Sequences and Limits

Transcription

1 Chapter Sequeces ad Limits Let { a } be a sequece of real or complex umbers A ecessary ad sufficiet coditio for the sequece to coverge is that for ay ɛ > 0 there exists a iteger N > 0 such that a p a q < ɛ holds for all itegers p ad q greater tha N This is called the Cauchy criterio Ay mootoe bouded sequece is coverget For ay sequece { } a the iferior limit ad the superior limit are defied by the limits of mootoe sequeces lim if a = lim if { a a + } ad lim sup a = lim sup { a a + } respectively Note that the iferior ad superior limits always exist if we adopt ± as limits A bouded sequece { a } coverges if ad oly if the iferior limit coicides with the superior limit World Scietific Publishig Co Pte Ltd

2 Problems ad Solutios i Real Aalysis Problem Prove that si!eπ) coverges to π as Problem Prove that the sequece ) + coverges to e/e ) as ) + + ) Problem 3 Prove that the sequece e /4 +)/ ) / coverges to as This was proposed by Cesàro 888) ad solved by Pólya 9) Problem 4 Suppose that a ad b coverge to α ad β as respectively Show that the sequece coverges to αβ as a 0 b + a b + + a b 0 Problem 5 Suppose that { a } 0 is a o-egative sequece satisfyig a m+ a m + a + C for all positive itegers m ad some o-egative costat C Show that a / coverges as This is essetially due to Fekete 93) I various places we ecouter this useful lemma i deducig the existece of limits World Scietific Publishig Co Pte Ltd

3 Sequeces ad Limits 3 Problem 6 For ay positive sequece { a } show that ) a + a + > e a for ifiitely may s where e is base of the atural logarithm Prove moreover that the costat e o the right-had side caot i geeral be replaced by ay larger umber Problem 7 For ay 0 < θ < π ad ay positive iteger show the iequality si θ + si θ + + si θ > 0 This was cojectured by Fejér ad proved by Jackso 9) ad by Growall 9) idepedetly Ladau 934) gave a shorter maybe the shortest) elegat proof See also Problem 59 Note that si θ = π θ = for 0 < θ < π which is show i Solutio 70 Problem 8 For ay real umber θ ad ay positive iteger show the iequality cos θ + cos θ cos θ + This was show by Rogosiski ad Szegö 98) Verblusky 945) gave aother proof Koumados 00) obtaied the lower boud 4/96 for Note that cos θ + = π θ si θ cos θ log si θ ) =0 for 0 < θ < π For the simpler cosie sum Youg 9) showed that m cos θ > = World Scietific Publishig Co Pte Ltd

4 4 Problems ad Solutios i Real Aalysis for ay θ ad positive iteger m Brow ad Koumados 997) improved this by replacig by 5/6 Problem 9 Give a positive sequece { a } 0 satisfyig a a 0 + ad a + a a for ay positive iteger show that a + a coverges as Show moreover that a θ coverges as where θ is the limit of the sequece a + /a This is due to Boyd 969) Problem 0 Let E be ay bouded closed set i the complex plae cotaiig a ifiite umber of poits ad let M be the maximum of Vx x ) as the poits x x ru through the set E where Vx x ) = x i x j ) i< j is the Vadermode determiat Show that M / )) coverges as This is due to Fekete 93) ad the limit τe) = lim M / )) is called the trasfiite diameter of E See Problem 59 World Scietific Publishig Co Pte Ltd

5 Sequeces ad Limits 5 Solutios for Chapter Solutio Let r ad ɛ be the itegral ad fractioal parts of the umber!e respectively Usig the expasio e = +! +! + +! + we have sice r =! +! +! + + )! ɛ = ) + ) + + < ɛ < ) + = Thus si!eπ) = siπɛ ) Note that this implies the irratioality of e Sice ɛ coverges to as we have lim siπɛ siπɛ ) ) = lim = π ɛ Hece si!eπ) coverges to π as Remark More precisely oe gets ɛ = ) + O ; 3 4 hece we have as si!eπ) = πɛ + 4π3 3 ɛ 3 + O ɛ 5 ) ) = π + ππ 3) + O 3 3 World Scietific Publishig Co Pte Ltd

