oostt polyomil hs fiitely my roots shod re oly outly my lgeri umers. Emple () shod tht re uoutly my trsedetl rel umers. It is quite diffiult to prove prtiulr umer is trsedetl. I umer theory ourse, it will e show tht d e re trsedetl. The followig re dditiol useful fts oerig outility. Theorem. () (Ijetio Theorem) Let f : A B e ijetive. If B is outle, A is outle. (Tig otrpositive, if A is uoutle, B is uoutle.) () (Surjetio Theorem) Let g : A B e surjetive. If A is outle, B is outle. (Tig otrpositive, if B is uoutle, A is uoutle.) (Resos. For the first sttemet, oserve tht the futio h : A f A defied y h f is ijetive (euse f is ijetive) d surjetive (euse h A f A ). So h is ijetio. If B is outle, f A is outle y the outle suset theorem, whih implies A is outle y the ijetio theorem. For the seod sttemet, oserve tht B g A g If A is outle, it is outle uio of A outle sets. By the outle uio theorem, B is outle.) A Fmous Ope Prolem i Mthemtis For two sets A d B is ommo to defie rd A rd B if d oly if eists ijetive futio f : A B This is wy to idite B hs t lest s my elemets s A Cotiuum Hypothesis. If S is uoutle, rd rd S (This mes every uoutle set hs t lest s my elemets s the rel umers.) I 94, Kurt Gödel shod tht the opposite sttemet would ot led to y otrditio. I 966, Pul Cohe wo the Fields Medl for showig the sttemet lso would ot led to y otrditio. So proof y otrditio my ot e pplied to every sttemet.

11 Chpter 4. Series Defiitios. A series is the summtio of outle set of umers i speifi order. If re fiitely my umers, the series is fiite series, otherwise it is ifiite series. The umers re lled terms. The sum of the first terms is lled the -th prtil sum of the series. A ifiite series is of the form st term d term rd term The first prtil sum is S. The seod prtil sum is S or my write it s. The th prtil sum is S Series re used frequetly i siee d egieerig to solve prolems or pproimte solutios. (E.g. trigoometri or logrithm tles re omputed usig series i the old dys.) 4 8 6? (S 4 We sy the series overges to, whih is lled the sum of the series. Emples.() ) () (S, S.) We sy the series diverges (to ). (). (S if is odd if is eve, S does t eist.) We sy the series diverges. Defiitios. A series I tht se, my write overges to umer S iff S S S d sy S is the sum of the series. A series diverges to iff the prtil sum S teds to ifiity s teds to ifiity. A series diverges iff it does ot overge to y umer. Remrs. () For every series, is sequee (of prtil sums) S. Coversely, if the prtil sum sequee S is give, fid the terms s follows: S, S S,, S S for. The S S S S S S. So S is the prtil sum sequee of. Coeptully, series d sequees re equivlet. So to study series, use fts out sequees. () Let N e positive iteger. overges to A if d oly if overges to B A euse B N So to see if series overges, my igore fiitely my terms. Theorem. If for y ostt overges to A d A B overges to B N A B N For simple series suh s geometri or telesopig series, fid their sums. A N N A

12 Theorem (Geometri Series Test). We hve r r r r r r r does t eist if r otherwise Emple. 999 So hs two deiml represettios! Theorem (Telesopig Series Test). We hve Emples. () () overges if d oly if is umer If series is ot geometri or telesopig, oly determie if it overges or diverges. This e doe most of the time y pplyig some stdrd tests. If the series overges, it my e etremely diffiult to fid the sum! Theorem (Term Test). If, the series (Reso. Suppose overges, my or my ot overge. overges to S The S. (If S d Term test is oly good for series tht re suspeted to e diverget! Emples. (). Here for ll, so () os () os os os diverges euse os os os os diverges euse si os otrditio.) 4 os d (4). Here 8 y the geometri series test. (5) times 4 times for ll, so 8 times os. We hve, the series S S. Series diverges. os. os (Otherwise, diverges to euse S S S d S hs it os os S S.) si si diverges.) If os The si (Term test does t pply!) Series overges (Term test does t pply.) Series

