Math 140A Elementary Analysis Homework Questions 3-1
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1 Math 0A Elemetary Aalysis Homework Questios -.9 Limits Theorems for Sequeces Suppose that lim x =, lim y = 7 ad that all y are o-zero. Detarime the followig limits: (a) lim(x + y ) (b) lim y x y Let s = ad for let s + = s +. (a) List the first four terms of (s ). (b) It turs out that (s ) coverges. Assume this fact ad prove that the limit is ( + ). 6 Let x = ad x + = x for. (a) Show that if a = lim x, the a = or a = 0. (b) Does lim x exist? Explai. (c) Discuss the apparet cotradictio betwee parts (a) ad (b). Assume all s = 0 ad that the limit L = lim exists. s + s (a) Show that if L <, the lim s = 0. Hit: Select a so that L < a < ad obtai N so that s + < a s for N. The show that s < a N s N for > N. (b) Show that if L >, the lim s = +. Hit: Apply (a) to the sequece t = s. Let p > 0. Use questio to show that a lim p = 0 if a + if a > does ot exist if a <.0 Mootoe Sequeces ad Cauchy Sequeces 6 (a) Let (s ) be a sequece such that s + s < N. Prove that (s ) is a Cauchy sequece ad hece a coverget sequece. (b) Is the result i (a) true if we oly assume that s + s < for all N. 7 Let S be a bouded oempty subset of R ad suppose sup S / S. Prove that there is a odecreasig sequece (s ) of poits i S such that lim s = sup S. 8 Let (s ) be a odecreasig sequece of positive umbers ad defie σ = (s + s + + s ). Prove that (σ ) is a odecreasig sequece. 9 Let s = ad s + = + s for. (a) Fid s, s ad s.
2 (b) Show that lim s exists. (c) Prove that lim s = 0. 0 Let s = ad s + = (s + ) for. (a) Fid s, s ad s. (b) Use iductio to show that s > for all (c) Show that (s ) is a oicreasig sequece. (d) Show that lim s exists ad fid lim s.. Subsequeces Cosider the sequeces defied as follows: a = ( ) b = c = d = (a) For each sequece, give a example of a mootoe subsequece. (b) For each sequece, give its set of subsequetial limits. (c) For each sequece, give its lim sup ad lim if. (d) Which of the sequeces coverges? diverges to +? diverges to? (e) Which of the sequeces is bouded? 6 Show that every subsequece of a subsequece of a give sequece is itself a subsequece of the give sequece. 7 Let (r ) be a eumeratio of the set Q of all ratioal umbers. Show that there exists a subsequece (r k ) such that lim k r k = +. 9 (a) Show that the closed iterval [a, b] is a closed set. Hit: a set is closed if it cotais all subsequetial limits of all sequeces i itself. (b) Is there a sequece (s ) such that (0, ) is its set of subsequetial limits? 0 Let (s ) be the sequece of umbers i the followig figure, listed i the idicated order. (a) Fid the set S of subsequetial limits of (s ). (b) Determie lim sup s ad lim if s......
3 Math 0A Elemetary Aalysis Homework Aswers -.9 Limits Theorems for Sequeces (a) lim(x + y ) = lim x + lim y = + 7 = 0. (b) lim y x y = lim lim y lim x (lim y ) = 7 7 = 8 9. Let s = ad for let s + = s +. (a) (s ) = (,, +, + +,...). (b) If lim s = s the takig the limit of both sides of s + = s + results i s = s + = s = s + = s = ( ± ) Give the s is clearly a positive sequece, it follows that s 0 ad so s = ( + ). 6 Let x = ad x + = x for. (a) If a = lim x, the takig limits of x + = x results i a = a, which has solutios a = ad a = 0. (b) Give that x = ad x + = x, we see that x is always. It is therefore impossible for lim x to equal either or 0. lim x does ot exist: ideed x +. For a proof, we use iductio to see that x for all : Clearly true for = sice x =. Suppose x for some. The x + = x +. Hece result. (c) There is o cotradictio. We eeded to assume the existece of a limit i (a) to the calculate what it would be. If we do t assume there is a limit, (a) is meaigless. (a) If L < the let a = +L L. Clearly L < a <. Now let ɛ =. Sice L = lim s +, s there exists N N such that N = s + s L < ɛ = ɛ + L < s + s < ɛ + L = s + < (ɛ + L) s = + L s = a s Iteratig this we immediately obtai s < a N s N. Sice a 0 for 0 < a <, we coclude that lim s = 0. (b) Writig t = s, we have L = lim t + t existig ad 0 < L <. Applyig part(a) to (t ) we obtai lim t 0, whece lim s = +. If s = a, the p s + s = a + p ( (+) p a = a p +) a. This is the limit L from questio. We immediately have lim a { p = 0 if a < + if a >
