REAL ANALYSIS 206/207 (MATH20) MARCUS TRESSL Lecture Notes Homepage of the course: http://persoalpages.machester.ac.uk/staff/marcus.tressl/teachig/realaalysis/idex.php Cotets. Sequeces.. Defiitio ad first examples of sequeces.2. Coverget sequeces 3.3. Scalar multiplicatio ad additio of sequeces 9.4. Bouded ad mootoe sequeces.5. Null sequeces 4.6. The algebra of limits 7.7. The Bolzao-Weierstraß Theorem ad Cauchy sequeces 23 2. Series 25 2.. Defiitio ad first examples 25 2.2. Covergece tests for series 3 Idex 35 Date: November 6, 206.
0 Real Aalysis - MATH20 Warmup questios. Most of you have see limits, series, cotiuous fuctios ad differetiable fuctios i school ad/or i calculus i some form. You might woder why we are talkig about them agai ow. Here are a couple of questios related to these topics that will be aswered i our course. The difficulty i aswerig them is mostly ot a lack of factual kowledge, but merely a lack of rigorous uderstadig of the fudametal otios limit, series, cotiuous fuctio ad differetiable fuctio. Aswers to the warmup questios will be give i due course, but you might ejoy tryig to tackle them right ow. I defiitely recommed tryig questio (8). () What is lim? What is lim ( + )? (2) A series is a ifiite sum of umbers. But what does that mea? For example: What is 2 + 3 4 + 5...? Is it actually meaigful to ask this questio? Oe might ask what is + + +...? How would you argue for the aswer this sum is? (3) A sail sits o oe ed of a rubber strap of 0 meters legth. After surise, the sail starts crawlig towards the other ed of the strap. By suset the sail made meter of the strap ad stops. Durig the ight, the sail rests ad the strap is stretched uiformly by 0 meters. O the ext day, the sail is crawlig meter agai ad the procedure repeats. Ca the sail ever reach the ed of the strap? (4) A popular explaatio of cotiuous fuctio reads: A fuctio is cotiuous if I ca draw its graph by a pe without takig the pe of the paper. This is ot a bad ituitio of cotiuity, but isufficiet i may cases. For example: (a) Is the fuctio f(x) = x si( x ) cotiuous at 0 whe we set f(0) = 0? Ca you decide this by drawig the graph of the fuctio? (b) Is the boudary of a sow flake (the graph of) a cotiuous fuctio? (5) Is every fuctio cotiuous somewhere? I terms of the popular explaatio of cotiuity i the previous item this asks: give a fuctio, ca you always draw at least a piece of the graph of the fuctio? (6) What is the derivative of x 3? If you have ot leared a rule for this derivative, apply the defiitio of derivative. (7) Give a example of a differetiable fuctio y = f(x) such that dy dx is strictly positive at the umber 2, but still the fuctio is ot icreasig ear 2. (8) Fially, a more subtle questio: What is a real umber? A popular aswer is somethig like A expressio of the form N.XXX..., where N is a iteger ad XXX... stads for a possibly ifiite strig of digits. If we accept this, the what is the differece betwee the umber 0.999... ad the umber?
Defiitio ad first examples of sequeces. Sequeces.. Defiitio ad first examples of sequeces. A sequece is a list of real umbers a, a 2, a 3, a 4,... idexed by atural umbers. Examples of sequeces: (i) 2, 4, 6, 8,.... Here a = 2, a 2 = 4, a 3 = 6 ad a 4 = 8. (ii), 2, 3, 4,.... Here a =, a 2 = 2, a 3 = 3 ad a 4 = 4. (iii) 0, 3 2, 2 3, 5 4, 4 5,... (iv), 2, 3 3, 4 4,... (v) si(), si(2), si(3), si(4),... The precise defiitio of a sequece is the followig:... Defiitio. A sequece is a fuctio N R, where N = {, 2, 3, 4,...} is the set of atural umbers. We write (a ) N for a sequeces which maps to a. The real umber a is called the th term of the sequece (a ) N. For example the 4 th -term i the sequece (iii) above, is 5 4. Sequeces i this sese are sometimes called ifiite sequeces. It is importat to otice that ifiite here does ot mea that the value set {a N} of the sequece (a ) N is ifiite. For example, the sequece,,,,... (more precisely: (( ) ) N ) is a ifiite sequece but its value set is {, }. Ofte, sequeces are commuicated by its geeral term. For example,,,,... is deoted by (( ) + ) N ad its geeral term is ( ) +. The formal defiitio applies to our examples above as follows: (i) The sequece 2, 4, 6, 8,... is a idicatio of the fuctio N R which maps a atural umber to 2 : 2 2 4 3 6. 2. So here the precise otatio of the sequece is (2 ) N ad 2 is its geeral term. The value set is the set of eve atural umbers. We will use this symbol to idicate a subtle fact, or to sigal a warig that here commo mistakes occur.
