ε > 0 N N n N a n < ε. Now notice that a n = a n.

Transcription

1 4 Sequees.5. Null sequees..5.. Defiitio. A ull sequee is a sequee (a ) N that overges to 0. Hee, by defiitio of (a ) N overges to 0, a sequee (a ) N is a ull sequee if ad oly if ( ) ε > 0 N N N a < ε..5.. Observatio. Let (a ) N be a sequee. (i) For every real umber r, (a ) N overges to r if ad oly if (a r) N is a ull sequee. This is true beause the defiitio of (a ) N overges to r is literally the same as oditio ( ) above, formulated for the sequee (a r) N. (ii) The sequee (a ) N is a ull sequee if ad oly if ( a ) N is a ull sequee, beause oditio ( ) for the sequee ( a ) N reads as Now otie that a = a. ε > 0 N N N a < ε. Null sequees are of partiular iterest for the followig reaso: Suppose we are give a sequee (a ) N ad we have a guess that it is overget with a speifi it r; the by.5.(i) we may work with the sequee (a r) N ad try to show that this is a ull sequee. I geeral this is easier to verify tha a = r Properties of ull sequees. Let (a ) N be a ull sequee. (i) If R the ( a ) N is a ull sequee. (ii) If (b ) N is aother ull sequee, the (a + b ) N is a ull sequee. (iii) (Sadwih rule for ull sequees) If (b ) N is a sequee with b a for all but fiitely may N, the (b ) N is a ull sequee. (iv) If (b ) N is a bouded sequees, the (a b ) N is a ull sequee. (v) If a 0 for all, ad p > 0 is a real umber, the (a p ) N is a ull sequee. Proof. (i) holds true by the salar multipliatio rule.3. ad (ii) holds true by the sum rule.3. (iii). Let ε > 0. We eed to fid some N N suh that b < ε wheever N. As (a ) N is a ull sequee by assumptio, there is some N N with a < ε for all N. Sie b a for all but fiitely may N, there is some N N with b a for all N. Hee if we take N to be greater or equal tha N ad N the both oditios apply: For eah N we have b as N N as N N a < ε. (iv). Sie (b ) N is bouded there is some B R with b B. By (i), (B a ) N is also a ull sequee. Now a b = a b a B = B a for all N. Thus by (iii) applied to (a b ) N ad (B a ) N, (a b ) N is a ull sequee.

2 a p < a k < (ε k ) k = ε for all N. Null sequees 5 (v). Let k N be suh that k < p. Let ε > 0 ad w.l.o.g. assume ε <. Sie ε k > 0, there is some N N with a < ε k for all N. As ε <, also a < for N ad so k < p implies ap < a k. We olude that.5.4. The stadard list of ull sequees. () = 0 for every p R with p > 0. p Proof. By..3() we kow that.5.3(v). = 0, hee () holds true by () = 0 for every R with >. Proof. By.5.(ii) we may also assume that 0, hee we may assume >. The ( ) N is stritly dereasig ad by.4.6, ( ) N overges to some r R. We apply the salar multipliatio rule ad obtai r = = = As > this meas r = 0. see questio 5 = = r. (3) p = 0 for all p R ad every R with >. Proof. Agai we may assume that >. Let k N be with p < k. Sie p < k (see the defiitio of p ) we may apply the sadwih rule for ull sequees ad replae p by k, hee we may assume that p N. Claim. The sequee ( ) N is bouded for all >. To see this, write = + x with x > 0 (so x = ). By the Biomial Theorem, = ( + x) = + ( ) x + ( ) x ( ) x ( ) x = ( ) x. Hee if x, the, i other words. Now x is equivalet to + x. This shows that the terms are bouded above by if + x. From this we learly get laim. Claim. = 0 for all >. To see this, apply laim to (whih is also > ). Thus ( ) N is bouded. Sie ( ) N is a ull sequee (by () applied to ) we see that = is a produt of a bouded ad a ull sequee. By.5.3(iv), this shows laim. Now we a prove (3) as follows: Reall from the begiig of the proof that we work with p N. We apply laim for p (whih also is > ). Thus = 0. By.5.3(v) the also p p = ( ) p = 0 p

3 6 Sequees (4)! = 0 for all R. Proof. Agai we may assume that > 0. We have! = We fix a atural umber K. For N with > K we the have! = K K! ( K)-times {}}{... (K + )... K K!. Sie ( ) N is a ull sequee ad K K! is a ostat, we see that K K! 0 as. By the sadwih rule for ull sequees, we get (4). Please read the assertios of.5.4 arefully. For example, item () says that for every d R with d <, the sequee (d ) N is a ull sequee. Now give a example of a sequee (a ) N that is ot a ull sequee with the property a < for all N.

