Sequence A sequence is a function whose domain of definition is the set of natural numbers.

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1 Chapter Sequeces Course Title: Real Aalysis Course Code: MTH3 Course istructor: Dr Atiq ur Rehma Class: MSc-I Course URL: wwwmathcityorg/atiq/fa8-mth3 Sequeces form a importat compoet of Mathematical Aalysis ad arise i may situatios The first rigorous treatmet of sequeces was made by George Cator (845-98) ad A Cauchy ( ) A sequece (of real umbers, of sets, of fuctios, of aythig) is simply a list There is a first elemet i the list, a secod elemet, a third elemet, ad so o cotiuig i a order forever I mathematics a fiite list is ot called a sequece (some authors cosidered it fiite sequece); a sequece must cotiue without iterruptio Formally it is defied as follows: Sequece A sequece is a fuctio whose domai of defiitio is the set of atural umbers Notatio: A ifiite sequece is usually deoted as { } s = or { s : N } or { s, s, s,} or simply as { } 3 But it is ot limited to above otatios oly The values eg i) { } = {,,3, } ii) s or by ( s ) s are called the terms or the elemets of the sequece { } =,,, 3 + ( ) =,,,, iii) { } { } iv) {,3,5,7,, }, a sequece of positive prime umbers Subsequece It is a sequece whose terms are cotaied i give sequece A subsequece of { s } is usually writte as { s } Icreasig Sequece s is said to be a icreasig sequece if s + s A sequece { } Decreasig Sequece s is said to be a decreasig sequece if s + s A sequece { } Mootoic Sequece s is said to be mootoic sequece if it is either icreasig or decreasig A sequece { } s

2 Ch : Sequeces - - Remars: A sequece { } s is mootoically icreasig if s s + s s is mootoically icreasig if + s, A positive term sequece { } A sequece { } s is mootoically decreasig if s s + A positive term sequece { s } is mootoically decreasig if s, s + Strictly Icreasig or Decreasig A sequece { s } is called strictly icreasig or decreasig accordig as s > + s or s < + s Examples: { } = {,,3,} is a icreasig sequece is a decreasig sequece { cos π } = {,,,,} is either icreasig or decreasig Questios: ) Prove that + is a decreasig sequece + ) Is is icreasig or decreasig sequece? + Bouded Sequece s is said to be bouded if there is a umber λ A sequece { } s < λ N For such a sequece, every term belogs to the iterval [ λ, λ] It ca be oted that if the sequece is bouded the its supremum ad ifimum exist If S ad s are the supremum ad ifimum of the bouded sequece { s }, the we write S = sup s ad s = if s

3 Ch : Sequeces Examples ( ) (i) { } u = is a bouded sequece v = si x is also bouded sequece Its supremum is ad ifimum is (ii) { } { } (iii) The geometric sequece { } bouded below by a π (iv) ta is a ubouded sequece ar, r > is a ubouded above sequece It is Covergece of the sequece The sequece,,,,,, is gettig closer ad closer to the umber We say that this sequece coverges to or that the limit of the sequece is the umber How should this idea be properly defied? The study of coverget sequeces was udertae ad developed i the eighteeth cetury without ay precise defiitio The closest oe might fid to a defiitio i the early literature would have bee somethig lie A sequece { s } coverges to a umber L if the terms of the sequece get closer ad closer to L However, this is too vague ad too wea to serve as defiitio but a rough guide for the ituitio, this is misleadig i other respects What about the sequece,,,,,,,,,? Surely this should coverge to but the terms do ot get steadily closer ad closer but bac off a bit at each secod step The defiitio that captured the idea i the best way was give by Augusti Cauchy i the 8s He foud a formulatio that expressed the idea of arbitrarily close usig iequalities Defiitio A sequece { s } of real umbers is said to coverget to limit s as, if for every real umber >, there exists a positive iteger, depedig o, so that s s < wheever > A sequece that coverges is said to be coverget A sequece that fails to coverge is said to diverget We will try to uderstad it by graph of some sequece Graphs of ay four sequeces is draw i the picture below

4 Ch : Sequeces Examples a) Prove that lim = (or coverges to ) Solutio: Let > be give By the Archimedea Property, there is a positive iteger = ( ) >, that is, < The, if, we have < < Thus we proved that for all >, there exists, depedig upo, = < wheever Hece coverges to poit b) Prove that lim = (by defiitio) + Solutio: Let > be give Now cosider = < < ( + > > ) + + Now if we choose <, the the above expressio gives us < wheever + Hece, we coclude that, lim = +

5 Ch : Sequeces c) Prove that lim = 3 (by defiitio) + Solutio: Let > be give Now cosider = + + = < < ( + > > ) + + Now if we choose <, the the above expressio gives us < wheever Hece, we coclude that lim = 3 + Questios: Use defiitio of the limits to prove the followigs: a) lim = b) lim = Review Triagular iequality: If a, b R, the a b a ± b a + b If a < for all >, the a = Theorem A coverget sequece of real umber has oe ad oly oe limit (ie limit of the sequece is uique) s coverges to two limits s ad t, where s t The for all >, there exists two positive itegers ad s s < > () ad s t < > () As () ad () hold simultaeously for all > max(, ) Thus for all > max(, ) we have s t = s s + s t s s + s t Suppose { } < + = As is arbitrary, we get s = t, that is, the limit of the sequece is uique