6 6 Problems ad Solutios i Real Aalysis Solutio Let { } d be ay mootoe icreasig sequece of positive itegers divergig to ad satisfyig d < for > We divide the sum ito two parts as follows ) ) ) d a = ) ) d b = + + First the sum a is roughly estimated above by d x dx = d ) ) < d ) Now usig the iequality log x) + x < 0 valid for 0 < x < we obtai 0 < a < e log d /) < e d which coverges to 0 as Next by usig Taylor s formula for log x) we ca take a positive costat c such that the iequality log x) + x c x holds for ay x / Thus for ay iteger satisfyig d / / we get log k ) + k c k c d for 0 k d Suppose further that d / coverges to 0 as For example d = [ /3] satisfies all the coditios imposed above Next take a positive costat c satisfyig e x c x for ay x Sice c d/ for all sufficietly large we have ek k ) = e log k/)+k c c d World Scietific Publishig Co Pte Ltd

7 Sequeces ad Limits 7 Dividig both sides by e k ad summig from k = 0 to d we get d k ) e k c c d d e k k=0 k=0 Hece which implies b d k=0 e k < ec c d e ) b e e e c c d e Therefore a + b coverges to e/e ) as + e d Solutio 3 Oe ca easily verify that the fuctio f x) = x log x satisfies all the coditios stated i Problem 5 7 Therefore the logarithm of the give sequece coverges to hece the limit is f ) f 0+) = 0 Solutio 4 Let M be a upper boud of the two coverget sequeces a ad b For ay ɛ > 0 we ca take a positive iteger N satisfyig a α < ɛ ad b β < ɛ for all itegers greater tha N If is greater tha N the ) a k b k αβ a k α)b k + αb k β) M + α )ɛ for ay iteger k i the iterval [ ] Therefore a k b k αβ a k b k αβ k k=0 + αβ + M ) [ ] + M + α )ɛ + αβ + M ) + World Scietific Publishig Co Pte Ltd

8 8 Problems ad Solutios i Real Aalysis We ca take so large that the last expressio is less tha M + α + )ɛ Solutio 5 For a arbitrary fixed positive iteger k we put = qk + r with 0 r < k Sice a = a qk+r qa k + C) + a r we have Takig the limit as we get a a k + C + a r k lim sup a a k + C k The sequece a / is therefore bouded Sice k is arbitrary we may coclude that lim sup a which meas the covergece of a / lim if k Solutio 6 Without loss of geerality we may put a = Suppose cotrary to the coclusio that there is a iteger N satisfyig ) + a+ e a k k a for all N Put s jk = exp j + + ) k for ay itegers j k Sice 0 < a + e / a we get successively 0 < a + s a 0 < a + s + a s ++ 0 < a +k+ s +k a s ++k s +k+k for ay o-egative iteger k Hece it follows that a > s + s s +k World Scietific Publishig Co Pte Ltd

9 Sequeces ad Limits 9 O the other had usig the iequality > exp s + j we get a > + j k j=0 ) dx = x + j + j which is a cotradictio sice the right-had side diverges to as k To see that the boud e caot be replaced by ay larger umber cosider the case a = log for The a + + ) log + ) log which coverges to e as ) = exp log + + O = exp + O log )) log Solutio 7 Deote by s θ) the left-had side of the iequality to be show Write θ for ϑ for brevity Sice s θ) = R e iθ + e iθ + + e ) iθ = cos + )ϑ si ϑ si ϑ we obtai the cadidates for extreme poits of s θ) o the iterval 0 π] by solvig the equatios cos + )ϑ = 0 ad si ϑ = 0 as follows: π + π 3π + 4π where the last two cadidates are )π/+) ad π if is eve ad )π/ ad π/ + ) if is odd I ay case s θ) vaishes at least at poits i the iterval 0 π) Sice s θ) ca be expressed as a polyomial i cos θ of degree ad cos θ maps the iterval [0 π] oto [ ] homeomorphically this polyomial possesses at most real roots i [ ] Therefore all these roots must be simple ad give the actual extreme poits of s θ) except for θ = π Clearly s θ) is positive i the right eighborhood of the origi ad the maximal ad miimal poits stad i lie alterately from left to right Thus s θ) attais its miimal values at the ))) World Scietific Publishig Co Pte Ltd