4 Additioally, if a =, we have the sequece or ( ) p both of which coverge to zero. If p a > the a = p a which therefore diverges to +. If a <, the a p = ( ) a p which p diverges by oscillatio. Hece result..0 Mootoe Sequeces ad Cauchy Sequeces 6 (a) Suppose m >, the s m s = s m s m + s m + s + s s m s m + + s + s < m + + j = j=0 Now let ɛ > 0 be give ad let N = l ɛ l. The m > > N = s m s < < N = ɛ. (s ) is therefore Cauchy. (b) If we oly assume that s + s < for all N, the the best we ca do is to coclude that s m s < m + + m. We caot coclude that this is bouded by ɛ. As a couterexample, cosider s = l. By the Mea Value Theorem applied to f (x) = l x o the iterval [, + ], l( + ) l + = x some x (, + ) Thus s + s <. However s +, so (s ) is ot Cauchy. 7 Let t = sup S. Sice t < sup S, the set S (t, sup S) is o-empty. Defie s to be the ifimum of this set. This is a odecreasig sequece (sice S (t +, sup S) is a subset of S (t, sup S) ad A B = if B if A). Moreover, 0 < sup S s sup S t = 0 hece lim s = sup S. 8 Cosider σ + σ. Sice s + s for all, we have that σ + σ = + (s + s + + s + s + ) (s + s + + s ) = ( + ) (s + + ( ( + ))(s + s + + s )) = ( + ) (s + (s + s + + s ))
5 ( + ) (s + (s + + s s + )) = 0 Thus (σ ) is a odecreasig sequece. 9 (a) s =, s = ( ) = 6 ad s = ( ) 6 = 8. (b) We claim that (s ) is oicreasig. Ideed, s + s = + s s = s + (s ). Sice s it follows that s s 0 = s. Cotiuig by iductio, we see that s + s 0 for all. (s ) is also a positive sequece ad is thus bouded below. (s ) thus coverges. (c) By (b), we kow that lim s = s for some s. Give this, s must satisfy s = lim s + = lim + s = s = s = 0,. Sice s = 6 choice. < ad (s ) is o-icreasig, it follows that s = 0 is the oly possible 0 (a) s =, s = 9 ad s = 7. (b) Certaily s = >. Now suppose that s > for some. The s + = (s + ) > ( ) + = Hece, by iductio, s > for all. ( ) = 0, thus (s ) is oicreasig. (c) s + s = ( s ) < (d) (s ) is oicreasig ad bouded below, thus lim s exists. Call the limit s. Clearly s satisfies s = lim s + = lim (s + ) = (s + ) = s = = s =.. Subsequeces (a) a k =. Here k = k. (b ), (c ) ad (d ) are already mootoe. For the last, observe that d = is decreasig (7 ) = (7 ) (b) (a ) has subsequetial limit set {, }. (b ) has subsequetial limit set {0}. (c ) has subsequetial limit set {+ }. (d ) has subsequetial limit set { 6 7 }.
6 (c) lim sup a =, lim if a =. lim sup b = 0 = lim if b. lim sup c = + = lim if c. lim sup d = 6 7 = lim if a. (d) (b ) ad (d ) coverge, (c ) diverges to +, ad (a ) diverges by oscillatio. (e) All but (c ) are bouded. 6 Here are two possible argumets. Use the defiitio of the subsequece (s k ) as a fuctio t = s σ, where σ(k) = k is a icreasig fuctio σ : N N. Thus a subsequece of (s k ) is a fuctio u = t µ, where µ(l) = k l is a icreasig fuctio µ : N N. But the, by the associativity of fuctioal compositio, we have u = (s σ) µ = s (σ µ), which defies (s kl ) as a subsequece of (s ). It should be clear that σ µ : N N is icreasig. View a subsequece as a ordered subset of a sequece usig the orderig s m s = m. If (s kl ) is a subsequece of (s k ), which is a subsequece of (s ), the we have the iclusio (s kl ) (s k ) (s ) as ordered sets. Sice the subset relatio preserves orderig ad set iclusio is trasitive, it is immediate that (s kl ) is a subsequece of (s ). 7 Let (r ) be a eumeratio of the set Q of all ratioal umbers. As observed i lectures, every real umber ad {+, } are subsequetial limits if (r ). Thus lim sup r = + whece, (Corollary.) there exists a mootoe subsequece (r k ) such that lim k r k = lim sup r = +. 9 (a) We must prove that the set of all limits of all sequeces takig values i [a, b] is itself [a, b]. Firstly, trivially ote that every t [a, b] is the limit of the sequece (t, t, t, t, t,...). Now let a s b. The ay subsequece of s with a limit s must satisfy a s b. Hece s [a, b], ad so [a, b] is closed. (b) First a cheatig aswer: Ay set of subsequetial limits must be closed. (0, ) is ope, hece there is o such sequece. We eed to do better tha this, for we have ot yet got a defiitio of ope, ad so caot categorize it as somehow (at least for bouded sets), a opposite cocept to closed. Suppose there is a sequece (s ) such that S = (0, ) is its set of subsequetial limits. The (Corollary. from book) there exists a mootoe subsequece whose limit is lim sup s. However (Theorem.7) we kow that lim sup s = sup S =. Thus is the limit of some subsequece of (s ). Cotradictio. 0 (a) Sice each term is repeated ifiitely ofte, the set S of subsequetial limits of (s ) is precisely { : N}. (b) lim sup s = ad lim if s = 0 as these are the supremum ad ifimum of the above set S. 6
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