Sequences and Series

Transcription

1 Sequeces ad Series Sequeces of real umbers. Real umber system We are familiar with atural umbers ad to some extet the ratioal umbers. While fidig roots of algebraic equatios we see that ratioal umbers are ot eough to represet roots which are ot ratioal umbers. For example draw the graph of y = x. We see that it cross the x-axis twice. The roots are such that their square is, but they caot be ratioal umbers accordig to the followig theorem. Theorem... Suppose that a 0,a,..., a ( ) are itegers such that a 0 0,a 0 ad that r satisfies the equatio a x + a x a x + a 0 =0. If r = p where p, q are itegers with o commo factors ad q 0. The q divides a q ad p divides a 0. This theorem tells us that oly ratioal cadidates for solutios of the above equatio have the form p q where p divides a 0 ad q divides a. Proof: Sice p q satisfies the equatio, we have a p + a p q a 0 q =0 i.e., a p = q(a p a 0 q ). This meas q divides a as p, q have o commo factors. O the other had we ca also write a 0 q = p(a p + a p a q ). Thus p divides a 0. /// Now we see that the possible ratioal roots of x =0are±, ±. But it is easy to check that ±, ± does ot satisfy x = 0. So the roots of x = 0 are ot ratioal umbers. This meas the set of ratioal umbers has gaps. So the atural questio to ask is: Ca we have a umber system without these gaps? The aswer is yes ad the complete umber system with out these gaps is the real lie R. We will ot look ito the developmet of R as it is ot easy to defie the real umbers. We assume that

2 there is a set R, whose elemets are called real umbers ad R is closed with respect to additio ad multiplicatio. That is, give ay a, b R, thesuma + b ad product ab also represet real umbers. Moreover, R has a order structure ad has o gaps i the sese that it satisfies the Completeess Axiom(see below). Let S be a o-empty subset of R. If S cotais a largest elemet s 0,thewecalls 0 the maximum of S. IfS cotais a smallest elemet s 0,thewecalls 0 the miimum of S. If S is bouded above ad S has least upper boud, the we call it the supremum of S. IfS is bouded below ad S has greatest lower boud, the we call it as ifimum of S. Ulike maximum ad miimum, sup S ad if S eed ot belog to the set S. A importat observatio is if α =sups is fiite, the for every ɛ>0, there exists a elemet s S such that s α ɛ. Note that ay bouded subset of Natural umbers has maximum ad miimum. Completeess Axiom: Every oempty subset S of R that is bouded above has a least upper boud. I other words, sup S exists ad is a real umber. The completeess axiom does ot hold for Q. That is, every o-empty subset of Q that is bouded above by a ratioal umber eed ot have ratioal least upper boud. For example {r Q : r }. Archimedea property: For each x R, there exists a atural umber N = N(x) such that x<n. Proof: Assume by cotradictio that this is ot true. The there is o N N such that x<n. i.e., x is a upper boud for N. The, let u be the smallest such boud of N. i.e., u R ad so u m for m N is ot a upper boud for N. Therefore, there exists k N such that u m <k, but the u<k+m, adk+m N. Cotradictio. /// Now it is easy to see the followig corollary Corollary: Let S = { : N}. Thew = if S =0. Proof: We ote that S is bouded below. Let ɛ>0beaarbitrary positive real umber. By above Archimedea property, there exists N such that >. The we have, ɛ Sice ɛ is arbitrary, we have w =0. (why?) 0 w <ɛ. Corollary: If y > 0 be a real umber, the there exists = (y) N such that y<. 3