6 Ch : Sequeces Questio Prove that if lim s = t, the lim s = t but coverse is ot true i geeral Review: For all a, b, c R, a b < c c b < a < c + b or c a < b < c + a Theorem (Sadwich Theorem or Squeeze Theorem) t be two coverget sequeces lim s Suppose that { } s ad { }, the the sequece { } = limt If s < u < t u also coverges to s Sice the sequece { s } ad { t } coverge to the same limit s (say), therefore for give > there exists two positive itegers, > s s < >, t s < > ie s < s < s + >, s < t < s + > Sice we have give s < u < t > s < s < u < t < s + > max(,, ) s < u < s + > max(,, ) ie u s < > max(,, ) ie limu = s Example Show that lim = ( + ) ( + ) ( ) Solutio Cosider s = ( + ) ( + ) ( ) As s < + + +, ( ) ( ) ( ) that is, times s < ( ) s < 4 times

7 Ch : Sequeces lim lim s < lim 4 lim s < lim s = Cauchy Sequece A sequece { s } of real umber is said to be a Cauchy sequece if for give umber >, there exists a positive iteger ( ) s sm < m, > Example The sequece is a Cauchy sequece Suppose s = ad > be give We choose a positive iteger = ( ) such that > The if m,, we have < ad similarly < Therefore, it m follows that if m,, the s sm = + < + = m m Sice > is arbitrary, we coclude that is Cauchy sequece Theorem A Cauchy sequece of real umbers is bouded s be a Cauchy sequece The for give umber >, there exists a positive iteger s sm < m, > Tae =, the we have Let { } Fix m = + the + + s s = s s + s s s + s + + sm < m, > < + s + >

8 Ch : Sequeces = λ >, ad λ s + = + ( chages as chages) Hece we coclude that { s } is a Cauchy sequece, which is bouded oe Note: (i) Coverget sequece is bouded (ii) The coverse of the above theorem does ot hold ie every bouded sequece is ot Cauchy s where s = ( ), It is bouded sequece because Cosider the sequece { } ( ) = < But it is ot a Cauchy sequece if it is the for = we should be able to fid a positive iteger s sm < for all m, > But with m = +, = + whe + >, we arrive at s s = ( ) ( ) m + + = + = < is absurd Hece { s } is ot a Cauchy sequece Also this sequece is ot a coverget sequece (it is a oscillatory sequece) Questio: Prove that every Cauchy sequece of real umber is bouded but coverse is ot true Theorem If the sequece { s } coverges to s the a positive iteger s > s for all > We fix = s > a positive iteger s s < for > s s < s Now s = s s s s s s + s s < ( ) s < s

9 Ch : Sequeces Theorem Let a ad b be fixed real umbers if { s } ad { } respectively, the as + bt coverges to as + bt (i) { } (ii) { } s t coverges to st t coverge to s ad t s (iii) coverges to s t t, provided t ad t Sice { s } ad { } t coverge to s ad t respectively, < s s > N t t < > N Also λ > s < λ (i) We have as + bt as + bt = a s s + b t t > ( { } ( ) ( ) ( ) ( ) a( s s) + b( t t) s is bouded ) < a + b > max (, ) =, where = a + b a certai umber This implies { as bt } (ii) (iii) + coverges to as + bt s t st = s t s t + s t st = s ( t t) + t ( s s ) s ( t t) + t ( s s ) < λ + t > max (, ) =, where = λ + t a certai umber This implies { } t t = t t t t s t coverges to st t t t t t t = < > max (, ) = = 3, where 3 t t t > t = a certai umber This implies t coverges to t s Hece s t = t coverges to s s t t = ( from (ii) )

10 Ch : Sequeces - - Theorem For each irratioal umber x, there exists a sequece { } umbers lim r = x Sice x ad x + are two differet real umbers a ratioal umber r x < r < x + Similarly a ratioal umber r r x < r < mi r, x + < x + Cotiuig i this maer we have x < r3 < mi r, x + < x + 3 x < r4 < mi r3, x + < x + 4 r of distict ratioal x < r < mi r, x + < x + r of the distict ratioal umber This implies that there is a sequece { } Sice Therefore Theorem lim x < r < x + x = lim x + = x ( ) lim r = x Let a sequece { } (i) If { } (ii) If { } s be a bouded sequece s is mootoically icreasig the it coverges to its supremum s is mootoically decreasig the it coverges to its ifimum Proof (i) Let S = sup s ad tae > Sice there exists s Sice { } S < s s is mootoically icreasig, therefore