13 the so f For oegtive series umer or equl to. So either or (i.e. for every ), hve S S S d overges to umer or.) For oegtive series, hve the followig tests. Theorem (Itegrl Test). Let f : [ f d. (Note i geerl, (Reso. This follows from f s show i the figures elow.) f derese to s f f f d ) S must eist s diverges to. (I short, either. The f d f f f S overges if d oly if f f () f () f () f () Emples. () Cosider the overgee or divergee of As overges. so Now () Cosider the overgee or divergee of As l d l diverges. Net l d l Theorem (p-test). For rel umer p p (Reso. For p to s. Sie d p p p Remrs. For eve positive iteger p l d d rt l their reiprols derese to Now l l So p p p 4 p d l l overges. terms re t lest, so the series diverges y term test. For p d p p p p if p d p if p, the itegrl test gives the olusio.) 4 So l l So overges if d oly if p. l the vlue of p ws omputed y Euler i 76. He got p dereses if p d B!

14 m where B d B B m for The vlues of 5 re uow. Oly i the m 98 s, R. Apery ws le to show ws irrtiol. Theorem (Compriso Test). Give u for every. If diverges, diverges. (Reso. u.) u. If Theorem (Limit Compriso Test). Give u, u d u overge) or (oth diverge to ). If u diverges diverges. (Seth of Reso. For lrge, u L. For L is umer, other lso overges. If oe diverges (to ), so does the other. For L evetully. So the lst two sttemets follow from the ompriso test.) Emples. Cosider the overgee or divergee of the followig series: () os () () (4) si 5. Solutios. () Sie os d overges, u is umer. If u overges. If u u, for every If is positive umer L either (oth u u u overges overges. If Lu L u. If oe series overges, the overges y p-test, u evetully. For L os overges. u () Sie for d diverges y the geometri series test, diverges. () Whe is lrge, 5. We ompute 5 Sie 5 overges y p-test, overges y the it ompriso test. 5 (4) Whe is lrge, is lose to, so si is lose to si euse (i.e. si s ). si We ompute si. Sie diverges y p-test, si diverges y the it ompriso test. For series with lterte positive d egtive terms, hve the followig test. Theorem (Altertig Series Test). If dereses to s (i.e. d 4 5 overges. ), 4

15 so (Reso. Sie, hve S S 4 S 6 S 5 S S Sie S S, the distes ete the prtil sums derese to d so S must eist.) Emples. Both l d e os overge y the ltertig series test euse s l d e l d e d os For series with ritrry positive or egtive term, hve the followig tests. Theorem (Asolute Covergee Test). If (Reso. From overges. The Defiitio. We sy overges, ut get overges solutely iff diverges. overges, Sie overges. overges.) overges. We sy Emples. Determie if the followig series overge solutely or oditiolly os () () os. os Solutios. (). Sie overges y p-test, it follows tht os ompriso test. So overges solutely y the solute overgee test. os () euse os d. l Hover, overges, so y the ompriso test, overges oditiollyiff os dereses to s os. So y the ltertig series test, overges. Therefore os overges oditiolly. Theorem (Rtio Test). If for every d eists, overges solutely my overge e g diverges 5 or diverge e g overges y the diverges.

16 so i.e. r, so r d (Seth of reso. Let r, for lrge,,,, r r r whih overges if r y the geometri series test d r diverges if r y the term test.) Theorem (Root Test). If (Seth of reso. Let r eists, overges solutely my overge e g diverges or diverge e g, for lrge, r So r, r.) Emples. Cosider the overgee or divergee of the followig series:! () (). Solutios. () Sie overges. Altertively, sie test, () Sie overges.!! e y the rtio test, y the rtio test,! overges., y the root Remrs. You my hve oserved tht i emple (), the it you got for pplyig the root test ws the sme s the it you got for pplyig the rtio test. This ws ot idet! Theorem. If for ll d r pplied to more series th the rtio test.) Emples. () Let So, r (This implies tht the root test e () Let! s ove. So! whe is lrge,! e e whih is simple versio of wht is lled Stirlig s formul. It is useful i estimtig! whe is lrge. For emple, sie log get! 6 whih hs out 57 digits. e e Theorem (Summtio y Prts). Let S j j S S 6 j d e