10 0 Problems ad Solutios i Real Aalysis poits lπ/ 0 π) whe 3 I the cases = ad = however s θ) has o miimal poits i 0 π) Now we will show that s θ) is positive o the iterval 0 π) by iductio o This is clear for = ad = sice s θ) = si θ ad s θ) = +cos θ ) si θ Suppose that s θ) > 0 for some 3 The the miimal values of s θ) are certaily attaied at some poits lπ/ i 0 π) whose values are s lπ ) = s lπ = s lπ ) si lπ + ) > 0 Therefore s θ) > 0 o the iterval 0 π) Remark Ladau 934) gave the followig elegat shorter proof usig mathematical iductio o Suppose that s θ) > 0 o 0 π) If s attais the o-positive miimum at some poit say θ the s θ ) = 0 implies si + ) θ = si θ ad hece cos + ) θ = ± cos θ Sice si θ = si + ) θ cos θ cos + ) θ si θ = si θ cos θ ± cos θ si θ beig equal either to 0 or si θ 0 accordig to the sig We are led to a cotradictio Solutio 8 The proof is substatially based o Verblusky 945) Write θ for ϑ for brevity Let c ϑ) be the left-had side of the iequality to be show It suffices to cofie ourselves to the iterval [0 π/] Clearly c ϑ) = cos ϑ/ / ad c ϑ) = 3 cos ϑ + cos ϑ World Scietific Publishig Co Pte Ltd

11 Sequeces ad Limits ad we assume that 3 Note that cos θ = si + )ϑ si )ϑ si ϑ = si + )ϑ si ϑ + si )ϑ si ϑ whose umerator is the secod differece of the positive sequece { si ϑ } Usig this formula we get c ϑ) = si ϑ k= si k + )ϑ si kϑ + si k )ϑ k + which ca be writte as si ϑ + si ϑ + + si )ϑ si ϑ 3 ) Hece we obtai c ϑ) 3 + cos ϑ 6 = 6 si ϑ 6 ) si ϑ + ) + si + )ϑ + + si + )ϑ si ϑ + ) si ϑ + si + )ϑ + ) si ϑ For ay ϑ satisfyig si + )ϑ 0 we obviously have c ϑ) /3 Moreover if ϑ belogs to the iterval 3π/ + ) π/) the usig Jorda s iequality si ϑ ϑ/π c ϑ) 3 + ) si3π/ + )) ) > Thus it suffices to cosider the iterval [π/ + ) π/ + )] I geeral we cosider a iterval of the form [ απ + βπ ] + For ay ϑ satisfyig si + )ϑ c o this iterval it follows that c ϑ) 6 si ϑ 6 c + ) si ϑ ) World Scietific Publishig Co Pte Ltd

12 Problems ad Solutios i Real Aalysis Now the right-had side ca be writte as /6 ϕsi ϑ) where ϕx) is a cocave fuctio; hece the maximum of ϕ is attaied at a ed poit of that iterval By usig we get ϕ si απ si ϑ 7ϑ si απ 7 απ ) = απ + 6 si + + Sice 3 the last expressio is less tha c + ) siαπ/ + )) απ) 6 + ) + c + ) + 7 siαπ/7) απ) 94 + c 7 siαπ/7) Similarly we get a estimate for aother ed poit For α = ad β = 4/3 we ca take c = 3/ so that the value of ϕ at the correspodig ed poit is less tha 039 ad 08 respectively Similarly for α = 4/3 ad β = we ca take c = so that the value of ϕ is less tha 034 ad 038 respectively Therefore the maximum of ϕ o the iterval [π/+) π/+)] is less tha /3 which implies that c ϑ) > / Solutio 9 We first show that a + a > + a0 ) by iductio o Whe = 0 this holds by the assumptio Put α = + / a 0 for brevity Suppose that ) holds for m We the have a k > α k a 0 for k m + Thus which is less tha a m+ a m+ a m+ a 0 a 0 k= a k+ a k a k a k m+ k= m+ α k < a k= 0 α ) = a0 a k World Scietific Publishig Co Pte Ltd

13 Sequeces ad Limits 3 Therefore a m+ a m+ > a a 0 a0 > + a0 ; thus ) holds also for = m + Let p > q be ay positive itegers I the same way a p+ a q+ a p a q a k+ a k a k a k p k=q+ p k=q+ a k which is less tha a q p q k= α k < a0 a q This meas that the sequece { } a + /a satisfies the Cauchy criterio sice aq diverges to as q Lettig p i the above iequalities we get a q+ θ a q a0 a q Multiplyig both sides by a q /θ q+ we have a q+ θ a q q+ θ q a0 θ q+ which shows that the sequece { a /θ } also satisfies the Cauchy criterio Solutio 0 Let ξ ξ + be the poits at which Vx x + ) attais its maximum M + Sice we have Vξ ξ + ) Vξ ξ ) = ξ ξ + ) ξ ξ + ) M + M ξ ξ + ξ ξ + Applyig the same argumet to each poit ξ ξ we get + similar iequalities whose product gives ) + M+ ξ i ξ j = M+ M i j World Scietific Publishig Co Pte Ltd

14 4 Problems ad Solutios i Real Aalysis Hece the sequece M / )) is mootoe decreasig World Scietific Publishig Co Pte Ltd