Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem

Transcription

1 Lecture 4: Cauchy sequeces, Bolzao-Weierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits of sequeces coverge. Defiitio We say that a sequece of real umbers {a } is a Cauchy sequece provided that for every ɛ > 0, there is a atural umber N so that whe, m N, we have that a a m ɛ. Example Let x be a real umber ad t (x) be the th trucatio of its decimal expasio as i Lectures 2 ad 3. The if, m N, we have that t (x) t m (x) 0 N, sice they share at least the first N places of their decimal expasio. Give ay real umber ɛ > 0, there is a N(ɛ) so that 0 N(ɛ) < ɛ. Thus we have show that the sequece {t (x)} is a Cauchy sequece. Example was cetral i our costructio of the real umbers. We got the least upper boud property by associatig to each sequece as i Example, the real umber x which is its limit. The class of Cauchy sequeces should be viewed as mior geeralizatio of Example as the proof of the followig theorem will idicate. Theorem Every Cauchy sequece of real umbers coverges to a limit. Proof of Theorem Let {a } be a Cauchy sequece. For ay j, there is a atural umber N j so that wheever, m N j, we have that a a m 2 j. We ow cosider the sequece {b j } give by b j = a Nj 2 j. Notice that for every larger tha N j, we have that a > b j. Thus each b j serves as a lower boud for elemets of the Cauchy sequece {a } occurig later tha N j. Each elemet of the sequece {b j } is bouded above by b +, for the same reaso. Thus the sequece {b j } has a least upper boud which we deote by L. We will show that L is the limit of the sequece {a }. Suppose that > N j. The a L < 2 j + a b j = 2 j + a b j 3(2 j ). For every ɛ > 0 there is j(ɛ) so that 2 j < ɛ ad we simply take N(ɛ) to N j(ɛ). The idea of the proof of Theorem is that we recover the limit of the Cauchy sequece by takig a related least upper boud. So we ca thik of the process of fidig the limit of the Cauchy sequece as specifyig the decimal expasio of the limit, oe digit at a time, as this how the least upper boud property worked. The coverse of Theorem is also true. Theorem 2 Let {a } be a sequece of real umbers covergig to a liit L. The the sequece {a } is a Cauchy sequece.

2 Proof of Theorem 2 Sice {a } coverges to L, for every ɛ > 0, there is a N > 0 so that whe j > N, we have a j L ɛ 2. (The reaso we ca get ɛ 2 o the right had side is that we put ɛ 2 i the role of ɛ i the defiitio of the limit.) Now if j ad k are both more tha N, we have a j L ɛ 2 ad a k L ɛ 2. Combiig these usig the triagle iequality, we get a j a k ɛ, so that the sequece {a j } is a Cauchy sequece as desired. Combiig Theorems ad 2, we see that what we have leared is that Cauchy sequeces of real umbers ad coverget sequeces of real umbers are the same thig. But the advatage of the Cauchy criterio is that to check whether a sequece is Cauchy, we do t eed to kow the limit i advace. Example 2 Cosider the series (that is, ifiite sum) S = = 2. We may view this as the limit of the sequece of partial sums a j = j = We ca show that the limit coverges usig Theorem by showig that {a j } is a Cauchy sequece. Observe that if j, k > N, we defiitely have a j a k It may be difficult to get a exact expressio for the sum o the right, but it is easy to get a upper boud. 2 ( ) =. The reaso we used the slightly wasteful iequality, replacig by 2 2 is that ow the sum o the right telescopes, ad we kow it is exactly equal to N. To sum up, we have show that whe j, k > N, we have a j a k N. 2