17 (Reso. Note S d S S for. So, S si Emple. Show tht overges. S S S S S S S Let si d Usig the idetity si m si os m os m S si si si os os si hve This implies S Now S si si solute overgee test, for every Applyig summtio y prts d otig tht si si si S S S si overges. S S y the telesopig series test. get So y the Comple Series Comple umers S S with S u i re sid to hve it d solutely overget d oditiol overget re the sme. The remrs d the si properties followig the defiitios of overget d diverget series re lso true for omple series. S S u i iff u u A omple series is series where the terms re omple umers. The defiitios of overget, The geometri series test, telesopig series test, term test, solute overgee test, rtio test d root test re lso true for omple series. For z iy hve z overges to z iy if d oly if overges to d Emples. () Note () If z z y overges to y So omple series e redued to rel series for study if eessry. z z i (otherwise d z i overges y p-test implies z z implies is otrditio). So z i diverges y term test. overges solutely. Hover, if z diverges y the rtio test. 7

18 let Defiitio. We sy Isertig Pretheses d Rerrgemets of Series. is otied from p : suh tht p (Note is the sum of p p terms.) Groupig Theorem. Let will overge to s Net, if (Reso. Let s d t e otied from y isertig pretheses iff is stritly iresig futio p is ouded d p p y isertig pretheses. If For the first prt, overges to s t p overges to s p p will overge to s s For the seod prt, let p p e ouded y M For positive iteger j p i j p i For r M p i r if p i r j r j if p i r j The s j t i j j i M j s j Emples. () Sie overges to, so y the theorem, j defie s ) () overges to, ut diverges y term test. So is importt. Also, overges to. Hover, the series without pretheses diverges (s S d S ) eve though the terms hve it. So ouded is importt. () Sie j j overges (y the it ompri- j j so test with Defiitio. j j ), so y the theorem, is rerrgemet of Emple. Give l j overges to the sme sum. iff is ijetio : suh tht 6. Oserve tht 4 4 (whih overges oditiolly). Cosider the rerrgemet 5 5 Riem s Rerrgemet Theorem. Let d is rerrgemet of suh tht l l l overge oditiolly. For y or,

19 the (Seth of reso. Let p if if d q if if The p q d Now oth p q must diverge to (If oth overges, their sum If oe overges d the other diverges to p lso.) Let u oegtive terms of e sequees of rel umers hvig its d u the order they our. Sie i the order they our d Q Q P Q differ from p p q will e fiite, otrditio. q will diverges to otrditio Now let P P e the e the solute vlue of the egtive terms i q oly y zero terms, they lso diverges to Let m e the smllest itegers suh tht P P m d P P m Q Q u Let m e the smllest itegers suh tht P P m Q Q P m P m d P P m Q Q P m P m Q Q u d otiue this wy. This is possile sie the sums of P d Q re Now if s t re the prtil sums of this series P P m Q Q whose lst terms re P m Q respetively, s P m d t u Q y the hoies of m Sie P Q hve it, so s t must hve it As ll other prtil sums re squeezed y s d t the series ostruted must hve it ) Dirihlet s Rerrgemet Theorem. If d overges to the sme sum s. (Reso. Defie p q s i the lst proof. Sie p q Sie p q my view zeros where For y positive iteger m overges solutely, every rerrgemet p q overge, sy to p d q p s rerrgemet of the oegtive terms of prtil sum s m m p prtil sum s m is lso iresig, hee p overges. Now, for every positive iteger As get p p Similrly, q q The p q ) Emple. 4 5 overges (solutely) to respetively. d isertig p p Sie p the p terms 4 terms 8 terms is rerrgemet of, so it lso overges to. p p Remrs. As osequee of the rerrgemet theorem, the sum of oegtive series is the sme o mtter how the terms re rerrged. 9