4 The algebra of its 7.6. The algebra of its..6.. Sadwih rule for sequees. Let (a ) N, (b ) N ad ( ) N be sequees with the followig properties. (a) Both (a ) N ad ( ) N overge to the same real umber r. (b) a b for all N. The also (b ) N overges to r. Proof. We show that (b a ) N is a ull sequee. The by the sum rule, b exists ad is equal to (a + b a ) = a + (b a ) = r + 0 = r, as desired. By (b) we have 0 b a a ad by the sadwih rule for ull sequees.5.3(iii) it suffies to show that ( a ) N is a ull sequee. However, by the salar multipliatio rule, ( a ) N overges with ( a ) = r ad so by the sum rule agai ( a ) = ( + ( a )) = + ( a ) = r r = 0, as desired..6.. Compatibility of its ad order. Let (a ) N ad (b ) N be overget sequees. (i) If a b for all but fiitely may, the a b. (ii) If a < b, the for all but fiitely may we have a < b. Proof. Let r = a ad s = b. We first show (ii). Let ε = s r all N, a r < ε, i partiular a < r + ε. There is some N N suh that for all N, b s < ε, i partiular s ε < b. Thus for all N := max{n, N } we have a < r + ε = s ε < b. Therefore, for all exept possibly for =,..., N we have a < b. (i). Suppose by way of otraditio that a b for all but fiitely may ad r > s. By (ii) with iterhaged role of (a ) N ad (b ) N we kow that for all, hee r + ε = r+s = s ε. There is some N N suh that for but fiitely may, a > b. But this otradits the assumptio a b for all but fiitely may.

5 8 Sequees.6.3. The algebra of its. Let (a ) N ad (b ) N be overget sequees. (i) (Sum rule) The sum (a + b ) N of the sequees (a ) N ad (b ) N is overget ad (a + b ) = a + b. (ii) (Mulipliatio rule) (a b ) N (alled the produt of the sequees (a ) N ad (b ) N ) is overget ad (a b ) = ( a ) ( b ). Observe that the salar multipliatio rule is a speial istae of the multipliatio rule: If R, the the salar multipliatio rule is the multipliatio rule applied to the ostat sequee () N. (iii) (Divisio rule) If b 0 for all N, ad (b ) N is ot a ull sequee, the also ( a b ) N is overget ad (iv) (Modulus rule) ( a ) N is overget ad (a ) = a. b b a = a. (v) (Root rule) If a 0 for all N ad p N the also ( p a ) N is overget ad p a = p a. Proof. (i) is the sum rule.3. ad is stated here agai for the sake of ompleteess. For the rest of the proof we write r = a ad s = b (ii). We must show that (a b ) N overges to r s ad it suffies to show that (a b r s) N is a ull sequee. Our strategy here is the followig: We shall estimate the modulus of the terms a b r s by terms of aother ull sequee ad the apply the sadwih rule for ull sequees.5.3(iii). We start with the estimatio: ( ) a b r s = (a b a s) + (a s r s) a b a s + a s r s = = a b s + a r s. Now we wat to see that the sequee with th term a b s + a r s is a ull sequee: (a) Sie a r ( ), ( a r ) N is a ull sequee ad by the salar multipliatio rule also ( a r s) N is a ull sequee.

6 The algebra of its 9 (b) Sie b s ( ), ( b s ) N is a ull sequee. Sie (a ) N is overget, it is bouded (f..4.). Now by.5.3(iv), the produt of a ull sequee ad a bouded sequee is itself a ull sequee. Thus ( a b s ) N is a ull sequee Applyig the sum rule to (a) ad (b) shows that the sequee ( a b s + a r s ) N is a ull sequee. We may therefore apply the sadwih rule for ull sequees.5.3(iii) i ombiatio with our omputatio ( ) to obtai that also ( a b r s ) N is a ull sequee. This is what we wated to show. (iii). So here s = b 0 by assumptio. We show that ( b s ) N is a ull sequee, i other words ( b ) N is overget ad ( b ) = b ; item (iii) the follows from the multipliatio rule applied to (a ) N ad ( b ) N. By the defiitio of overgee of (b ) N ad the hoie ε = s, there is some K N suh that b s < s for all K. We ow apply the derived triagle rule..(iv) ad get for K: b = s + (b s) derived triagle rule I partiular if K, the b s ad s b s as b s < s b s = s b ( ) b s = s b s s b b s s = s s b. Sie (b ) N overges to s, s b 0 ( ). By the salar multipliatio rule (or by (ii)), also s s b 0 ( ). So by the sadwih rule for ull sequees.5.3(iii) applied to ( ), also ( b s ) N is a ull sequee. (iv). We must show that a = r. By the derived triagle iequality we kow for eah N: a r a r. Sie the right had side here teds to 0 as we a apply the sadwih rule for ull sequees.5.3(iii) ad get that a r 0 as. This is what we had to show. (v). If (a ) N is a ull sequee, the we a apply.5.3(v) ad see that also ( p a ) N is a ull sequee, whih ofirms (v). Therefore we may assume that (a ) N is ot a ull sequee, i other words For eah N we have r = a 0. (+) ( p a p r) ( p a p + p a p p r p a p r p + p r p ) = a r. Sie a, r 0 we a estimate (++) p r p p a p + p a p p r p a p r p + p r p. By ombiig (+) ad (++) we obtai p a p r p r p a r for all N. s.

7 0 Sequees Sie r 0 this a be writte as p a p r p r p a r for all N. Hee by the sadwih rules for ull sequees.5.3(iii) applied to ( p a p r) N ad ( p r p (a r)) N, also ( p a p r) N is a ull sequee. This shows that ( p a ) N overges to p r Example. Let a = We wat to deide whether (a 3 ) N is overget ad if it is overget to fid its it. We aot apply the divisio rule diretly, beause either the sequee ( 3 ) N or the sequee ( ) N is overget. The idea is to apply the followig strategy: We idetify the domiat term i the expressio , i.e. the term that grows fastest. So here this is 3. We divide umerator ad deomiator by the domiat term. So here this gives Now we a apply the algebra of its together with the stadard list of ull a = = ( 3 ) 3 ( ) = 3 sequees to see that the ew umerator 3 overges to (ovie yourself that this is true) ad the ew deomiator overges to. Hee by the divisio rule, the quotiet (whih is our origial a ) overges to =. Let us do aother example of this type. Let a = 3! !. The domiat term is! (this a be see from the stadard list of ull sequees, see.5.4 (4), (3)). Therefore 3! + 4 a = 5 + 6! = 3 + 4! 5! + 6 From the stadard list of ull sequees, we kow that 4! ad! overge to 0. Hee by the algebra of its, a overges to 3 6 =. Be aware that the reipe of.6.4 is ot always appliable i a obvious way. For example oe a use this strategy to ompute the it of the sequee ( + ) N by writig ( + ) ( + + ) + = + + ad the applyig the reipe of.6.4. This will be arried out i the examples lasses. O the other had the reipe of.6.4 is ot always suessful, for example look at the sequees (! ) N. The it is ot lear what the domiat term is. It is ot diffiult to see that (! ) N is a ull sequee (this is treated i the examples lasses). Or, look at the sequee ( ) N where > 0. We laim that ( ) N overges to. 3

8 The algebra of its To see this, we first assume that. the ( ) N is bouded below by ad bouded above by. Moreover ( ) N is dereasig ad so by the mootoe overgee theorem, it overges to some r R. By the ompatibility of its ad order (f..6.), r. We must show that r =. We have r = by the root rule = = Now for every overget sequee (a ) N, also the sequee (a ) N is overget ad a = a. Applyig this to our sequee ( ) N gives = = r. So r = r ad r, whih implies r =. This shows the laim if. If <, the > ad from the previous ase we get =. Now we apply the divisio rule to obtai = = = = =. Other sequees eve eed more hard work. For example look at the sequee ( ) N. Is it overget? We takle this questio ow. First a lemma Lemma. For all k 0 we have ( ) k k k! with equality if ad oly if k = 0 or k =. Proof. If k = 0, the ( ) k = = 0 k!. If k =, the ( ) k = = k k!. So it remais to show that ( ) k < k k! for k >. We have ( )! = k k! ( k)! = k!! ( k)! = k! ( )... ( k+) < k-times k! {}}{..., sie the last fator k + is < for k >. Thus ( k We a ow show that.6.6. ( ) N is overget with =. Proof. For all 3 we have by.6.5: ( ) ( + ) = + < k k= < + k k = + = + +, k= k= ) < k k!. k a strit iequality ours whe k = thus ( + ) +, i other words + + for all 3.

9 Sequees By the mootoe overgee theorem ( ) N overges to some r ad we have to show that r =. We have r = = = ( ) = ( ) = = ( ) ( ). Now = (see the omputatio before.6.5) ad by the root rule of.6.3, =. We obtai r = = r. Hee r = r ad r = as desired.

10 The Bolzao-Weierstraß Theorem ad Cauhy sequees 3.7. The Bolzao-Weierstraß Theorem ad Cauhy sequees. This setio osist of two further fudametal statemets (.7. ad.7.) at the begiig of real aalysis. They are ot examiable ad iluded for the iterested reader. Surprisigly the followig is true..7.. Theorem. (Bolzao-Weierstraß) Every bouded sequee has a overget subsequee. Proof. Let (a ) N be our bouded sequee, thus for some B R, B > 0, every a satisfies B a B. Defie for k N: A k := {a k} s k := sup(a k ). Observe that A is the value set of (a ) N. Sie A k A for all k ad A is bouded by assumptio, also A k is bouded. So by the ompleteess axiom for R (.4.3), A k has ideed a supremum i R, whih justifies the defiitio of s k. As A k [ B, B], also s k [ B, B]. Sie A A A 3... we have s s... Hee (s k ) k N is a mootoe ad bouded sequee. So by.4.6, (s k ) k N overges to some real umber r. We ow defie a subsequee (a i ) i N of (a ) N by defiig < <... as follows: Choie of. Sie s k r (k ) there is some K N suh that r s k for all k K. Sie s K is the supremum of A K, there is some {K, K +,...} with s K < a s K. I partiular a s K. We a ow estimate the distae of a to r by a r = a s K + s K r a s K + s K r + =. Choie of i, provided we have already hose i. Sie s k r (k ) there is some K N suh that r s k i for all k K. We may ertaily hoose K > i if eessary. Sie s K is the supremum of A K, there is some i {K, K +,...} with s K i < a i s K. I partiular s K a i i. Agai we estimate the distae of a i to r by a i r = a i s K + s K r a i s K + s K r i + i = i. Thus we have defied < < 3 <... ad we kow that for all i N, a i r i. By the sadwih rule for ull sequees.5.3(iii), this meas that a i r 0 (i ), i.e. (a i ) i N overges to r..7.. Defiitio. (Augusti-Louis Cauhy, ) A sequee (a ) N is alled a Cauhy sequee (prooued like o she ), if for every ε > 0 there is some N N suh that for all, k N we have a a k < ε.

11 4 Sequees Hee i a Cauhy sequee, the terms get loser ad loser to eah other. However, oe has to uderstad the expressio get loser ad loser to eah other i the sese of the defiitio above, beause there are sequees (a ) N with a + a 0 as (so here also i some sese the terms get loser ad loser to eah other), suh that (a ) N is ot Cauhy. We will see a example of this type later, whe we are dealig with series. For the impatiet reader: hoose a = Every overget sequee (a ) N (with it r) is Cauhy: Let ε > 0 ad hoose N N with a r < ε for all N. The for all, k N we have a a k = a r + r a k a r + r a k = a r + a k r < ε + ε = ε Theorem. Every Cauhy sequee overges. Proof. Let (a ) N be our Cauhy sequee. Our strategy i the proof is the followig: We first show that (a ) N is bouded. By Bolzao-Weierstraß (.7.) we the kow that (a ) N possesses a overget subsequee (a i ) i N. Let R be the it of that subsequee. Fially we will show that (a ) N overges to r. Applyig the defiitio of a Cauhy sequee with ε = says that for some N N we have a a m < ε wheever, m N. If we set m = N we see that all elemets a with N are otaied i the iterval (a N ε, a N + ε). It follows that (a ) N is bouded (also see questio 7). By Bolzao-Weierstraß (.7.) we ow have a overget subsequee (a i ) i N, whih has a it r. We show that (a ) N overges to r. Let ε > 0. Sie i a i = r there is some I N suh that for all i I we have a i r < ε. Sie (a ) N is a Cauhy sequee there is some N N suh that for all, m N we have a a m < ε. Fix N with N. Take i N with i I ad i N. The ( ) a r = a a i + a i r a a i + a i r. Sie i, N we have a a i < ε. Sie i I we have a i r < ε. Hee from ( ) we get a r < ε + ε = ε. Hee a sequee overges if ad oly if it is Cauhy. The advatage of the otio of a Cauhy-sequee is that we do ot eed to kow the it of the sequee if we wat to ofirm overgee. We will see i the ext setio that some its are very hard to ompute, whereas it is ofte easy to see why a partiular sequee is overget.