3 Fially, we have the followig desity theorem Theorem... Let x, y are real umbers such that x<y. The there exists a ratioal umber q such that x<q<y. Proof: W.l.g. assume that x>0. Now let N be such that y x>. Otherwise for all N. Now cosider the set < y x S = {m N : m >x}. The S is o-empty (by Archimedea property). By well -orderig of N, S has miimal elemet say m 0.Thex< m 0. By the miimality of m 0, we see that m 0 x. The, m 0 x + <x+(y x) =y. Therefore, x< m 0 <y.. Sequeces ad their limit Defiitio... A sequece of real umbers is a fuctio from N to R. Notatio. It is customary to deote a sequece as {a } =. Examples... (i) {c} =,c R, (ii) { ( )+ } =, (iii) { } = ad (iv) { } =. Defiitio..3. A sequece {a } = coverges to limit L if for every ɛ>0 (give) there exists a positive iteger N such that N = a L <ɛ. Notatio. L = lim a or a L. Examples..4. (i) It is clear that the costat sequece {c} =,c R, has c as it s limit. (ii) Show that lim =0. Solutio. Let ɛ>0 be give. I order to show that / approaches 0, we must show that there exists a iteger N N such that for all N, 0 = <ɛ. But / < ɛ >/ɛ. Thus, if we choose N N such that N>/ɛ, the for all N, / < ɛ. 4

4 (iii) Cosider the sequece {( ) + } =. It is ituitively clear that this sequece does ot have a limit or it does ot approach to ay real umber. We ow prove this by defiitio. Assume to the cotrary, that there exists a L R such that the sequece {( ) + } = coverges to L. The for ɛ =, there exists a N N such that ( ) + L <, N. (.) For eve, (.) says L <, N. (.) while for odd, (.) says L <, N. (.3) which is a cotradictio as = + L + +L <. Lemma..5. If {a } the L = M. is a sequece ad if both lim a = L ad lim a = M holds, Proof. Suppose that L M. The L M > 0. Let ɛ = L M. As lim a = L, there exists N N such that a L <ɛfor all N. Also as lim a = M,there exists N N such that a M <ɛfor all N. Let N =max{n,n }. The for all N, a L <ɛad a M <ɛ.thus L M a L + a M < ɛ = L M, which is a cotradictio. /// Theorem..6 (Sadwich theorem for sequeces). Let {a }, {b } ad {c } be three sequeces such that a b c for all N. If lim a = L ad lim c = L, the lim b = L. Proof. Let ɛ>0 be give. As lim a = L, there exists N N such that Similarly as lim c = L, there exists N N N = a L <ɛ. (.4) N = c L <ɛ. (.5) Let N =max{n,n }. The, L ɛ<a (from (.4)) ad c L + ɛ ( from (.5)). Thus L ɛ<a b c L + ɛ. Thus b L <ɛfor all N. Hece the proof. /// 5

5 Examples..7. { cos (i) Cosider the sequece cos theorem lim =0. } =. The cos. Hece by Sadwich (ii) As 0 ad 0 as, also coverges to 0 by Sadwich theorem. (iv) If b>0, the lim b =. Solutio. First assume that b>. Leta = b. As b>, a > 0 for all N. Further, b =(+a ) +a. The 0 a b. Thus a 0, i.e., b as. Now if b<, the take c = ad it is easy to show the result. /// b Examples..8. (i) lim =. (ii) If x>0 the lim x =0. (+x) (iii) If p>0, the lim log() p =0. Solutio. (i) Let a =. The 0 a for all N. Further, =(+a ) ( ) = a. Thus 0 a ( ). As 0 as, by Sadwich theorem, ( ) ( ) a 0, i.e., as. (ii) Let k be a iteger such that k>x, k>0. The for >k, Hece, ( + p) > C k p k = 0 < As x k<0, x k 0. Thus! k!( k)! pk = pk k! x k k p k Π [ i +]> k k!. i= ( + x) < k k! x k x k (>k). x 0 as. ( + x) 6

6 (iii) For ay N there exists m N such that m p < (m +) or equivaletly m p <(m +) p. Let ɛ>0. Sice as, there exists N N such that (e ɛ,e ɛ ), N (or) log log ( ɛ, ɛ), N. That is 0 as. This implies that log(m +) lim =0. m p m As p < log p < p log(m +). Now the coclusio follows from Sadwich theorem. m Defiitio..9. (Subsequece): Let {a } be a sequece ad {,,...} be a sequece of positive itegers such that i>jimplies i > j. The the sequece {a i } i= is called a subsequece of {a }. Theorem..0. If the sequece of real umbers {a }, is coverget to L, the ay subsequece of {a } is also coverget to L. Proof. Let { i } i= be a sequece of positive itegers such that {a i } i= forms a subsequece of {a }. Let ɛ>0begive. As{a } coverges to L, there exists N N such that a L <ɛ, N. Choose M N such that i N for i M. The a i L <ɛ, i M. Hece the proof. /// Defiitio... (Bouded sequece): A sequece {a } is said to be bouded above, if there exists M R such that a M for all N. Similarly, we say that a sequece {a } is bouded below, if there exists N R such that a N for all N. Thus a sequece {a } is said to be bouded if it is both bouded above ad below. Lemma... Every coverget sequece is bouded. Proof. Let {a } be a coverget sequece ad L = lim a.letɛ =. The there exists N N such that a L < for all N. Further, a = a L + L a L + L<+L, N. Let M =max{ a, a,..., a, + L }. The a M for all N. Hece {a } is bouded. /// 7