20 4 the whih i.e. ut Chpter 5. Rel Numers Deiml represettios d poits o lie re possile wys of itroduig rel umers, ut they re ot too oveiet for provig my theorems. Isted will itrodue rel umers y its importt properties. Aiomti Formultio. There eists set (lled rel umers) stisfyig the followig four ioms: () (Field Aiom) is field (i.e. hs two opertios d suh tht for y,,, (i),, (ii),, (iii),, (iv) re uique elemets, with suh tht,, (v) for eh, is uique elemet suh tht ; if is uique elemet suh tht (vi).) (This iom llows us to do lger with equtios. Defie to me to me to me Also, defie ) () (Order Aiom) hs (orderig) reltio suh tht for y, (i) etly oe of the followig,, is true, (ii) if,,, (iii) if,, (iv) if d,. (This iom llows us to wor with iequlities. For emple, usig (ii) d (iii), see tht if d d d euse d Also, get (for otherwise would imply y (iii) tht implies y (iv) tht otrditio). Now defie to me to me or et. Also, defie losed itervl [ : ope itervl : et. Prt (i) of the order iom implies y two rel umers e ompred. We defie m to e the mimum of d similrly for miimum. Also, defie m The d i.e. Net if d oly if d Filly, ddig d y y y get y y y whih is the trigle iequlity y y ) () (Well-orderig Aiom) is ll-ordered suset of (i.e. for y oempty suset S of, is m S suh tht m for ll S. This m is the lest elemet (or the miimum) of S). (This iom llows us to formulte the priiple of mthemtil idutio lter.) Defiitios. For oempty suset S of, S is ouded ove iff is some M suh tht M for ll S. Suh M is lled upper oud of S. The supremum or lest upper oud (deoted y sup S or lu S) of S is upper oud M of S suh tht M M for ll upper ouds M of S. Emples. () For S : upper ouds of S re ll M So sup S S () For S :, the upper ouds of S re ll M So sup S S (4) (Completeess Aiom) Every oempty suset of whih is ouded ove hs supremum i. (This iom llows us to prove results tht hve to do with the eistee of erti umers with speifi properties, s i the itermedite vlue theorem.) Defiitios. is the turl umers (or positive itegers), is the itegers, m : m d is the rtiol umers d : is the irrtiol umers. Remrs (Eerises). The first three ioms re lso true if is repled y Hover, the ompleteess iom is flse for For emple, S : is ouded ove i it does ot hve supremum i

21 let y As ove, defie S to e ouded elow if is some m suh tht m for ll S. Suh m is lled lor oud of S. The ifimum or gretest lor oud (deoted y if S or gl S) of S is lor oud m of S suh tht m m for ll lor ouds m of S. if S sup S lor ouds re here S upper ouds re here If S is ouded ove d elow, S is ouded. (Note sup S, if S my or my ot e i S. Also, if S is ouded, for ll S M m sup S if S ) Remrs (Eerises). () (Completeess Aiom for Ifimum) Every oempty suset of whih is ouded elow hs ifimum i. This follows from osiderig B : B (This is the refletio of B out.) If B is ouded elow, B is ouded ove d if B sup B Similrly, if B is ouded ove, B is ouded elow d sup B if B -B B ( ) if(-b) sup(-b) ifb A supb ifa supa () For set B, if it is ouded ove d let B : B (This is the slig of B y ftor of ) We hve sup B sup B If A B, if B if A wheever B is ouded elow d sup A sup B wheever B is ouded ove. uits ifb B supb if(+b) +B sup(+b) () For B : B (This is trsltio of B y uits.) It follows tht B hs supremum if d oly if B hs supremum, i whih se sup B sup B The ifimum sttemet is similr, i.e. if B if B More geerlly, if A d B re ouded, lettig A B y : A B hve sup A B sup A sup B d if A B if A if B Simple Cosequees of the Aioms. Theorem (Ifiitesiml Priiple). For, y, y for ll if d oly if y (Similrly, y for ll if d oly if y.) Proof. If y, for ll, y y y y (iv) of the field iom d (iii) of the order iom. Coversely, if y for ll ssumig y y y Sie y The other sttemet follows from the first sttemet sie y is the sme s y get y y (iii) of the order iom. Let lso hve y These otrdit (i) of the order iom. So Remrs. Tig y, see tht for ll if d oly if. This is used whe it is diffiult to show two epressios re equl, ut it my e esier to show for every Theorem (Mthemtil Idutio). For every, A is (true or flse) sttemet suh tht A is true d for every, A is true implies A is lso true. The A is true for ll. Proof. Suppose A is flse for some. The S : A is flse is oempty suset of. By the ll-orderig iom, is lest elemet m i S The A m is flse. Also, if A is flse, m. Tig otrpositive, this mes tht if m A is true.

