4.4 Monotone Sequences and Cauchy Sequences

Transcription

1 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES Mootoe Sequeces ad Cauchy Sequeces 4.4. Mootoe Sequeces The techiques we have studied so far require we kow the limit of a sequece i order to prove the sequece coverges. However, it is ot always possible to fid the limit of a sequece by usig the defiitio, or the limit rules. This happes whe the formula defiig the sequece is too complex to work with. It also happes with sequeces defied recursively. Furthermore, it is ofte the case that it is more importat to kow if a sequece coverges tha what it coverges to. I this sectio, we look at two ways to prove a sequece coverges without kowig its limit. We begi by lookig at sequeces which are mootoe ad bouded. These terms were defied at the begiig of this chapter. You will recall that i order to show that a sequece is icreasig, several methods ca be used.. Direct approach, simply show that a + a for every.. Equivaletly, show that a + a 0 for every. 3. Equivaletly, show that a + positive. a for every if both a ad a + are 4. If a = f ), oe ca show a sequece a ) is icreasig by showig that f is icreasig that is by showig that f x) By iductio. We ow state ad prove a importat theorem about the covergece of icreasig sequeces. Theorem 344 A icreasig sequece a ) which is bouded above coverges. Furthermore, lim a = sup {a }. Proof. We eed to show that give ɛ > 0, there exists N such that N = a A < ɛ where A is the limit. Let ɛ > 0 be give. Sice a ) is bouded above, by theorem 58, {a } has a supremum. Let A = sup {a }. Let ɛ > 0 be give. the, A ɛ < A. By theorem 5, there exists a elemet of {a }, call it a N, such that A ɛ < a N A. Sice a ) is icreasig, we have a a N for every N. Therefore, if N, A ɛ < a A ɛ < a A 0 0 A a < ɛ = a A < ɛ

2 3 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Corollary 345 A decreasig sequece a ) which is bouded below coverges. Furthermore, lim a = if {a }. Proof. The proof is similar to the proof of the theorem. Theorem 346 A mootoe sequece coverges if ad oly if it is bouded. Proof. The proof follows from results prove i the previous sectio ad the previous theorem ad its corollary. Suppose we have a mootoe sequece x ). If we assume that x ) coverges, the it follows that it is bouded by theorem 3 that it is bouded. Coversely, if we assume that x ) is bouded the it is both bouded above ad below. Sice x ) is mootoe, the it is either icreasig or decreasig. If it is icreasig, it will coverge by the previous theorem sice it is also bouded above. If it is decreasig, the it will coverge by the corollary sice it is bouded below. Example 347 Prove that the sequece whose geeral term is a = k=0 k! coverges. We try to establish this result by showig that this sequece is o-decreasig ad bouded above. a ) is icreasig: a = ! !. Therefore, a + a = + )! > 0 a ) is bouded above. For this, we use the fact that! for every. The proof of this fact is left as a exercise. Therefore, ! ! < ) 0 ) < ) < + < + < 3 ) ) Sice a ) is bouded above ad icreasig, it must coverge. We will call the limit e.

3 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 33 Example 348 Fid lim a where a ) is defied by: a = a + = a + 6) If we kew the limit existed, fidig it would be easy. We must first establish that it exists. We do this by showig that this sequece is icreasig ad bouded above. This part is left as a exercise. Oce this fact has bee established, the we kow the sequece must coverge. Let L be its limit, we must fid L. Before proceedig, we will recall the followig fact: lim a + = lim a. Therefore, if for every we have a + = a + 6), the we must also have: lim a + = lim a + 6) L = lim a + 6) L = L + 6) L = L + 6 L = 6 Example 349 Let a = + ). Show lim a exists. This limit is i fact the umber e, but we wo t show that. Agai, to show that a ) coverges, we show that it is icreasig ad bouded above. a ) bouded above. To establish this, we use the followig fact: If k is a iteger such that < k, the ) )... k + ) k = ) )... k ) The proof of this is left as a exercise. Usig the biomial theorem, we have: a = + ) ) k ) ) )... k + ) !! k!! = + + ) ) )... k ) ! k!! < + +! ! < e where e is the limit foud i the exercise above. Thus, a ) is bouded by e. ) ) )...