2 Sequeces (ii) The sequece, 2, 3, 4,... is a idicatio of the fuctio N R which maps a atural umber to : 2 2 3 3.. So here the precise otatio of the sequece is ( ) N ad is its geeral term. The value set is the set { N}. Note (agai) that the terms a do ot eed to be differet for differet. I particular, for every real umber r we have the costat sequece with value r: r, r, r, r,.... So here a = r for every N. We ca display sequeces, more precisely the graph of a sequece (a ) N, as follows: R a = a 8 a 4 = a 5 a 3 a 6 = 0 N a 2 a 7 = Elemets (, a ) of the graph of (a ) N are idicated by a. The value set {a N} is the set of all elemets o the vertical lie idicated by a...2. Defiitio. A subsequece of a sequece (a ) N is ay sequece of the form (a k ) k N, where < 2 < 3 <... itself is a strictly icreasig sequece of atural umbers. It is importat to otice that i this case k k for all k (as follows from a straightforward iductio o k).
Coverget sequeces 3 This defiitio looks more complicated as it is. A subsequece of a a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9... ca be thought of what is left from the origial sequece by deletig some of its terms. For example a a 3 a 5 a 6 a 9... might be the begiig of a subsequece of (a ) N. The oly requiremet is that the remaiig list is ot fiite (that is, we have ot deleted all but fiitely may terms from the origial list). The facy double idex otatio of the defiitio, i the example above would be =, 2 = 3, 3 = 5, 4 = 6, 5 = 9... ; that is, the list, 2, 3,... simply is a eumeratio of the remaiig idices after the deletio...3. Examples. Clearly, every sequece is a subsequece of itself. The followig are all subsequeces of the sequece (a ) N : (i) (a 2 ) N, so here k = 2k. This looks like a typo, however, (a 2 ) N is exactly the same sequece as (a 2k ) k N. (ii) (a 2 ) N, so here k = 2 k. (iii) Take m N 0, where N 0 is defied as N 0 := N {0} = {0,, 2, 3,...}. The the sequece (a +m ) N is a subsequece of (a ) N. Here k = m+k. Aother otatio for (a +m ) N is (a ) >m. The followig are all subsequeces of the sequece ( ) N:.2. Coverget sequeces. (! ) N, ( ) N ad ( 3 + ) N. We start with a remider of elemetary properties of the modulus of real umbers..2.. Rules about the absolute value. Recall that for a real umber a, the absolute value, or the modulus of a is defied as { a if a 0 a = a if a < 0. We have the followig rules for all a, b R: (i) a b a b ad a b, by defiitio of a. (ii) The absolute value is multiplicative, i.e. a b = a b. This is clear. (iii) The triagle iequality holds: a + b a + b, because a + b a + b as b b a + b as a a
4 Sequeces ad (a + b) = a + ( b) a + b as b b Replacig b by b i the triagle iequality gives a b a + b. (iv) The followig derived triagle iequality holds: a b a b, because a + b. as a a a = (a b) + b a b + b, so a b a b, ad by symmetry ( a b ) = b a b a = a b. Replacig b by b i the derived triagle iequality gives a b a + b. (v) For all ε > 0 we have a ε b a + ε a b ε, because a ε b meas a b ε ad b a + ε meas (a b) = b a ε. We are iterested i the behavior of the ters of a sequece (a ) N for very large. I particular we wat to kow whe the terms cluster at some real umber. The ext defiitio is the most importat oe i this course..2.2. Mai defiitio. Let (a ) N be a sequece ad let r R be a real umber. We say that (a ) N coverges to r if for every real umber ε > 0 there is some atural N N such that for every N we have a r < ε. If this is the case we write a r ( ) ad also say a has limit r as teds to ifiity, or a teds to r as teds to ifiity or r is the limit of (a ) N. A sequece (a ) N is called coverget if there is some r R with a r ( ). Otherwise, (a ) N is called diverget..2.3. Discussio of the mai defiitio. Let us have a closer look at this defiitio. Let r R be a real umber ad let (a ) N be a sequece. (a) We claim that (a ) N coverges to r if ad oly if for every real umber ε > 0, the th -term a lies i the ope iterval (r ε, r + ε) for all but fiitely may N. r ε r r + ε To see this, ote first that the expressio a r < ε i.2.2 is equivalet to a (r ε, r + ε) (cf..2.(v)). Now, the expressio I. For all but fiitely may N, a is i (r ε, r + ε) is equivalet to the expressio II. There is some N N such that for all N, a is i (r ε, r + ε). Covice yourself that I. ad II. are ideed equivalet. The idea here is, that the fiite list of terms a,..., a N cotais the exceptioal terms, i.e. those terms for which a is ot i (r ε, r + ε).