22 let so 5] S so i.e. so y, Now A is true, so m implies A is true. By hypothesis, A d m imply m. So m Let m A m is true, otrditio. m m Theorem (Supremum Property). If set S hs supremum i d is S suh tht sup S sup S. Proof. Sie sup S sup S, sup S is ot upper oud of S. The is S suh tht sup S. Sie sup S is upper oud of S, sup S. Therefore sup S sup S. Theorem (Ifimum Property). If set S hs ifimum i d is S suh tht if S if S. Proof. Sie if S if S, if S is ot lor oud of S. The is S suh tht if S. Sie if S is lor oud of S, if S. Therefore if S if S. Theorem (Arhimede Priiple). For y, is suh tht. Proof. Assume eists suh tht for ll hve The : hs upper oud. By the ompleteess iom, hs supremum i. By the supremum property, is suh tht sup, whih yields the otrditio sup. Questio. How is otied i? How is otied i? Below will show tht is dese i i the sese tht ete y two distit rel umers how lose, is rtiol umer. Similrly, is dese i First eed lemm. Lemm. For every o mtter eists lest iteger greter th or equl to (I omputer siee, this is lled the eilig of d is deoted y ) Similrly, eists gretest iteger less th or equl to (This is deoted y [] I omputer siee, this is lso lled the floor of d is deoted y ) Proof. By the Arhimede priiple, is suh tht The By (iii) of the order iom, The set S : is oempty suset of euse S By the ll-orderig iom, is lest positive iteger m The m is the lest iteger greter th or equl to So the eilig of every rel umer lwys eist. Net, to fid the floor of less th or equl to e the lest iteger greter th or equl to is the gretest iteger Theorem (Desity of Rtiol Numers). If y, is m suh tht m y. Proof. By the Arhimede priiple, is suh tht y So y d hee y Let m [] m [] [] m So m m y y Theorem (Desity of Irrtiol Numers). If y, is suh tht y. Proof. Let. By the desity of rtiol umers, is m suh tht m y (If m pi other rtiol umer ete d y So my te m ) Let m y d Emples. () Let S 4 S is ot ouded elow d so S hs o ifimum. O the other hd, S is ouded ove y 5 d every upper oud of S is greter th or equl to 5 S So sup S 5 () Let S : 4 I the emples followigthe defiitio of supremum, sw sup S Here will show if S (Note S.) Sie for ll is lor oud of S. So y the ompleteess iom for ifimum, if S must eist. Assume S hs lor oud t By the Arhimede priiple, is suh tht t The t S otrditio to t eig lor oud of S So is the gretest lor oud of S () Let S [ 6 Sie 6 for every S hs s lor oud d 6 s upper oud. We will show if S d sup S 6 (Note S d 6 S.) Sie S every lor oud t stisfy t Therefore if S For supremum, ssume is upper oud u 6 Sie S u By the desity of rtiol umers, is r suh tht u r 6 The r [ 6 S As u r otrdits u eig upper oud of S every upper oud u 6 Therefore, sup S 6

23 y it the the ]) the how the Chpter 6. Limits Limit is the most importt oept i lysis. We will first disuss its of sequees, its of futios. Defiitios. A (ifiite) sequee i set S (e.g. S or S [ is list of elemets of S i speifi order. Briefly it is deoted y (Mthemtilly it my e vied s futio : S with for ) We sy the sequee is ouded ove iff the set is ouded ove. (Bouded elow d ouded sequees re defied similrly.) We will lso write sup for the supremum of the set d if for the ifimum of the set CAUTION: Sie seldom tl out set with oe elemet from ow o, so ottios lie will deote sequees uless epliitly stted otherwise. For diste ete d y is ommoly deoted y d y whih equls y Below will eed qutittive mesure of wht it mes to e lose for disussio of the oept of it. For ope itervl is lled the -eighorhood of Note if d oly if d i.e. every umer i hs diste less th from Limit of sequee is ofte eplied y syig it is the umer the s re loser d loser to s gets lrger d lrger. There re two d poits out this epltios. () Beig lose or lrge is feelig! It is ot ft. It ot e proved y logil rgumet. () The effet of eig lose umulte to yield lrge seprtio! If two umers hvig diste less th or equl to re osidered lose, is lose to d is lose to d is lose to, 99 is lose to, ut is quite fr from. So wht is the meig of lose? How it e defied so it e heed? Ituitively, sequee gets lose to umer if d oly if the diste d goes to. This hppes if d oly if for every positive the diste d evetully eomes less th The followig emple will try to me this more preise. Emple. As gets lrge, ituitively my thi gets lose to. For soo (tht is, for wht ) will the diste d e less th? (Wht if? Wht if? Wht if is ritrry positive umer?) Solutio. Cosider d Solvig for get If 9 So s soo s 6 diste ete d will e less th (If 99 So 8 will do. If 999 So 55 will do. If [ ] will do. If sie for every so will do. So for every is K so tht s soo s K diste d will e less th ) Note the vlue of K depeds o the vlue of the smller is, the lrger K will e. (Some people write K to idite K depeds o ) Defiitio. A sequee overges to umer (or hs it ) iff for every, is K suh tht for every K implies d (whih mes K K K Let us ow do few more emples to illustrte how to show sequee overges y heig the defiitio. Lter, will prove some theorems tht will help i estlishig overgee of sequees. Emples. () Let For every, let K, K implies. So overges to () Let For every implies, eists iteger K K. So overges to (y the Arhimede priiple). The K