7 .3 Operatios o coverget sequeces Theorem.3.. Let {a } ad {b } be two sequeces such that lim a lim = M. The (i) lim (a + b )=L + M. (ii) lim (ca )=cl, c R.. (iii) lim (a b )=LM. ( ) a (iv) lim = L if M 0. M b = L ad Proof. (i) Let ɛ>0. Sice a coverges to L there exists N N such that a L <ɛ/ N. Also, as b coverges to M there exists N N such that b M <ɛ/ N. Thus (a + b ) (L + M) a L + b N M <ɛ N =max{n.n }. (ii) Easy to prove. Hece left as a exercise to the studets. (iii) Let ɛ>0. Sice a is a coverget sequece, it is bouded by M (say). Also as a coverges to L there exists N N such that a L <ɛ/m N. Similarly as b coverges to M there exists N N such that Let N =max{n,n }.The b M <ɛ/m N. a b LM = a b a M + a M LM a (b M) + M(a L) = a b M + M a L <ɛ/+ɛ/ =ɛ 8

8 (iv) I order to prove this, it is eough to prove that if lim a = L, L 0,the lim /a =/L. Without loss of geerality, let us assume that L>0. Let ɛ>0 be give. As {a } forms a coverget sequece, it is bouded. Choose N N such that a >L/ for all N. Also, as a coverges to L, there exists N N such that a L <L ɛ/ for all N.LetN =max{n,n }.The Examples.3.. (i) Cosider the sequece 5 0 0=0. (ii) Cosider the sequece N = a L = a L a L < L L ɛ { } 5 5. The lim = lim 5 = ɛ. /// =5 lim lim = { } 3 6. Notice that = 3 6/ 5+4/ 3/5..4 Diverget sequece ad Mootoe sequeces Defiitio.4.. Let {a }be a sequece of real umbers. We say that a approaches ifiity or diverges to ifiity, if for ay real umber M>0, there is a positive iteger N such that N = a M. If a approaches ifiity, the we write a as. A similar defiitio is give for the sequeces divergig to. I this case we write a as. Examples.4.. (i) The sequece {log(/)} diverges to. I order to prove this, for ay M>0, we must produce a N N such that log(/) < M, N. But this is equivalet to sayig that >e M, N. Choose N e M. The, for this choice of N, log(/) < M, N. Thus {log(/)} diverges to. Defiitio.4.3. If a sequece {a }does ot coverge to a value i R ad also does ot diverge to or, we say that {a }oscillates. 9

