Monotonic Sequences

Definition2.1Monotonic sequence ... A sequence sn s n of real numbers is called monotonic if one of the following is true: For all n∈N, n ∈ N , we have sn≤sn+1 ... Skiptomaincontent\(\newcommand{\doubler}[1]{2#1} \newcommand{\lt}{} \newcommand{\amp}{&} \) FrontMatter 1BoundedSequences 2MonotonicSequences 3CauchySequencesAuthoredinPreTeXt Anotherclassofsequenceswhosebehavioriswellregulatedistheclassofsequenceswhich"donotchangedirection."Thesearethemonotonicsequences. Definition2.1Monotonicsequence Asequence\(s_n\)ofrealnumbersiscalledmonotonicifoneofthefollowingistrue: Forall\(n\in\mathbb{N}\text{,}\)wehave\(s_n\leqs_{n+1}.\) Forall\(n\in\mathbb{N}\text{,}\)wehave\(s_n\geqs_{n+1}.\) Inthefirstcase,wesaythesequenceisincreasing.Inthesecondcase,wesaythesequenceisdecreasing.Ifeitherinequalityisastrictinequality(\(\lt\)or\(\gt\)),thenwesaythesequenceis"strictly"increasingordecreasingrespectively.Monotonicityaloneisnotsufficienttoguaranteeconvergenceofasequence.Indeed,manymonotonicsequencesdivergetoinfinity,suchasthenaturalnumbersequence\(s_n=n\text{.}\)But,ifwecanforceamonotonicsequencetoremaintrappedbetweenaconstantceilingandfloor,wecanguaranteeitwillconverge.Thisisthemonotoneconvergencetheorem. Theorem2.2Monotoneconvergencetheorem Let\(s_n\)beanincreasing(respectively,decreasing)sequence.If\(s_n\)isboundedabove(respectively,boundedbelow),then\(s_n\)isconvergentand \begin{align*} \lim_{n\to\infty}s_n\amp=\sup\{s_n\}\amp\amp\text{(respectively,}\quad\inf\{s_n\}.) \end{align*} Inotherwords,anincreasingsequencewhichisboundedabovewillconvergetoitssupremum;adecreasingsequencewhichisboundedabovewillconvergetoitsinfimum.Proof Wewillshowtheprooffortheincreasingcase.Thedecreasingcaseisanalogousbyreversinginequalitysymbolsandtradingsupremumforinfimum. Let\(s_n\)beanincreasingsequencethatisboundedabove.Thecompletenessaxiomoftherealnumbersguaranteesthatanybounded-aboveandnonemptysetofrealnumbershasaleastupperbound,so\(S=\sup\{s_n\}\)exists.Wewillshowthatthesequence\(s_n\)convergesto\(S\text{.}\) Let\(\epsilon\gt0\)bearbitrarilychosen. Since\(S=\sup\{s_n\}\)istheleastupperboundofthesetofalltermsofthesequence,weknowbyaresultfromapreviousquizthatthereexistsatleastonetermofthesequencewhichisgreaterthan\(M-\epsilon\text{.}\) Define\(N\in\mathbb{N}\)tobetheindexofthisterm,i.e.choose\(N\)suchthat \begin{gather} s_N>M-\epsilon.\tag{2.1} \end{gather} Nowlet\(n\geqN\)bearbitrarilychosen. Becausethesequenceisincreasing,weknowthat\(n\geqN\)implies\(s_n\geqs_N\text{.}\)But(2.1)thenguaranteesthat \begin{align} s_n\geqs_N\amp>S-\epsilon.\tag{2.2} \end{align} Furthermore,since\(S\)isanupperboundfor\(\{s_n\}\)wehave\(s_n\leqS\text{.}\)Thisguaranteesthat\(s_n-S\leq0\)andhence\(|s_n-S|=S-s_n.\)Forthisreason,(2.2)showsthat \begin{align*} |s_n-S|\amp=S-s_n\amp\ltS-(S-\epsilon)=\epsilon, \end{align*} completingtheproof. Themonotoneconvergencetheoremprovidesapowerfulone-twopunchthatissufficienttoprovethatasequenceconvergesbyprovingtwo(probably)simplerproperties:thatitismonotonic,andthatitisbounded.Thefollowingexampleillustrateshowthemonotoneconvergencetheoremmightbeappliedtoaconcreteexampleofasequence. Example2.3Arecursively-definedsequence Provethatthesequencewhichisdefinedrecursivelyby \begin{align*} s_1=1\amp\amps_{n+1}=\sqrt{1+s_n} \end{align*} converges. Proof:Thesequenceismonotonic:Ifthisisindeedtrue,thenallitstermswillfollowthepatternsuggestedbyitsfirsttwo,namelythat \begin{align*} s_1=1\amp\amp\lt\amp\amp\sqrt{2}=s_2. \end{align*} Sowewilltrytoprovethat\(s_n\)isastriclyincreasingsequence,i.e.we'llprovethat \begin{align} s_{n+1}\amp\gts_n\amp\amp\foralln\geq1\tag{2.3} \end{align} Basecase:\(n=1\text{.}\)Wehave \begin{align*} s_{1+1}=s_n\amp=\sqrt{2}\amp\gt1=s_1 \end{align*} establishingthebasecase. Inductionhypothesis:Assumethat(2.3)istrue. Inductioncase:Wemustshowthat \begin{align*} s_{(n+1)+1}\amp\gts_{n+1}\amp\amp\text{i.e.,that}\amps_{n+2}\gts_{n+1}. \end{align*} Butbydefinitionofthissequence, \begin{align*} s_{n+2}\amp=\sqrt{1+s_{n+1}}\amp\\ \amp\gt\sqrt{1+s_n}\amp\text{bytheinductionhypothesis}\knowl{./knowl/mdn-4.html}{\text{(2.3)}}\\ \amp=s_{n+1}\amp\text{bydefinitionofthissequence.} \end{align*} Thuswehaveproventheclaimthatthissequenceisincreasing. Thesequenceisboundedabove:Inspectingthefirstfewtermsofthesequence,wemightconjecturethatthewholesequenceisboundedaboveby2.Thisiswhatwe'llprove,againbyinduction.We'llshowthat \begin{align} |s_n|\amp\leq2\amp\amp\foralln\geq1.\tag{2.4} \end{align} Basecase:\(n=1\text{.}\)Wehave \begin{align*} |s_1|=1\amp\leq2. \end{align*} Inductionhypothesis:Assumethat(2.4)istrue. Inductioncase:Wemustshowthat \begin{gather*} |s_{n+1}|\leq2. \end{gather*} Bydefinitionofthissequence, \begin{align*} |s_{n+1}|\amp=s_{n+1}\amp\text{(sincethesequenceisnonnegative)}\\ \amp=\sqrt{1+s_n}\amp\\ \amp\leq\sqrt{1+2}\amp\text{bytheinductionhypothesis}\knowl{./knowl/mdn-5.html}{\text{(2.4)}}\\ \amp=\sqrt{3}\leq2. \end{align*} Thisprovestheclaimthatthissequenceisboundedabove. Since\(s_n\)isincreasingandboundedabove,bythemonotoneconvergencetheoremitisconvergent.