3 Sice we ca make the right had side arbitrarily small by takig N sufficietly large, we see that {a j } is a Cauchy sequece. This example gives a idicatio of the power of the Cauchy criterio. You would ot have foud it easier to prove that the limit exists if I had told you i advace that the series coverges to π2 6. Let {a } be a sequece of real umbers. Let { k } be a strictly icreasig sequece of atural umbers. We say that {a k } is a subsequece of {a }. We will ow prove a importat result which helps us discover coverget sequeces i the wild. Theorem 3(Bolzao-Weierstrass) Let {a } be a bouded sequece of real umbers. (That is, suppose there is a positive real umber B, so that a j B for all j.) The {a } has a coverget subsequece. Proof of Bolzao-Weierstrass All the terms of the sequece live i the iterval I 0 = [ B, B]. We cut I 0 ito two equal halves( which are [ B, 0] ad [0, B]). At least oe of these cotais a ifiite umber of terms of the sequece. We choose a half which cotais ifiitely may terms ad we call it I. Next, we cut I ito two halves ad choose oe cotaiig ifiitely may terms, callig it I 2. We keep goig. (At the jth step, we have I j cotaiig ifiitely may terms ad we fid a half, I j+ which also cotais ifiitely may terms.) We defie the subsequece {a jk } by lettig a jk be the first term of the sequece which follows a j,..., a jk ad which is a elemet of I j. We claim that {a jk } is a Cauchy sequece. Let s pick k, l > N. The both a jk ad a jl lie i the iterval I N which has legth B. Thus 2 N a jk a jl B 2 N. We ca make the right had side arbitrarily small by makig N sufficietly large. Thus we have show that the subsequece is a Cauchy sequece ad hece coverget. A questio you might ask yourselves is: How is the proof of the Bolzao Weierstrass theorem related to decimal expasios? Our fial topic for today s lecture is the Squeeze theorem. It is a result that allows us to show that limits coverge by comparig them to limits that we already kow coverge. Theorem 4(Squeeze theorem) Give three sequeces of real umbers {a }, {b }, ad {c }. If we kow that {a } ad {b } both coverge to the same limit L ad we kow that for each we have a c b, the the sequece {c } also coverges to the limit L. 3

4 Proof of Squeeze theorem Fix ɛ > 0. There is N > 0 so that whe > N, we have a L ɛ. There is N 2 > 0 so that whe > N 2, we have b L ɛ. We pick N to be the larger of N ad N 2. For > N, the two iequalities above, we kow that a, b (L ɛ, L + ɛ). But by the iequality a c b, we kow that c [a, b ]. Combiig the two facts, we see that c (L ɛ, L + ɛ), so that c L ɛ. Thus the sequece {c } coverges to L as desired. Example 3 Calculate lim ( + + ). The limit above seems a little complicated so we ivoke the squeeze theorem. We observe that the iside of the paretheses is betwee ad 2. (Actually it is gettig very close to 2 as gets large. Thus ( + + ) 2. Thus we will kow that provided we ca figure out that ad lim ( + + ) =, lim =. lim 2 =. The first limit is easy sice every term of the sequece is. It seems to us that the th roots of two are gettig closer to, but how do we prove it. Agai, it seems like a job for the squeeze theorem. Observe that ( + ) 2, 4

5 sice + are the first two terms i the biomial expasio. Thus We kow that ad perhaps we also kow that sice 2 +. lim =, lim + =, becomes arbitrarily small as gets large. Thus by the squeeze theorem, we kow lim 2 =, ad hece lim ( + + ) =. Example 3 is a reasoable illustratio of how the squeeze theorem is always used. We might begi with a very complicated limit, but as log as we kow the size of the terms cocered, we ca compare, usig iequalities to a much simpler limit. As of yet, we have ot said aythig about ifiite limits. We say that a sequece {a } of positive real umbers coverges to ifiity if for every M > 0, there is a N so that whe > N, we have a > M. Here M takes the role of ɛ. It is measurig how close the sequece gets to ifiity. There is a versio of the squeeze theorem we ca use to show limits go to ifiity. Theorem 5(ifiite squeeze theorem) Let {a } be a sequece of positive real umbers goig to ifiity. Suppose for every, we have b a. The the sequece {b } coverges to ifiity. Proof of the ifiite squeeze theorem For every M, there exists N so that whe > N, we have a > M. But sice b a, it is also true that b > M. Thus {b } goes to ifiity. Example 4 = =. 5

6 We will prove this by comparig each reciprocal to the largest power of two smaller tha it. Thus > Combiig like terms, we get > O the right had side, we are summig a ifiite umber of 2 s. Thus the sum is ifiite. Somethig to thik about: Ofte oe shows that the harmoic series diverges by comparig it to the itegral of x which is a logarithm. Are there ay logarithms hidig i Example 4? 6