9 Lemma.4.4. Let {a }ad {b } be two sequeces. (i) If {a }ad {b } both diverges to, the the sequeces {a + b } ad {a b } also diverges to. (ii) If {a }diverges to ad {b } coverges the {a + b } diverges to. Example.4.5. Cosider the sequece { + } =. We kow that + ad both coverges to. But the sequece { + } = diverges to 0. To see this, otice that, for a give ɛ>0, + <ɛif ad oly if <ɛ +ɛ. Thus, if N is such that N> 4ɛ, the for all N, + <ɛ.thus + coverges to 0. This example shows that the sequece formed by takig differece of two divergig sequeces may coverge. Defiitio.4.6. A sequece {a } of real umbers is called a odecreasig sequece if a a + for all N ad {a } is called a oicreasig sequece if a a + for all N. A sequece that is odecreasig or oicreasig is called a mootoe sequece. Examples.4.7. (i) The sequeces { /}, { 3 } are odecreasig sequeces. (ii) The sequeces {/}, {/ } are oicreasig sequeces. Lemma (i) A odecreasig sequece which is ot bouded above diverges to (ii) A oicreasig sequece which is ot bouded below diverges to. Example.4.9. If b>, the the sequece {b } diverges to. Theorem.4.0. (i) A odecreasig sequece which is bouded above is coverget. (ii) A oicreasig sequece which is bouded below is coverget. Proof. (i) Let {a }be a odecreasig, bouded above sequece ad a =supa. Sice N the sequece is bouded, a R. We claim that a is the limit poit of the sequece {a }. Ideed, let ɛ>0 be give. Sice a ɛ is ot a upper boud for {a }, there exists N N such that a N >a ɛ. As the sequece is odecreasig, we have a ɛ<a N a for all N. Also it is clear that a a for all N. Thus, a ɛ a a + ɛ, N. Hece the proof. The proof of (ii) is similar to (i) ad is left as a exercise to the studets. /// 0

10 Examples.4.. (i) If 0 <b<, the the sequece {b } coverges to 0. Solutio. We may write b + = b b<b. Hece {b } is oicreasig. Sice b > 0 for all N, the sequece {b } is bouded below. Hece, by the above theorem, {b } coverges. Let L = lim b. Further, lim b + = lim b b = b lim b = b L. Thus the sequece {b + } coverges to b L. O the other had, {b + } is a subsequece of {b }. Hece L = b L which implies L =0as b. (ii) The sequece {(+/) } is coverget. ( )( ) k Solutio. Let a =(+/) =.Fork =,,...,, the (k +) th k k=0 term i the expasio is ( )( ) ( k +) k = k k! ( )( ) ( k ). (.6) Similarly, if we expad a +, the we obtai ( +) terms i the expasio ad for k =,, 3,..., the (k +) th term is ( )( ) ( k ). (.7) k! It is clear that (.7) is greater tha or equal to (.6) ad hece a a + which implies that {a }is odecreasig. Further, a =(+/) = k=0 ( )( ) k < + k k= k! < +e<3. Thus {a }is a bouded mootoe sequece ad hece coverget. Theorem.4.. Every sequece has a mootoe subsequece. Proof. Pick x N such that x x N for all >N. We call such x N as peak. If we are able to pick ifiitely may x N i s,the{x Ni } is decreasig ad we are doe. If there are oly fiitely may x N s ad let x be the last peak. The we ca choose such that x x. Agai x is ot a peak. So we ca choose x 3 such that x 3 x. Proceedig this way, we get a o-decreasig sub-sequece. The followig theorem is Bolzao-Weierstrass theorem. Proof is a cosequece of Theorem.4.

11 Theorem.4.3. Every bouded sequece has a coverget subsequece. Theorem.4.4. Nested Iterval theorem: Let I =[a,b ], be o-empty closed, bouded itervals such that I I I 3... I I +... ad lim (b a )=0. The =I cotais precisely oe poit Proof. Sice {a }, {b } [a,b ], {a }, {b } are bouded sequeces. By Bolzao-Weierstrass theorem, there exists sub sequeces a k,b k ad a, b such that a k a, b k b. Sice a is icreasig a <a <... a ad b >b >... b. Itiseasytoseethata b. Also sice 0 = lim a b = a b, wehavea = b. It is easy to show that there is o other poit i =I. Remark.. closedess of I caot be dropped. for example the sequece {(0, )}. The =(0, )= because there caot be ay elemet x such that 0 <x< else Archimedea property fails. Corollary.4.5. R is ucoutable. Proof. It is eough to show that [0, ] is ucoutable. If ot, there exists a oto map f : N [0, ]. Now subdivide [0, ] ito 3 equal parts so that choose J such that f() J. Now subdivide J ito 3 equal parts ad choose J so that f() J.Cotiue the process to obtai J so that f() J. These J satisfy the hypothesis of above theorem, so = = {x} ad x [0, ]. By the costructio, there is o N such that f() =x. cotradictio to f is oto..5 Cauchy sequece Defiitio.5.. A sequece {a }is called a Cauchy sequece if for ay give ɛ>0, there exists N N such that, m N = a a m <ɛ. Example.5.. Let {a }be a sequece such that {a }coverges to L (say). Let ɛ>0 be give. The there exists N N such that Thus if, m N, we have a L < ɛ N. a a m a L + a m L < ɛ + ɛ = ɛ.