Coverget sequeces 5 (b) The atural umber N i.2.2 depeds o ε. This becomes clear from the picture above. If we shrik ε the we possibly have to icrease N so that the terms a with N are all cotaied i (r ε, r + ε). (c) Let us illustrate (a) ad (b) with the sequece ( ) N. We claim that ( ) N coverges to 0. How to prove this? Let us first write dow the defiitio.2.2 of covergece for this sequece: For each ε > 0 we must fid a atural umber N N such that for all N we have 0 < ε. Note that r = 0 here. We uwid this statemet by first tryig special values of ε: ε = 2. The we have to fid a atural umber N N such that for all N, < 2. Now we get a coditio that we ca work with, amely the coditio: < 2. We kow that this is true whe 3 ad so here we have N: we choose N = 3. ε = 7 The we have to fid a atural umber N N such that for all N, < 7. Agai we get a coditio that we ca work with, amely the coditio: < 7. We kow that this is true whe 8 ad so here we choose N = 8. We see that the choice of N depeds o ε. Here the complete proof that ( ) N coverges to 0: Let ε > 0. We ow do ot specify a particular value of ε, but we thik of ε as a give ad fixed real umber. We must fid a atural umber N N (depedig o ε) such that for all N we have 0 < ε. As i the special cases ε = 2, ε = 7 we obtai a coditio that has to be satisfied, amely for all N we have 0 < ε. We rephrase 0 < ε to obtai the correct choice for N: 0 < ε is equivalet to < ε, that is, to ε <. Hece if we choose N as a atural umber with N > ε, the N implies N > ε ad thus 0 < ε. Agai, the depedecy of N o ε ca be see by the choice i this proof: we choose N as a atural umber with N > ε. After this example, it is a good idea to read agai items (a), (b) above! A proof of the covergece of a sequece directly from the defiitio as i item (c) above is called a proof from first priciples. (d) I compact form, defiitio.2.2 says that a teds to r as teds to if ad oly if ε > 0 N N N a r < ε. If we wat to stress the depedecy of N o ε we also write ε > 0 N(ε) N N(ε) a r < ε..2.4. Remarks o wrog ad imprecise formulatios of covergece Let (a ) N be a sequece ad let r be a real umber. Look at the followig two statemets which are frequetly used to describe covergece of (a ) N with limit r. (a) the terms a of the sequece get progressively closer to the umber r (b) For every real umber ε > 0, all but fiitely may a lie i the ope iterval (r ε, r + ε) Both statemets do ot describe covergece:
6 Sequeces (a) The mai problem here is that the clause is imprecise ad therefore caot be applied whe we wat to check covergece. Look at the followig examples: Let a = +, so the sequece looks like 2, 3 2, 4 3, 5 4,... The terms a get closer ad closer to 0, but the sequece does ot coverge to 0. Let a = ( ), so the sequece looks like, 2, 3, 4, 5, 6, 7, 8,.... The the terms a also get closer ad closer to 0 i the sese that we ca fid terms of arbitrary small distace to 0. However the sequece doses ot coverge to 0 (or i fact to ay umber). Now let a = +( ), so the sequece looks like 0,, 0, 2, 0, 3, 0, 4,... The the limit is 0, but each odd term is closer to the limit tha is ay eve term. So if we would take (a) as a defiitio, the (a ) N would ot coverge. (b) This is flawed i the sese that it is ot made clear what all but fiitely may a meas: If all but fiitely may a meas all but fiitely may values a, the the statemet does ot describe covergece. Couterexample: a = ( ) which looks like,,,,... Here the set of all a is {, }; hece all but fiitely may of the a lie i every(!) iterval. So if we d choose this defiitio, the every real umber would be a limit of this sequece. O the other had, if all but fiitely may a lie i... meas for all but