24 (whih () Let os Show tht overges to y heig the defiitio. Solutio. For every os eists iteger K y the Arhimedi priiple. The K implies os os K So overges to (4) Let z Show tht z overges to y heig the defiitio. Solutio. (Let u z z By the iomil theorem, so tht u implies z ) For every eists iteger K (y the Arhimede priiple). The K u K So z overges to. z u u u u u Remrs. (i) From the defiitio, see tht overges to overges to d overges to re equivlet euse i the defiitio, is the sme s (ii) To show overges to mes for every hve to fid K s i the defiitio or show suh K eists. O the other hd, if re give tht overges to, for every eve hoose for our oveiee,) is K s i the defiitio for us to use. Theorem (Uiqueess of Limit). If overges to d y, y (d so my write ). Proof. For every, will show y. (By the ifiitesiml priiple, will get y.) Let By the defiitioof overgee, re K, K suh tht K d K y. Let K m K K. By the trigle iequlity, y K K y K K y. Boudedess Theorem. If overges, is ouded. Proof. Let. For, is K suh tht K Let M m,, K, for every M (i.e. [ M M]).. Remrs. The overse is flse. The sequee is ouded, ut ot overget. I geerl, ouded sequees my or my ot overge. Theorem (Computtio Formuls for Limits). If overges to d y overges to y, (i) y overges to y, respetively, i.e. y y, (ii) y overges to y, i.e. y y, (iii) y overges to y, provided y for ll d y. Proof. (i) For every, re K K suh tht K d K y y. Let K m K K The K implies K d K So for these s, y y (ii) We prove lemm first. y y y y Lemm. If is ouded d,. Proof. Sie is ouded, is M suh tht M for ll. For every, sie M d overges to, is K suh tht K M M. 4

25 so i.e. z z To prove (ii), write y y y y y y y y is ouded y the oudedess theorem. So y (i) d the lemm, y y y y y y y y. Sie overges, y y y (iii) Note y Sie y overges to y is K suh tht K implies y y y By the trigle iequlity, y y y y y y y for K The for every y m mi y y K y Net will show ( y (ii), y y y y y y ). For every let m y Sie y y, is K suh tht K y y K y y y y y y m y The Remrs. () As i the proofs of the uiqueess of it d prt (i) of the omputtio formuls, whe hve K d K my s ll te K m K K to sy K d from ow o. () By mthemtil idutio, show tht the omputtio formuls lso hold for fiitely my sequees. Hover, the umer of sequees must sty ostt s the followig emple shows terms Sdwih Theorem (or Squeeze Limit Theorem). If y for ll d z. Proof. For y, is K suh tht K z d y z, i.e., y z y so z, i.e. z. y z. Sie Emple. Let [ ] for every (Note ) The Sie y the sdwih theorem, Theorem (Limit Iequlity). If for ll d,. Proof. Assume. The for, is K suh tht K, otrditio., whih implies Remrs. By the it iequlity ove, if y,, y y, tig y get the it y y Also, if d i.e. [ d imply [ (This is flse for ope itervls s ) Supremum Limit Theorem. Let S e oempty set with upper oud There is sequee i S overgig to if d oly if sup S. sup S Proof. If sup S for, y the supremum property, is S suh tht sup S. Sie, the sdwih theorem implies sup S. Coversely, if sequee i S overges to sup S implies sup S Sie is upper oud of S sup S Therefore sup S 5