12 Thus {a }is Cauchy. Lemma.5.3. If {a }is a Cauchy sequece, the {a }is bouded. Proof. Sice {a }forms a Cauchy sequece, for ɛ = there exists N N such that a a m <,, m N. I particular, a a N <, N. Hece if N, the a a a N + a N < + a N, N. Let M =max{ a, a,..., a N, + a N }. The a M for all N. Hece {a }is bouded. /// Theorem.5.4. If {a }is a Cauchy sequece, the {a }is coverget. Proof. Let a k be a mootoe subsequece of the Cauchy sequece {a }. The a k is a bouded, mootoe subsequece. Hece {a k } coverges to L(say). Now we claim that the sequece {a } itself coverges to L. Letɛ>0. Choose N,N such that, k N = a a k <ɛ/ k N = a k a <ɛ/. The, k max{n,n } = a a a a k + a k a <ɛ. Hece the claim. /// Therefore, we have the followig Criterio: Cauchy s Criterio for covergece: A sequece {a } coverges if ad oly if for every ɛ>0, there exists N such that a a m <ɛ m, N. Problem: Let {a }be defied as a =,a + =+ a. The show that {a }is Cauchy. Solutio: Note that a > ada a = a +>. The a + a = a a a a a a a a,. 3

13 Hece a m a a m a m + a m a m a + a a a α α,α= So give, ɛ>0, we ca choose N such that N < ɛ. Ideed the followig holds, Theorem.5.5. Let {a } be a sequece such that a + a <α a a for all N for some N ad 0 <α<. The {a } is a Cauchy sequece. Theorem.5.6. For ay sequece {a } with a > 0 provided the limit o the right side exists. a + lim a/ = lim a Proof. Let ɛ>0 be arbitrary. Suppose the secod limit exists (say l), the there exists N N such that l ɛ< a + <l+ ɛ, N. a Takig = N,N +,..., m ad multiplyig we get (l ɛ) m N < a m a N < (l + ɛ) m N, m N + equivaletly, (l ɛ) N m a m N < (a m) m < (l + ɛ) N m a m N, m N +. Now the result follows from the fact that lim m (l ± ɛ) N/m a /m N = l ± ɛ. /// a + Corollary: If a > 0 ad lim = l<, the lim a =0 a a + Corollary: If a > 0 ad lim = l>, the a. a Problems: (i) lim a / =, if a>0. (ii) lim α x =0, if x < adα IR. Solutio: a (i) Takea = a, the lim + a =. (ii) Ifx 0,takea = α x a, the lim + a = lim( + )α x = x. 4

14 .6 Limit superior ad limit iferior Defiitio.6.. Let {a }be a bouded sequece. The limit superior of the sequece {a }, deoted by lim sup a, is defied as lim sup a := if k N sup k a. Similarly limit iferior of the sequece {a }, deoted by lim if a,isdefiedas lim if a := sup if a. k N k Example.6.. (i) Cosider the sequece {a } = {0,, 0,,...}. The β =sup{a m,m } =ad α = if{a m,m } =0. Therefore, lim if a =0, lim sup a =. (ii) Cosider the sequece {a } = {,,, 3,...}. The for large k sup{a m,m k} lim k k 0 < if{a m,m k} lim k The by sadwich theorem, we see that lim sup a =ad lim if a =0. Lemma.6.3. (i) If {a }is a bouded sequece, the lim sup a lim if a. (ii) If {a }ad {b } are bouded sequeces of real umbers ad if a b for all N, the lim sup a lim sup b ad lim if a lim if b. (iii) Let {a }ad {b } are bouded sequeces of real umbers. The lim sup(a + b ) lim sup a + lim sup b ad lim if (a + b ) lim if a + lim if b. 5