4 34 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES a ) icreasig. The kth term i the expasio of a is ) k! The kth term i the expasio of a + is ) )... k! + + j Sice + < j, it follows that j So, a ) is icreasig. + > j. Hece, a + > a. Sice a ) is icreasig ad bouded above, it follows that a ) coverges. We defie this limit to be the umber e Cauchy Sequeces Defiitio 350 Cauchy Sequece) A sequece x ) is said to be a Cauchy sequece if for each ɛ > 0 there exists a positive iteger N such that m, N = x m x < ɛ. We begi with some remarks. Remark 35 These series are amed after the Frech mathematicia Augusti Louis Cauchy ). Remark 35 It is importat to ote that the iequality x m x < ɛ must be valid for all itegers m, that satisfy m, N. I particular, a sequece x ) satisfyig x + x < ɛ for all N may ot be a Cauchy sequece. Remark 353 A Cauchy sequece is a sequece for which the terms are evetually close to each other. Remark 354 I theorem 33, we proved that if a sequece coverged the it had to be a Cauchy sequece. I fact, as the ext theorem will show, there is a stroger result for sequeces of real umbers. We ow look at some examples. Example 355 Cosider x ) where x =. Prove that this is a Cauchy sequece. Let ɛ > 0 be give. We wat to show that there exists a iteger N > 0 such that m, < N = x m x < ɛ. That it we would like to have x x m < ɛ m < ɛ Sice m < + m If we make + m < ɛ, the result will follow. This will happe if both < ɛ that is whe > ɛ ad m < ɛ that is whe >. So, we see that if N is a ɛ iteger larger tha ɛ the m, > N = x m x < ɛ. )... k + k ). ).

5 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 35 Example 356 Cosider x ) where x = k k= Let ɛ > 0 be give. We wat to show that there exists a iteger N > 0 such that m, < N = x m x < ɛ without loss of geerality, let us assume that > m). That it we would like to have m x x m < ɛ k k < ɛ k= k= < ɛ k k=m+ Rememberig telescopig sums ad the fact that x x m = = < k k=m+ k k=m+ k=m+ = m k <, we see that k k ) k k ) < m + So, as we saw i the previous example, if N is a iteger larger tha ɛ, the m, < N = x m x < ɛ. We ow look at importat properties of Cauchy sequeces. Theorem 357 Every Cauchy sequece is bouded. Proof. See problems. Theorem 358 A sequece of real umbers coverges if ad oly if it is a Cauchy sequece. Proof. We have already prove oe directio. Let x ) be a sequece of real umbers. If x ) coverges, the we kow it is a Cauchy sequece by theorem 33. Assume x ) is a Cauchy sequece. We must prove that it coverges. By theorem 357, we kow that {x } is bouded. Therefore, by the completeess axiom, for each, the umber a = if {x, x +, x +,...}

6 36 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES exists. Furthermore, the sequece a ) is icreasig ad bouded why?). Therefore, by theorem 346, a ) coverges. Let a = lim a. We prove that lim x = a. Let ɛ > 0 be give. Sice x ) is a Cauchy sequece, we ca fid N > 0 such that m, N = x x m < ɛ. It follows that [ both x ) ad a ) are cotaied i the iterval x N ɛ, x N + ɛ ] why). Thus, the limit a is also i this iterval. Therefore, for all N, we have: x a x x N + x N a < ɛ This proves that lim x = a, thus x ) coverges. Remark 359 The key to this theorem is that we are dealig with a sequece of real umbers. The fact that a Cauchy sequece of real umber coverges is liked to the fact that R is complete. I fact, this is sometimes used as a defiitio of completeess. Some texts say that a set is complete if every Cauchy sequece coverges i that set. It is possible to fid a Cauchy sequece of ratioal umbers which does ot coverge i Q. We fiish this sectio with a importat theorem. Defiitio 360 A ested sequece of itervals is a sequece {I } of itervals with the property that I + I for all. Theorem 36 Nested Itervals) If {[a, b ]} is a ested sequece of closed itervals the there exists a poit z that belogs to all the itervals. Furthermore, if lim a = lim b the the poit z is uique. Proof. We provide a sketch of the proof ad leave the details as homework. First, prove that {a } ad {b } are mootoe ad bouded. Thus, they must coverge. Let a = lim a ad b = lim b. The, explai why a b. Coclude from this Exercises. Prove that! for every. Prove that if k is a iteger such that < k, the ) )... k + ) k = ) 3. Prove that the sequece give by a = a + = a + 6) )... k ) is icreasig ad bouded above by 6. hit: use iductio for both).

7 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES Show that the sequece defied by a = a + = 3 a is icreasig ad satisfies a < 3 for all. The, fid its limit. 5. Let a ad b be two positive umbers such that a > b. Let a be their arithmetic mea, that is a = a + b. Let b be their geometric mea, that is b = ab. Defie a + = a + b ad b + = a b. a) Use mathematical iductio to show that a > a + > b + > b. b) Deduce that both a ) ad b ) coverge. c) Show that lim a = lim b. Gauss called the commo value of these limits the arithmetic-geometric mea. 6. Prove theorem Aswer the why? parts i the proof of theorem Fiish provig theorem 36.

8 38 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES 4.5 Subsequeces Discuss subsequeces, the Bolzao-Weierstrass theorem, limit sup ad if Exercises