fiitely may, a lies i..., the this is the correct defiitio. Coverget sequeces i everyday life: (a) If we wat to measure surface areas (e.g. a ellipse) we approximate the area by small rectagles, write dow the sum of the areas of these rectagles ad hope that the sequece of these umbers progressively approximate the area. (b) Give atural umbers, m, we have a good ituitio about what 2 m is. However, what shall we thik of 2 2? Or better: How is 2 2 defied? We will aswer this i due course. (c) The decimal expasio of umbers is commuicated by spellig out a sequece, e.g. π is 3, 459265358... Which sequece do we actually mea here? If we would be able to write dow the decimal expasio of π, our descriptio would look like this: a = 3 a 2 = 3, a 3 = 3, 4 a 4 = 3, 4 a 5 = 3, 45 a 6 = 3, 459. However, where does this sequece come from? Istead, we ca do the followig: For each > 0 the iterval (π, π) cotais a ratioal umber a. The sequece (a ) N obtaied i this way, coverges to π. I fact, this argumet has othig to do with the special choice of the real umber π ad is valid for all real umbers. I this sese we gave a precise descriptio of
Coverget sequeces 7 a sequece covergig to π which does ot tell us what π is! We ll come back to the precise defiitio of π later..2.5. Uiqueess of limits. Let (a ) N be a sequeces that coverges to r R ad to s R. The r = s. Proof. Suppose r s. W.l.o.g. we may assume that r < s. We choose ε = s r 2, which is half the distace of r to s. I a picture: r r + ε = s ε s Sice (a ) N coverges to r, there is some N N such that r ε < a < r + ε for all N. Sice (a ) N coverges to s, there is some N 2 N such that s ε < a < s + ε for all N 2. Hece for N ad N 2 we have s ε < a < r + ε. But this is ot possible, sice s ε = r + ε. By the uiqueess of limits we may ow also write if (a ) N is coverget with limit r. lim a = r.2.6. Fiite modificatio rule. Let (a ) N be a sequeces that coverges to r R. If (b ) N is aother sequece with a = b for all but fiitely may, the also (b ) N coverges to r. Proof. Firstly, we observe that the coditio says precisely the same as a = b for all but fiitely may there is some K N with a = b for all K. I order to show that (b ) N coverges to r we look ito the defiitio of covergece.2.2: We have to pick some real umber ε > 0 ad we have to fid some N N (depedig o ε) such that, ( ) for all N : b r < ε. So let us pick a real umber ε > 0. Sice (a ) N coverges to r we already kow that there is some N N such that for all N we have a r < ε. Now if i additio is bigger tha K, the we kow b = a. So we choose N = max{n, K} ad obtai ( ).
8 Sequeces.2.7. Propositio. Every subsequece of a coverget sequece (a ) N coverges to lim a. Proof. If (a k ) k N is a subsequece of (a ) N, the k k for every k. Hece if ε > 0 ad N N with a r < ε for all N, the also a k r < ε for all k N. Notice that.2.7 does ot oly say that every subsequece of a coverget sequece coverges. The statemet is that all subsequeces of a coverget sequece (a ) N coverge, ad they all coverge to the same umber, amely to lim a. The assertio i.2.7 is may times used i its cotrapositive way. I.e., if we wat to check that a sequece is diverget, we are sometimes able to spot a ocoverget subsequece, or, two subsequeces that coverge, but ot to same the same limit. A prime example is the sequece (a ) N = (( ) ) N. You have doe questio 3 where you have proved from first priciples that (( ) ) N is diverget. Usig.2.7, we ca re-cofirm this: The subsequece (a 2 ) N of (a ) N is the costat sequece, which coverges to, whereas the subsequece (a 2+ ) N of (a ) N is the costat sequece, which coverges to -. Hece by.2.7, the origial sequece (a ) N must be diverget.