15 Example.6.4. Cosider the sequeces {( ) } ad {( ) + }. Here a =( ) ad b =( ) +. Also lim sup a = lim sup b =.Buta + b =0for all N ad hece lim sup (a + b )=0. Thus a strict iequality may hold i (iii) the above Lemma. Theorem.6.5. If {a }is a bouded sequece, the there exists subsequeces {a k } ad {b k } such that lim sup a = lim a k ad lim if a = lim b k. Proof. Sice {a }is bouded, lim sup a = α exists. The from the defiitio, for each k N there exits a k such that α k <a k <α+ k. Therefore, a k α as k. Similarly, oe ca obtai b k. /// Theorem.6.6. If there exists a subsequece a k t. The t s := lim sup a. Proof. Suppose NOT. The choose ɛ>0 such that t ɛ>s.the we ca fid N such that N = a <t ɛ Therefore a t >ɛfor all N. Hece such a sequece caot have a coverget subsequece. /// Remark.. From the above two theorems we ca say that the limsup is the supremum of all limits of subsequeces of a sequece. Remark.3. I case of ubouded sequeces, either lim sup or lim sup or both ca approach. Eve i this case, oe ca show the existece of subsequeces that approach ifiity. Remark.4. If we ca fid the limits of all subsequeces of {a }. The lim sup is othig but the supremum of all these limits. Similarly, lim if is the ifimum of all these limits. Problem Fid lim sup ad lim if of {a }where a = ( ) +( ) +. Solutio: The sequece {a }is bouded ad has two coverget subsequeces { } ad {( + ) }. So the two limits are ad. Therefore, lim sup a = ad lim if a =. Theorem.6.7. If {a }is a coverget sequece, the lim if a = lim a = lim sup a. 6

16 Proof. Let L = lim a. The give ɛ>0 there exists N N such that a L <ɛ, N. Equivaletly L ɛ<a <L+ ɛ, for all N. Thus,if N, L + ɛ is a upper boud for the set {a k k N}. Ifα k := sup{a k k }, the we ote that L ɛ <α N L+ɛ ad α N+ <L+ ɛ,...,α <L+ ɛ for all N (As α is decreasig). Also a >L ɛ, N = α L ɛ, N. Therefore, lim α = L. Hece lim sup a = L. Similarly, oe ca prove the lim if a = L. /// Theorem.6.8. If {a }is a bouded sequece ad if lim sup the {a }is a coverget sequece. Proof. Notice that lim sup a = lim (sup{a k k }) ad lim if a = lim (if{a k k }). Give that L = lim sup a.thusforɛ>0, there exists N N such that sup{a,a +,...} L <ɛ, N. a = lim if a = L, L R, This implies a <L+ ɛ, N (.8) Similarly there exists N N such that if{a,a +,...} L <ɛ, N. This implies L ɛ<a, N (.9) Let N =max{n,n }. The from (.8)ad (.9) we get Examples.6.9. lim a L <ɛ, N. Thus the sequece {a }coverges. /// ( + ) = e. Assume that e = lim k=0. k! 7

17 Solutio. Let a = k! ad b = b = k=0 ( k C k =+ ) k=0 ( + ). Now, k! k= k Π i= ( i ) a.(see (.6) This implies Further, if m, the b = ( + ) = k=0 lim sup C k ( ) k b lim sup a = e. m ( k C k =+ ) k=0 m k! k= k Π i= ( i ). Keepig m fixed ad lettig, we get lim if b m k! k=0 which implies a lim if b. Hece e = lim if a lim if b. Fially we have the followig more precise versio of theorem.6.6 Theorem.6.0. Let {a } be ay sequece of ozero real umbers. The we have lim if a + a lim if a / lim sup a / lim sup a + a. Proof. The iequality i the middle is trivial. Now we show the right ed iequality. Let L = lim sup a + a. W.l.g assume L<. Letɛ>0. The there exists N N such that a + a <L+ ɛ N. The for ay >N,wecawrite a = a a a a... a N+ a N a N < (L + ɛ) N a N =(L + ɛ) ((L + ɛ) N a N ). 8

18 Now takig a = ((L ɛ) N a N ), we have, a / < (L + ɛ)a / for > N. Sice lim a / =, we coclude that lim sup a / (L + ɛ). Sice ɛ is arbitrary, we get the result. Similarly, we ca prove the first iequality. 9