11 Ch : Sequeces - - S < s < s < S < S + for > S < s < S + for > s S < for > lim s = S (ii) Let s = if s ad tae > Sice there exists s Sice { } s < s + s is mootoically decreasig, therefore s < s < s < s < s + for > s < s < s + for > s s < for > Thus lim s = s Questio: Let { } s be a sequece ad lim s = s The prove that lim s + = s Prove that a bouded icreasig sequece coverges to its supremum 3 Prove that a bouded decreasig sequece coverges to its ifimum Recurrece Relatio A sequece is said to be defied recursively or by recurrece relatio if the geeral term is give as a relatio of its precedig ad succeedig terms i the sequece together with some iitial coditio Example: Let t > ad let { } t be defied by t (i) Show that { t } is decreasig sequece (ii) It is bouded below (iii) Fid the limit of the sequece Sice t > ad { } t is defied by t t > Also t t+ = t + t ( ) = + t for = + t ; t t + t = = > t t t > t + This implies that t is mootoically decreasig

12 Ch : Sequeces - - Sice t >, t is bouded below Sice t is decreasig ad bouded below therefore t is coverget Let us suppose limt = t The lim t+ = lim t lim = limt t t = t = t t = t t t ( t ) = t = t t + = Questio: Let { t } be a positive term sequece Fid the limit of the sequece if 4t+ = 3t for all 5 u be a sequece of positive umbers The fid the limit of the sequece Let { } = + for if u+ u u 4 The Fiboacci umbers are: F = F =, ad for every 3, F is defied by F the recurrece relatio F = F + F Fid the lim (this limit is ow as F golde umber) Theorem Every Cauchy sequece of real umbers has a coverget subsequece s is a Cauchy sequece Let > the a positive iteger s s <,, =,,3, Suppose { } Put b = ( s ) ( ) ( ) + s s + + s s b = ( s ) ( ) ( ) + s s + + s s ( ) + ( ) + + ( ) s s s s s s < = ( ) = =

13 Ch : Sequeces b < This gives { } b is bouded Sice b = b + ( s + s ) { } where + b is coverget b = s s s = b + s, coverget s is a certai fix umber therefore { } s which is a subsequece of { s } is Theorem (Cauchy s Geeral Priciple for Covergece) A sequece of real umber is coverget if ad oly if it is a Cauchy sequece s be a coverget sequece, which coverges to s The for give > a positive iteger, s s < > Now for > m > s sm = s s + s sm s s s s = s s + s s Let { } + m m < + = s is a Cauchy sequece This shows that { } s is a Cauchy sequece the for >, there exists a positive iteger m s sm <, m > m (i) s is a Cauchy sequece, Coversely, suppose that { } Sice { } therefore it has a subsequece { s } covergig to s (say) This implies there exists a positive iteger m s s < > m (ii) Now s s = s s + s s this shows that { } s s + s s < + = > max( m, m), s is a coverget sequece

14 Ch : Sequeces Example Prove that is diverget sequece 3 t be defied by Let { } For m, N, t = > m we have t tm = m + m + > ( m times) = ( m) = m I particular if = m the t tm > t is ot a Cauchy sequece therefore it is diverget This implies that { } Theorem (ested itervals) I Suppose that { I } is a sequece of the closed iterval I [ a, b ] I, ad ( b a ) I + poit as the Sice I + I a < a < a < < a < a < b < b < < b < b < b { } 3 3 =, cotais oe ad oly oe a is icreasig sequece, bouded above by b ad bouded below by a b is decreasig sequece bouded below by a ad bouded above by b b both are coverget Ad { } { } Suppose { } a ad { } a coverges to a ad { } a b = a a + a b + b b But b coverges to b a a + a b + b b as a = b ad a < a < b

15 Ch : Sequeces Limit Iferior of the sequece Suppose { s } is bouded below the we defie limit iferior of { } limif s = limu, where u = if { s : } If s is ot bouded below the limif s = Limit Superior of the sequece Suppose { s } is bouded above the we defie limit superior of { } limsup s = limv, where v = sup { s : } If s is ot bouded above the we have limsup s = + s as follow s as follow Note: (i) A bouded sequece has uique limit iferior ad superior (ii) Let { s } cotais all the ratioal umbers, the every real umber is a subsequecial limit the limit superior of s is + ad limit iferior of s is (iii) Let { s } = ( ) + the limit superior of s is ad limit iferior of s is (iv) Let s = + cosπ u = if s : The { } = if + cos π, + cos( + ) π, + cos( + ) π, cosπ if is odd = + cos( + ) π if is eve + lim if s = limu = ( ) Also v = sup { s : } + cos( + ) π if is odd + = + cosπ if is eve lim sup s = limv = ( )

16 Ch : Sequeces Theorem s is a coverget sequece the If { } ( ) ( ) lim s = lim if s = lim sup s Let lim s = s the for a real umber >, a positive iteger s s < (i) ie s < s < s + v = sup s : s < v < s + s < limv < s + (ii) If { } The from (i) ad (ii) we have s = lim sup { s } We ca have the same result for limit iferior of { } u if { s : } s by taig = Refereces: W Rudi, Priciple of Mathematical Aalysis, 3 rd Editio, McGraw-Hill, Ic, 976 RG Bartle ad DR Sherbert, Itroductio to Real Aalysis, 4 th Editio, Joh Wiley & Sos, Ic, 3 BS Thomso, JB Brucer ad AM Brucer, Elemetary Real Aalysis, Pretice Hall (Pearso), URL:

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca