3. Sequences. 3.1 Basic definitions

Transcription

1 3. Sequeces 3.1 Basic defiitios Defiitio 3.1 A (ifiite) sequece is a fuctio from the aturals to the real umbers. That is, it is a assigmet of a real umber to every atural umber. Commet 3.1 This is the first time we meet the otio of a fuctio, which will be the cetral cocept of the ext chapter. As for ow, we take this (very otrivial) otio as evidet. Notatio: Sequeces, like ay other fuctios, are labeled by letters. We may refer, for example, to the sequece a. The value that a returs for, say, the iput 3 is deoted by a 3, rather tha a(3). More geerally, we deote by a the value that the fuctio a returs for the iput. The subscript i a is called the idex (28$1*!) of that elemet. But sequeces are very special fuctios as their domai of defiitio (%9$#%.&(;) is a iductive set. Thus, we ca refer to the first elemet ad to a successor or a predecessor of a certai elemet. A more commo otatio for the sequece a is as follows, (a ) =1. Note, however, that the idex i this otatio is a dummy variable (892 %1;:/). We could have as well writte (a k ) k=1. Whe there is o risk of cofusio, we will deote the sequece simply by (a ) (which should ot be cofused with its -th elemet a ). Sequeces ca be defied i various ways. The most commo way of defiig a fuctio is by providig a formula for the -th elemet of that sequece (i.e., a rule for calculatig a give ). Aother way of defiig a fuctio is based o the

2 50 Sequeces iductive property of N: the first elemet is specified alog with the formula for calculatig a +1 give a (or, more geerally, give a 1,a 2,...,a ). Such a defiitio is called recursive. Examples 3.1 (a) The costat sequece: a = 5. (b) The sequece of aturals: a =. (c) A alteratig sequece: b = ( 1). (d) The harmoic sequece: c = 1. (e) The sequece of primes (d ) = (2,3,5,7,11,...). Note that we do ot have a explicit formula for d. (f) The sequece of digits of p: (e ) = (3,1,4,1,5,9,...). (g) The Fiboacci sequece: f 1 = 1, f 2 = 1, f +1 = f + f 1. Commet 3.2 It is very commo to refer, say, to the harmoic sequece as the sequece 1". While very ituitive, this way of referece is problematic. How does it differ, for example, from the sequece 1k or the sequece 1m? O the other had, the otatio (1) =1 is uambiguous (here is agai a dummy variable). Commet 3.3 We have to distiguish a sequece (a ) =1 from the set of values that the sequece assumes, {a N}. For the example, if the the set of values it assumes is (a ) =1 = (0,1,0,0,1,1,0,0,0,1,1,1,...), {a N} = {0,1}. I a set, every elemet appears oce ad there is o order amog elemets. 3.2 Limits of sequeces Cosider the harmoic sequece, a = 1. Its elemets are positive ad decreasig (every elemet is smaller tha its successor). While o elemet equals zero, we uderstad o a ituitive level that the sequece teds toward zero". I this sectio we will defie formally what it meas for a

3 3.2 Limits of sequeces 51 sequece to ted to some real umber (there is othig special about tedig to zero). Let s start to costruct a defiitio to the statemet the sequece (a ) teds to the real umber a". Very iformally, we would say that this meas that whe is very large, a is very close to a". This is, of course, ot a mathematical statemet. What does very large" mea? Ad what does very close to a" mea? Let s start by makig the very close to a" clause more rigorous. How ca we measure a distace from a? The distace of a umber x from a is the absolute value x a. Whe we say that the distace of x from a is less tha some r > 0, we mea that x a < r. Defiitio 3.2 Give a R ad r > 0, we defie the ope ball ((&;5 9&$,) of radius r about a by B(a,r) = {x R x a < r}. The term ball" is atural whe we thik about the same defiitio i threedimesioal space. Equipped with this ew defiitio, we will try to refie our defiitio of a sequece tedig to a umber. (a ) teds to a if for every r > 0 ad sufficietly large, a B(a,r). This is still ot good eough. What do we mea ow by sufficietly large? Thik of the harmoic sequece: o matter how small r is, from some owards, all the elemets of the sequece are i B(0,r). This observatio motivates the followig defiitio: Defiitio 3.3 Let P be a sequece of logical propositios, which ca be either True or False. We say that the propositios hold for sufficietly large, if there exists a idex N N such that for all > N, P = True. I formal otatio, ( N N)( > N)(P = True). With that we ca fially defie what it meas for a sequece to ted to a umber: Defiitio 3.4 A sequece (a ) coverges (;21,;/) to a R if for every e > 0, the elemets of the sequece are i B(a,e) for sufficietly large. Formally, or ( e > 0)( N N)( > N)(a B(a,e)), ( e > 0)( N N)( > N)(a a < e). We call the real umber a a limit (-&"#) of the sequece, ad deote the fact that

4 52 Sequeces a is a limit of the sequece (a ) by Other popular otatio are lim a = a. a a or a a. Commet 3.4 Note the a limit rather tha the limit. We do t yet kow that a limit of a sequece, if it exists, is uique. Defiitio 3.5 A sequece is called coverget (;21,;/) if it has a limit; otherwise it is called diverget (;9$";/). Example 3.1 The simplest example to start with is the costat sequece a = a. It seems obvious that this sequece teds to a. We have to be careful, ad make sure that a is the limit accordig to the defiitio. That is, we have to prove that ( e > 0)( N N)( > N)(a a < e). Substitutig the value of a, we have to prove that ( e > 0)( N N)( > N)( a a 0 < e). This is trivially true. Give ay e > 0 we may take N = 1. Ideed, for every > N, a a = 0 < e. Example 3.2 Cosider ext the harmoic sequece a = 1. We wat to show that lim a = 0, amely that ( e > 0)( N N)( > N)(1 < e). Take for example e = All the elemets a of the sequece are i B(0,0.01) from = 101. More geerally, let e > 0 be give. Take N = 1e. The, for all > N, a = 0 = 1 < 1 N = 1 1e < e, which completes the proof. Example 3.3 Let a = +1, or (a ) = ( 2 1, 3 2, 4 3,...).

5 3.2 Limits of sequeces 53 If you evaluate the elemets of this sequece you ll quickly guess that lim a = 0. The questio is whether we ca prove that ( e > 0)( N N)( > N)( +1 < e)? We will use the followig algebraic trick, a = ( +1 )( +1+ ) 1 = < I order to have a 0 < e we ca take to be greater tha (12e) 2. Therefore, give e > 0, we take N = 1 2 2e. The for all > N, which completes the proof. a 0 < 1 2 < 2e 2 = e, Example 3.4 Cosider ext the sequece a = Start with ituitio. As becomes very large, the umerator is domiated by the 3 3 term, whereas the deomiator is domiated by the 4 3 term. It makes sese to guess that as becomes larger ad larger, the sequece approaches a costat, To prove it we eed to show that lim a = 3 4. ( e > 0)( N N)( > N)(a 34 < e). This requires some work. Cosider the differece, a 3 4 = 4( ) 3( ) 4( ) = ( ) < = For > 3, 2 > 8, which implies that > 6 2, hece for all > 3, a 3 4 < = 7 6.

6 54 Sequeces We ca ow close the proof. Give e > 0, let N = max(3,6e7). The, for every > N, a 3 4 < 7 6 < e. Example 3.5 Let a R. We will show that there exists a sequece (r ) of ratioal umbers that coverges to a. The idea is very simple. For every cosider the ope ball B(a,1). By the desity of the ratioals, there exists a ratioal umber r B(a,1). Pick oe. This costructs a sequece (which we do t care to kow explicitly). This sequece coverges to a, because give e > 0, let N = 1e. The, for every > N, r B(a,1) B(a,e). 3.3 Uiqueess of the limit A covergig sequece has a limit. The questio is whether it is possible to coverge to two differet limits. We will show that the limit is uique, thus justifyig the referece to the limit of a covergig sequece. The ratioale behid the proof is very simple. If a sequece (a ) coverges to a, the for ay (small) iterval aroud a, the sequece must evetually be withi this iterval. If the sequece also coverges to b, the for ay (small) iterval aroud b, the sequece must evetually be withi this iterval. We ca take those itervals sufficietly small so that they are disjoit (.*9'), leadig to a cotradictio. Let s proceed step by step: Lemma 3.1 Let a,b R, a b. The there exists a e > 0 such that B(a,e) ad B(b,e) are disjoit. Proof. Suppose, without loss of geerality (a otio we have to discuss), that a < b. Let e = (b a)2. The, B(a,e) = 3 2 a 1 2 b, 1 2 (a +b) ad B(b,e) = 1 2 (a +b), 3 2 b 1 2 a, ad these two ope segmets are ideed disjoit.

7 3.3 Uiqueess of the limit 55 Lemma 3.2 Let P ad Q be two sequeces of propositios (assumig values True ad False). If P holds for large eough ad Q holds for large eough, the P Q holds for large eough. Proof. It is give that ad ( N 1 )( > N 1 )(P = True), ( N 2 )( > N 2 )(Q = True). Let N = max(n 1,N 2 ). The, for all > N, both > N 1 ad > N 2, hece P = True ad Q = True. Theorem 3.3 Uiqueess of the limit. Let (a ) be a coverget sequece. If a R ad b R are limits of (a ), the a = b. Proof. Assume, by cotradictio that a b. By Lemma 3.1 there exists a e > 0 such that B(a,e) B(b,e) =. By the defiitio of the limit, a B(a,e) for large eough ad a B(b,e) for large eough. It follows from Lemma 3.2 that a B(a,e) B(b,e) for large eough, which is impossible. We coclude this sectio by discussig diverget sequeces. A sequece is diverget if it does ot have a limit. I other words, This requires some elaboratio. Sice it follows that Thus, a R a is ot a limit of (a ). a is a limit of (a ) ( e > 0)( N N)( > N)(a a < e), a is a ot limit of (a ) ( e > 0)( N N)( > N)(a a e). (a ) is diverget ( a R)( e > 0)( N N)( > N)(a a e). Example 3.6 Cosider the sequece of atural, a =. This sequece is diverget, for let a R. Take e = 1. For all N N, there exists a > N such that a a 1.

8 56 Sequeces Example 3.7 Cosider the alteratig sequece a = ( 1). We first claim that 1 is ot a limit is this sequece, take e = 2. For every N N there exists a > N, such that a = ( 1), i.e., a 1 2 or a B(2,1). We ext claim that o a 1 ca be a limit of (a ), for let e > 0 be such that B(a,e) B(1,e) =. For every N N there exists a > N such that a = 1, amely, a B(a,e). 3.4 Bouds ad order Defiitio 3.6 A sequece (a ) is said to be upper bouded (-*3-/ %/&2() if there exists a M R such that ( N)(a M). If is said to be lower bouded (39-/ %/&2() if there exists a m R such that ( N)(m a ). It is said to be bouded if it is both upper bouded ad lower bouded. Commet 3.5 The sequece (a ) is upper (resp. lower) bouded if ad oly if the set of values it assumes, {a N} is upper (resp. lower) bouded. The property of beig bouded does ot see" the order withi the sequece. Example The sequece of aturals, a =, is lower bouded by ot upper bouded. 2. The harmoic sequece is bouded. 3. The sequece a = ( 1) is either upper or lower bouded. Theorem 3.4 A coverget sequece is bouded. Proof. Let (a ) be a sequece that coverges to a limit a. We eed to show that there exist L 1,L 2 such that By defiitio, settig e = 1, L 1 a L 2 N. ( N N)( > N)(a B(a,1)),

9 3.4 Bouds ad order 57 or, Sice ( N N)( > N)(a 1 < a < a +1). {a 1 N} is a fiite set, there exist M = max{a 1 N} m = mi{a 1 N}. The for all, mi(m,a 1) a max(m,a +1). Propositio 3.5 Suppose that (a ) ad (b ) are coverget sequeces, lim a = a ad lim b = b. Suppose that a > b. The there exists a N N, such that b > a for all > N, i.e., the sequece (b ) is evetually greater (term-by-term) tha the sequece (a ). Proof. By Lemma 3.1, there exists a e > 0 such that A = B(a,e) ad B = B(b,e) are disjoit. I fact, every elemet i A is smaller tha every elemet i B. By Lemma 3.2 there exists a N N such that for all > N, a A ad b B, which implies that a < b. Corollary 3.6 Let a,b R with a < b. Let (a ) be a coverget sequece with limit a. The, a < b for large eough. Proof. Apply the previous propositio with the costat sequece b = b. Propositio 3.7 Suppose that (a ) ad (b ) are coverget sequeces, lim a = a ad lim b = b, ad there exists a N N, such that a b for all > N. The a b.

10 58 Sequeces Proof. This is a immediate corollary of the previous propositio (reverse implicatio of the egatios). Commet 3.6 If istead, a < b for all > N, the we still oly have a b. Take for example the sequeces a = 1 ad b = 2. Eve though a < b for all, both coverge to the same limit. Theorem 3.8 Sadwich. Suppose that (a ) ad (b ) are sequeces that coverge to the same limit `. Let (c ) be a sequece for which there exists a N N such that a c b for all > N. The lim c = `. Proof. By the give assumptios ad Lemma 3.2, ( e > 0)( N N)( > N)( e < a ` ad b ` < e ad a c b ). Sice, e < a ` ad b ` < e ad a c b e < c ` < e It follows that ( e > 0)( N N)( > N)(c ` < e). Example 3.9 Sice for all, it follows that 1 < 1+1 < = 1+1, lim 1+1 = Limit arithmetic Suppose we have two sequeces (a ) ad (b ). We ca form ew sequeces, such as (c ) give by c = a +b, a (d ) give by d = a b. If the elemets of b are o-zero, the we ca also form a sequece (e ), give by e = 1 b. Suppose that (a ) ad (b ) are both coverget sequeces with limits a ad b. Ca we ifer the covergece ad limits of the sequeces (a +b ), (a b ) ad (1b )?

11 3.5 Limit arithmetic 59 Lemma 3.9 Let (a ) be a sequece. The the followig statemets are equivalet: 1. (a ) coverges to a. 2. (a a) coverges to zero. 3. (a a) coverges to zero. Proof. Sice a a = (a a) 0 = a a 0, it follows that ( e > 0)( N N)( > N)(a a < e) ( e > 0)( N N)( > N)((a a) 0 < e) ( e > 0)( N N)( > N)(a a 0 < e) are equivalet statemets. Propositio 3.10 If (a ) coverge to a, the (a ) coverges to a. Proof. From the reverse triagle iequality, 0 a a a a. Sice (a a) coverges to zero, we ca ivoke the sadwich theorem. Commet 3.7 Note that the coverse is ot true. Set a = ( 1), the (a ) coverges to 1, but (a ) is diverget. Lemma 3.11 If x B(a,r) ad y B(b,r) the x +y B(a +b,2r). Proof. This is immediate. We kow that a r < x < a +r b r < y < b +r. It oly remai to add" the two iequalities. Theorem 3.12 Limits of sums of sequeces. Let (a ) ad (b ) be coverget sequeces. The the sequece c = a +b is also coverget, ad lim c = lim a + lim b.

12 60 Sequeces Proof. Deote the limits of (a ) ad (b ) by a ad b. Let e > 0 be give. From the defiitio of the limit ad Lemma 3.2, a B(a,e2) ad b B(b,e2) for large eough. Ivokig Lemma 3.11, we obtai that for large eough, which implies that c B(a +b,e) lim c = a +b. Commet 3.8 Note that the coverse is ot true. (a + b ) may be coverget, whereas (a ) ad (b ) are ot. Commet 3.9 The theorem about the limit of a sum of two sequeces ca be readily exteded to ay fiite sum of sequeces. Commet 3.10 By a similar argumet we may show that provided that the right-had side exists. lim (a b ) = lim a lim b, Lemma 3.13 Let e > 0 ad let a,b R. If e x Ba,mi1, 2(b+1) ad y Bb,mi1, e 2(a+1), the xy B(ab,e). Proof. Start with xy ab = (x a)y+a(y b). Usig the triagle iequality, xy ab x ay+ay b. Sice y < b+1. x a < e2(b+1) ad y b < e2(a+1), it follows that e xy ab < (b+1) 2(b+1) +a e 2(a+1) < e, which cocludes the proof.

13 3.5 Limit arithmetic 61 Theorem 3.14 Limits of products of sequeces. Let (a ) ad (b ) be coverget sequeces. The the sequece c = a b is also coverget, ad lim c = lim a lim b. Proof. Deote the limits of (a ) ad (b ) by a ad b. Let e > 0 be give. From the defiitio of the limit ad Lemma 3.2, e a Ba,mi1, 2(b+1) ad b Bb,mi1, for large eough. Ivokig Lemma 3.13, we obtai that for large eough, which implies that c B(ab,e) e 2(a+1) lim c = ab. Corollary 3.15 Let (a ) be a covergece sequece ad let b R. The, the sequece (ba ) is coverget with lim (ba ) = b lim a. Proof. Apply Theorem 3.14 with the costat sequece b = b. I remais to prove a sequece arithmetic theorem regardig the ratio of sequeces. Lemma 3.16 Let b 0 ad y Bb,mi b 2, b2 e 2. The, y 0 ad 1 y B 1 b,e. Proof. It is give that y b < b 2. Sice y = b (b y), it follows from the triagle iequality that y b b y > b 2,

14 62 Sequeces which proves that y 0. The, 1 y 1 b = b y yb < b2 e2 b2b = e, which cocludes the proof. Theorem 3.17 Let (b ) be a coverget sequece whose limit is ot zero. The, the sequece c = 1b is well-defied for large eough. Furthermore, it is coverget, ad lim c = 1 lim b. Proof. Deote the limit of b by b. Let e > 0 be give. From the defiitio of the limit b Bb,mi b 2, b2 e 2 for large eough. It follows from Lemma 3.16 that b 0 ad c B 1 b,e for large eough, which cocludes the proof. Corollary 3.18 Let (a ) be a coverget sequece, ad let (b ) be a coverget sequece whose limit is ot zero. The, the sequece c = a b is well-defied for large eough. Furthermore, it is coverget, ad lim c = lim a. lim b Example 3.10 Use limit arithmetic to calculate the limit of a = Theorem 3.19 Let (a ) be a bouded sequece ad let (b ) be a sequece that coverges to zero. The lim a b = 0. Proof. Let M be a boud for (a ), amely, ( N)(a M).

15 3.6 Covergece of meas 63 Sice (b ) coverges to zero, ( e > 0)( N N)( > N)b < e M. Thus, ( e > 0)( N N)( > N)(a b Mb < e), which implies that the sequece (a b ) coverges to zero. Example 3.11 The sequece coverges to zero. a = si 3.6 Covergece of meas Let (a ) be a sequece. We defie a ew sequece (s ) as follows, s 1 = a 1 s 2 = 1 2 (a 1 +a 2 ) s 3 = 1 3 (a 1 +a 2 +a 3 ) etc. For the geeral term, s = 1 a k. k=1 Theorem 3.20 Cezaro. If (a ) is coverget, the so is (s ) ad lim s = lim a. Proof. Deote by a the limit of (a ). Note that ad by the triagle iequality, s a = 1 (a k a), s a 1 a k a. Recall that a coverget sequece is bouded, let M be a boud for {a N}. By the triagle iequality, for all N, a a a +a M +a.

16 64 Sequeces Give e > 0, there exists a N N, such that for every > N, The, for every > N, N a a < e 2. s a 1 k a+ a 1 k=n+1 < N N (M +a)+ N (M +a)+ e 2. e 2 a k a Let The for every > N, N e = maxn, 2N(M +a). s a < e 2 + e 2 = e. 3.7 Geeralized limits A sequece is diverget if it does ot have a limit. There are two types of diverget sequeces: some just do t have a limit", whereas other grow idefiitely without bouds", or decrease idefiitely without bouds". Defiitio 3.7 Let (a ) be a sequece. We say that it teds to ifiity (;5!&: 4&21*!-) if ( M R)( N N)( > N)(a > M). We write lim a =. Commet 3.11 Recall that ifiity is ot a real umber. Likewise: Defiitio 3.8 Let (a ) be a sequece. We say that it teds to mius ifiity if ( M R)( N N)( > N)(a < M). We write lim a =. Commet 3.12 If a sequece teds to plus or mius ifiity we say that it coverges i a wide sese ("(9% 0"&/"). A sequece that teds to plus or mius ifiity is still diverget.

17 3.7 Geeralized limits 65 We ow start to ivestigate properties of sequeces that coverge i a wide sese. Propositio 3.21 A sequece that teds to ifiity is ot bouded from above. Similarly, a sequece that teds to mius ifiity is ot bouded from below. Proof. If (a ) is bouded from above, ( M R)( N)(a < M). It is the ot true that ( M R)( N)(a M). A fortiori, it is ot true that ( M R)( N N)( > N)(a > M). Propositio 3.22 Let (a ) ad (b ) be sequeces satisfyig a b for sufficietly large. If the lim a =, lim b =. Proof. Let M > 0 be give. By defiitio, ad usig Lemma 3.2, there exists a N N, such that for all > N, a > M ad a b. If follows that for all > N, b > M, which cocludes the proof. Propositio 3.23 Let (a ) be a sequece of o-zero elemets, satisfyig lim a = 0. The 1 lim a =.

18 66 Sequeces Proof. By defiitio, ( M > 0)( N N)( > N)(0 < a < 1M), hece which cocludes the proof. ( M > 0)( N N)( > N)(1a > M), Commet 3.13 Note that lim a = 0. does ot implies that (1a ) coverges a wide sese. Propositio 3.24 Let (a ) be a sequece satisfyig lim a =. The lim 1 = 0. a Proof. By defiitio, ( e > 0)( N N)( > N)(a > 1e), hece which cocludes the proof. ( e > 0)( N N)( > N)(0 < 1a < e), 3.8 Mootoe sequeces Defiitio 3.9 A sequece a is called icreasig (%-&3) if a +1 a for all. It is called strictly icreasig (:// %-&3) if a +1 > a for all. We defie similarly decreasig (;$9&*) ad strictly decreasig (:// ;$9&*) sequeces. Ay oe of those sequeces is called mootoe. Example The sequece (a ) = is strictly icreasig. 2. The sequece (b ) = 1 is strictly decreasig. 3. The sequece (c ) = ( 1) is ot mootoe. 4. The sequece (d ) = a is both icreasig ad decreasig.

19 3.8 Mootoe sequeces 67 Theorem 3.25 Let (a ) be a icreasig sequece. If it is bouded from above, the it is coverget. Otherwise, it teds to ifiity. Proof. The secod statemet is easier to prove. Suppose that (a ) is icreasig ad ot bouded from above. The, for every M R there exists a N N such that Sice the sequece is icreasig, a N > M. ( > N)(a > M), which proves that the sequece teds to ifiity. Suppose ow that (a ) is bouded from above. This implies the existece of a least upper boud. Set a = sup{a N}. (Note that a supremum is a property of a set, i.e., the order i the set does ot matter.) By the defiitio of the supremum, Sice the sequece is o-decreasig, ad i particular, Similarly, ( e > 0)( N N)(a e < a N a). ( e > 0)( N N)( > N)(a e < a N a a), ( e > 0)( N N)( > N)(a a < e). Theorem 3.26 Let (a ) be a decreasig sequece. If it is bouded from below, the it is coverget. Otherwise, it teds to mius ifiity. Corollary 3.27 Every mootoe sequece coverges i a wide sese. Example 3.13 Cosider the sequece a = We will first show that a < 3 for all. From the biomial formula, (a+b) = k ak b k,

20 68 Sequeces follows that 1+ 1 = k k k 1 k 1 1 = k! 1 1 ( j) = k! j=0 k 1 j=0 1 j. This sequece is icreasig as the larger, the more terms there are, ad each grows. Moreover, k! = k! k < 3. k=3 k=3 It follows that (ad equals to e). lim 1+ 1 exists 3.9 Cator s lemma Cosider the sequece of segmets, I = [0,1]. For every N, I +1 I. Moreover, the legth of the segmets teds to zero. If we look at the itersectio of all the I s, we fid out that it cotais a sigle poit, =1 I = {0}. If we rather used ope, or semi-ope segmets, J = (0,1], It still holds that J +1 J, ad that the legth of the segmets teds to zero. Yet, =1 J =. Theorem 3.28 Cator s lemma. Let (I ) be a sequece of closed segmets satisfyig I +1 I ad lim I = 0, where I deotes the segmet s legth. The there exists a uique real umber c such that A = I = {c}. =1

21 3.10 Subsequeces ad partial limits 69 Proof. Let I = [a,b ]. Sice I +1 I, it follows that (a ) is icreasig ad (b ) is decreasig. Sice a 1 a < b b 1, it follows that (a ) is bouded from above ad (b ) is bouded from below. By Theorem 3.25, both sequeces are coverget. Deote, a = lim a ad b = lim b. Recall that for mootoe sequece, a = sup{a N} ad b = if{b N}. Sice the legth of the segmets teds to zero, if follows from limit arithmetic that hece a = b. Furthermore, sice 0 = lim I = lim (b a ) = b a, a = sup{a N} = if{b N}, it follows that a I for all, amely a A. It remais to prove that A cotais a uique poit. Let g A. The, for every, 0 g a (b a ), ad by the sadwich theorem, g = a Subsequeces ad partial limits Defiitio 3.10 Let (a ) be a sequece. A subsequece (%9$2 ;;) of (a ) is ay sequece a 1,a 2,..., such that 1 < 2 <. More formally, (b ) is a subsequece of (a ) if there exists a strictly icreasig sequece of atural umbers ( k ) k=1, such that b k = a k. Commet 3.14 Every sequece is its ow subsequece for k = k. Commet 3.15 The sequece (a ) =1 is the same as (a k) k=1, but uless k = k, it is ot the same as the sequece (a k ) k=1. Example 3.14 The sequece b = 12 is a subsequece of the harmoic sequece a = 1, for the choice k = 2k. Ideed, b k = 1 2k = a 2k.

22 70 Sequeces Example 3.15 Let (a ) be the sequece of atural umbers, amely a =. The subsequece (b ) of all eve umbers is b k = a 2k, i.e., k = 2k. The followig lemma makes a umber of obvious statemets: Lemma If ( k ) is a icreasig sequece of idexes the k k. 2. Let ( k ) be a icreasig sequece of itegers. Let (P ) be a sequece of propositios. If P holds for sufficietly large, i.e., ( N N)( > N)(P = True), the (P k ) holds for sufficietly large k. i.e., ( K N)( k > K)(P k = True). 3. Let ( k ) be a icreasig sequece of itegers. Let (P ) be a sequece of propositios. If (P k ) holds for sufficietly large k, i.e., ( K N)( k > K)(P k = True). the (P ) holds for ifiitely may s, i.e., ( N N)( > N)(P = True). 4. Every sub-subsequece is a subsequece. 5. If A N is a ifiite set, the there exists a sequece ( k ) of idexes such that k A for all k. Defiitio 3.11 Let (a ) be a sequece. A real umber a is called a partial limit (*8-( -&"#) of (a ) if it is the limit of a subsequece of (a ). That is, if there exists a strictly icreasig sequece of itegers ( k ), such that a = lim k a k. Similarly, we defie partial limits i the wide sese. Example 3.16 A costat sequece a = c oly has oe subsequece, ad oly oe partial limit, c. More geerally, every limit is also a partial limit. Example 3.17 The sequece a = ( 1) has two partial limits, 1 ad 1. It is easy to show that these are its oly partial limits.

23 3.10 Subsequeces ad partial limits 71 Example 3.18 Every atural umber is a partial limit of the sequece, (1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,...). Propositio 3.30 If (a ) is coverget with limit a, the every subsequece of (a ) coverges to a, ad i particular, a is the oly partial limit. Proof. Let ( k ) be a icreasig sequece of idexes. Give e > 0, let P = (a a < e). This clause holds for sufficietly large, hece by Lemma 3.29(2), P k = (a k a < e). holds for sufficietly large k. Corollary 3.31 If a sequece has two partial limits the it is ot coverget. Partial limits ca be characterized with o referece to a particular subsequece: Propositio 3.32 A real umber a is a partial limit of a sequece (a ) if ad oly if every eighborhood of a cotais ifiitely may elemets of that sequece. Proof. Suppose first that a is a partial limit of (a ). If follows that there exists a icreasig sequece of idexes ( k ) such that Let e > 0 be give, ad let lim a k = a. k P = (a B(a,e)). The, P k holds for sufficietly large k, ad by Lemma 3.29(3), P holds for ifiitely may s. Suppose ext that every eighborhood of a cotais ifiitely may elemets of (a ). Cosider the set I 1 = { N a B(a,1)} Sice this set is ot empty, there exists a 1 I 1, i.e., a 1 B(a,1). Cosider ext the set I 2 = { N a B(a,12)}{ N 1 }. This set is ot empty, hece it cotais a elemet 2, which, by defiitio, satisfies 2 > 1 ad a 1 B(a,12).

24 72 Sequeces We proceed iductively, settig I k+1 = { N a B(a,1(k +1))}{ N k }. This set is ot empty, hece it cotais a elemet k+1, which, by defiitio, satisfies k+1 > k ad a k+1 B(a,1(k +1)). We have thus costructed a subsequece a k. Sice 0 a k a < 1 k, it follows from the sadwich theorem" that (a k ) coverges to a. Propositio 3.33 is a partial limit of (a ) if ad oly if (a ) is ot bouded from above. Similarly, is a partial limit of (a ) if ad oly if (a ) is ot bouded from below. Proof. The proof is essetially the same. We ext prove this very importat theorem: Theorem 3.34 Bolzao-Weierstrass. Every bouded sequece has a covergig subsequece. Proof. Suppose that M > 0 is a boud for the sequece, amely, ( N)(a [ M,M]). We costruct recursively a sequece of segmets (I ) satisfyig: 1. I +1 I. 2. I +1 = 1 2 I. 3. I cotais ifiitely may elemets of (a ). This sequece is costructed usig bisectio (%**7(). Specifically, let I cotai ifiitely may elemets of (a ), which meas that A = {k N a I } is a ifiite set of idexes. Partitio I ito two closed segmets of equal size, which oly itersect at oe poit, ad defie I = I R I L, A R = {k N a I R } ad A L = {k N a I L }.

25 3.10 Subsequeces ad partial limits 73 Sice A = A R A L is a ifiite set, either A R or A L must be ifiite. The, set I +1 = I R I L I R = otherwise. By Cator s lemma, there exists a uique umber a i the itersectio of all the I. We will prove that a is a partial limit of (a ). Ideed, give e > 0, let be large eough such that I < e. The, B(a,e) I. Sice I cotais ifiitely may elemets of (a ) so does B(a,e), ad by Propositio 3.32, a is a partial limit of (a ). The Bolzao-Weierestrass ca be proved i a completely differet way: it is a immediate corollary of the followig lemma: Lemma 3.35 Ay sequece cotais a subsequece which is either decreasig or icreasig. Proof. Let (a ) be a sequece. Let s call a umber a peak poit (!*: ;$&81) of the sequece a if a m < a for all m >. a peak poi! a peak poi! There are ow two possibilities. There are ifiitely may peak poits: If 1 < 2 < are a sequece of peak poits, the the subsequece a k is decreasig. There are fiitely may peak poits: The let 1 be greater tha all the peak poits. Sice it is ot a peak poit, there exists a 2, such that a 2 a 1. Cotiuig this way, we obtai a o-decreasig subsequece. Commet 3.16 There is a fudametal differece betwee the two proofs. The first proof ca be geeralized with little modificatio to bouded sequeces i R. The secod proof relies o the fact that R is a ordered set, hece the possiblity to defie mootoe sequeces.

26 74 Sequeces Corollary 3.36 Every sequece has a subsequece that coverges i the wide sese. Proof. Either the sequece of bouded, i which case this is a cosequece of the Bolzao-Weierstrass theorem, or it is ot bouded, ad this is a cosequece of Propositio Propositio 3.37 Let (a ) be a sequece that does ot coverge i the wide sese. The, it has at least two partial limits (i the wide sese). Proof. Let a be a partial limit of (a ) i the wide sese. Suppose first that a R. Sice, by assumptio, a is ot a limit of (a ) there exists a e > 0 such that a B(a,e) for ifiitely may s. Thus, we ca costruct a subsequece a k such that ( k N)(a k B(a,e)). By Corollary 3.36, this subsequece has a partial limit (i the wide sese) b. Sice b B(a,e), it differs from a. The proof is similar if a = ±. Corollary 3.38 A sequece (a ) coverges i the wide sese to a if ad oly if a is its oly partial limit The expoetial fuctio We have see that the sequece a = 1+ 1 is bouded ad mootoically icreasig, hece covergig. The limit, which we deoted by e = lim is a umber betwee 2 ad 3. Likewise, for every x R we may defie the sequece As for the case x = 1, a = k xk k = 1 k! a = 1+ x. x k k 1 k j=1 ( j) = k! x k k 1 j=1 1 j.

27 3.11 The expoetial fuctio 75 This sequece is icreasig as the larger the more terms there are, ad the k-th term is larger. Also, a k!. Let N = 2x, i.e., xn < 12. The, for > N, N a k! + x k N x k k=n+1 = k! + N x k k=n+1 = k! + xn N! N x k k! + xn N! N x k k! + xn 2N!. x k x k k! x N x N N! (N +1)... k=n+1 k=n+1 x N (N +1) N The right-had side is idepedet of, which meas that (a ) is a bouded sequece, hece coverges. Sice the sequece depeds o x, so does the limit. We defie exp(x) = lim 1+ x. I particular, exp(1) = e. I the previous chapter, we defie the otio of powers with real-valued expoets. Thus, for every x R, we ca defie a umber e x, whose defiitio, we recall, is We ow claim that Theorem 3.39 For every x R, e x = sup{e r Q r x}. exp(x) = e x. That is, where lim 1+ x = sup{e r Q r x}, e = lim 1+ 1.

28 76 Sequeces Proof. We will first show that this idetity holds for every x N. Set x = m N, ad cosider the sequece a = 1+ m. Sice it coverges to exp(m), every subsequece coverges to exp(m) as well. Set k = mk. The, a k = 1+ m mk mk = 1+ 1 m k k. Sice it follows from limit arithmetic that i.e., lim 1+ 1 k k k = e, lim a k = lim 1+ 1 k m k k k = e m, lim 1+ m = e m. Next suppose that x = pq, p,q N, ad cosider the sequece The, cosider the sequece a = 1+ p q. a q = 1+ p q q. This sequece is a subsequece (every q-th term) of a sequece that coverges to e p, i.e., lim aq = e p. Agai, by limit arithmetic, which implies that or equivaletly, q lim aq = lim a, (exp(pq)) q = e p, exp(pq) = e pq. It remais to deal with the case x R. Note that both e x ad exp(x) are icreasig fuctios of x. Cosider the sets, A = {exp(r)q r x} ad B = {exp(r)q r x}.

29 3.12 Limit iferior ad limit superior 77 We already kow that A = {e r Q r x} ad B = {e r Q r x}. For the latter case, we kow that e x is the uique umber separatig the sets A ad B. Sice exp(x) also separates A ad B, it follows that exp(x) = e x Limit iferior ad limit superior Not taught this year Cauchy sequeces I may cases, we would like to kow whether a sequece is coverget eve if we do ot kow what the limit is. We will ow provide such a covergece criterio. Defiitio 3.12 A sequece (a ) is called a Cauchy sequece if ( e > 0)( N N)( m, > N)(a a m < e). Commet 3.17 A commo otatio for the coditio satisfied by a Cauchy sequece is lim,m a a m = 0. Theorem 3.40 A sequece coverges if ad oly if it is a Cauchy sequece. Proof. Oe directio is easy 1. If a sequece (a ) coverges to a limit a, the By the triagle iequality, ( e > 0)( N N)( > N)(a a < e2). ( e > 0)( N N)( m, > N)(a a m a a+a m a < e), i.e., the sequece is a Cauchy sequece. Suppose ext that (a ) is a Cauchy sequece. We first show that the sequece is bouded. Takig e = 1, ( N N)( > N)(a a N+1 < 1). 1 There is somethig amusig about callig sequeces satisfyig this property a Cauchy sequece. Cauchy assumed that sequeces that get evetually arbitrarily close coverge, without beig aware that this is somethig that ought to be proved.

30 78 Sequeces The, for every > N, whereas for N, a < a N+1 +1, a max k N a k, which proves that the sequece is bouded. By the Bolzao-Weierstrass theorem, it follows that (a ) has a covergig subsequece. Deote this subsequece by b k = a k ad its limit by b. We will show that the whole sequece coverges to b. By the Cauchy property ( e > 0)( N N)( m, > N)(a a m < e2), whereas by the covergece of the sequece (a k ), Combiig the two, ( N N)( K N)( k > K)(b k b < e2). ( e > 0)( N N)( k N k > N)( > N)(a b a b k +b k b < e). This cocludes the proof. Commet 3.18 Limits of sequeces ca be defied for oly for sequeces i R. Limits ca be defied for sequeces i ay metric space, which is a set S o which a distace fuctio d is defied. A sequece (a ) i S coverges to a S if ( e > 0)( N N)( > N)(d(a,a) < e). I ay metric space we ca defie a Cauchy sequece: (a ) is a Cauchy sequece if ( e > 0)( N N)(,m > N)(d(a,a m ) < e). It is ot geerally true that a Cauchy sequece i a metric space coverges. Metric spaces i which every cauchy sequece coverges are called complete. This is the fact the more geeral defiitio of completeess for a ordered field.