26 ] y ] respetively. the ] Ifimum Limit Theorem. Let S e oempty set with lor oud There is sequee i S overgig to if d oly if if S. Emples. () Let S : Sie for ll the sequee i S overges to By the ifimum it theorem, if S () Let S y : [ lor oud of S Now the sequee Sie y for ll ] () Let A d B e ouded i Prove tht if A B : A is lor oud of the set S Now d y [ i S overges to By the ifimum it theorem, if S B sup A B is sup A if B Solutio. Sie A d B re ouded, sup A d if B eist y the ompleteess iom. For A B hve for some A d B So sup A if B Hee sup A sup B is upper oud for A B By the supremum it theorem, is sequee A suh tht overges to sup A By the ifimum it theorem, is sequee B suh tht overges to if B The is sequee i A B d overges to sup A if B y the omputtio formuls for its. By the supremum it theorem, fore sup A B sup A if B Defiitio. A susequee of is sequee j, where j d. Emples. For the sequee if set j j get the susequee If j j get the susequee If j is the j-th prime umer, get the susequee 5 7 Remrs. () Tig j j see tht every sequee is susequee of itself. A susequee lso e thought of s otied from the origil sequee y throwig wy possily some terms. Also, susequee of susequee of is susequee of () By mthemtil idutio, hve j j for ll j euse d j j j implies j j Susequee Theorem. If, j for every susequee j of. (The overse is j trivilly true euse every sequee is susequee of itself.) Proof. For every, is K suh tht K. The j K j K j. Questio. How tell if sequee overges without owig the it (espeilly if the sequee is give y reurree reltio)? For erti types of sequees, the questio hs esy sr. mootoe stritly mootoe iff iresig deresig Defiitios. is iff stritly iresig is stritly deresig iresig or deresig is stritly iresig or deresig respetively. Mootoe Sequee Theorem. If is iresig d ouded ove, sup. (Similrly, if is deresig d ouded elow, if.) Proof. Let M sup, whih eists y the ompleteess iom. By the supremum property, for y, is K suh tht M K M. The j K M K j M j M M j Remr. Note the ompleteess iom ws used to show the it of eists (without givig the vlue). Emples. () Let d The d So y the mootoe sequee theorem, hs it Now Tig its d usig the susequee theorem, get So or. Sie will derese to the it. it is. Similrly, if 6

27 sie so so. () Does represet rel umer? Here hve ested rdil defied y d The questio is whether overges to rel umer. (Computig few terms, suspet tht is iresig. To fid upper oud, oserve tht if implies ) Now y mthemtil idutio, show tht (If 4 tig squre roots, get ) By the mootoe sequee theorem, hs it We hve The or. Sie Aother ommo type of sequees is otied y miig deresig sequee d iresig sequee ito oe of the form I the et emple, will hve suh situtio d eed two theorems to hdle these id of sequees. Nested Itervl Theorem. If I [ ] is suh tht I I I, I [ If, I otis etly oe umer. [ [ [ ] ] ]... where Proof. I I I implies is iresig d ouded ove y d is deresig d ouded elow y. By the mootoe sequee theorem, overges to sup d overges to if. Sie for every, tig its, hve. Cosequetly, [ ] (i.e. ) for ll if d oly if So I [ If d I Remrs. Note i the proof, the mootoe sequee theorem ws used. So the ested itervl theorem lso impliitly depeded o the ompleteess iom. Itertwiig Sequee Theorem. If m d m overge to lso overges to Proof. For every m overges to is K suh tht m K m Sie m lso overges to is K suh tht m K m Now if K m K K either m K m or m K m Emple. Does represet umer? Here hve otiued frtio defied y d We hve 4 5 Plottig these o the rel lie suggests for ll This e esily estlished y mthemtil idutio. (If Tig reiprol d pplyig the reurree reltio, hve Repetig these steps oe more, get 4 ) Let I [ m m ] I I I Now m Usig this, get By the sdwih theorem, m 4 9 m m m m m m 4 9 m m So y the ested itervl theorem, I for some