Scalar multiplicatio ad additio of sequeces 9.3. Scalar multiplicatio ad additio of sequeces. It would be a tedious udertakig to prove covergece from first priciples for all sequeces. Istead we are lookig for geeral priciples which allow to deduce covergece (ad to compute the limits) from kow sequeces after applyig certai operatios. I this sectio we itroduce the simplest rules of this sort..3.. Scalar multiplicatio rule. If (a ) N is a coverget sequeces ad c R, the also (c a ) N is a coverget sequece ad lim (c a ) = c lim a. Proof. A special case here is c = 0. I this case the sequece (c a ) N is equal to the costat sequece of value 0, which coverges to 0. We may therefore assume that c 0. We write r = lim a ad we must show that (c a ) N coverges to c r. Accordig to the defiitio of covergece.2.2 we have to pick some real umber ε > 0 ad we have to fid some N N (depedig o ε) such that for all N, ( ) c a c r < ε. How do we fid such a N? Let us first write out what we kow by our assumptios o the covergece of the give sequece: As (a ) N coverges to r, there is some N N (depedig o ε) such that for all N, It follows a r < ε. c a c r = c (a r) = c a r as c 0 < c ε. What ow? Recall that we started with ε ad we still have to fid some N N such that for all N, c a c r < ε. O the other had we were able to fid the atural umber N (depedig o ε) with the property that for all N, c a c r < c ε. The idea ow is to apply the N -argumet for ε c istead of ε!! The the argumet above shows that we ca fid a atural umber N with the property that for all N, ε c a c r < c c = ε. Thus the atural umber N, defied as N := N (the symbol := stads for defied as ) that we foud for ε c is suitable to solve our iitial problem ( ) for ε..3.2. Sum rule. If (a ) N, (b ) N are coverget sequeces, the also their sum (a + b ) N is a coverget sequece ad lim (a + b ) = lim a + lim b.
0 Sequeces Proof. We write r = lim a, s = lim b ad we must show that (a + b ) N coverges to r + s. Accordig to the defiitio of covergece.2.2 we have to pick some real umber ε > 0 ad we have to fid some N N (depedig o ε) such that for all N, ( ) a + b (r + s) < ε. How do we fid such a N? Let us first write out what we kow by our assumptios o the covergece of the two give sequeces: As (a ) N coverges to r, there is some N N (depedig o ε) such that for all N, a r < ε. As (b ) N coverges to s, there is some N 2 N (depedig o ε) such that for all N 2, b s < ε. Hece if we take N 3 as the maximum of N ad N 2, the we kow for all N 3 : a r < ε ad b s < ε. Now by the triagle iequality (cf..2.(iii)), we get for all N 3 : a + b (r + s) = (a r) + (b s) a r + b s ε + ε = 2ε. What ow? Recall that we started with ε ad we still have to fid some N N such that for all N, a + b (r + s) < ε. O the other had we were able to fid the atural umber N 3 (depedig o ε) with the property that for all N 3, a + b (r + s) < 2ε. The idea ow is to apply the N 3 -argumet for ε 2 istead of ε!! Hece the argumet above shows that we ca fid a atural umber N 3 with the property that for all N 3, a + b (r + s) < 2 ε 2 = ε. Thus the atural umber N := N 3 that we foud for ε 2 is suitable to solve our iitial problem ( ) for ε.
Bouded ad mootoe sequeces.4. Bouded ad mootoe sequeces..4.. Defiitio. For a subset S of R ad a elemet r R we write S r if s r for every s S ad S < r if s < r for every s S. Similarly the otatio r S ad r < S has to be uderstood. (i) Give a subset S of R we call every elemet r R with S r a upper boud for S. S is called bouded from above if there is a upper boud for S. Similarly every elemet r R with the property r S is called a lower boud for S. S is called bouded from below if there is a lower boud for S. S is called bouded if S is bouded from above ad from below. For example the set of all ratioal umbers q with q 2 < 2 is bouded from above by 3 2 ad bouded from below by 2. (ii) A sequece (a ) N is called bouded from above, bouded from below or bouded, if its value set has this property..4.2. Propositio. Every coverget sequece is bouded. Proof. Let (a ) N be our coverget sequece with limit r. We choose ε = ad exploit the defiitio of covergece: Hece we kow that there is some N N such that for all N, a r < ; i other words r < a < r +. Hece the set of all a with N is bouded above by r + ad bouded below by r. However, the set {a,..., a N } is fiite ad therefore bouded. Sice the uio of two bouded sets is bouded, the value set of (a ) N is bouded. The followig caot be proved ad is a axiom for real umbers:.4.3. Completeess axiom of R. Every oempty subset S of R which has a upper boud, has a least upper boud..4.4. Rules o least upper bouds ad greatest lower bouds. If S R is bouded from above, the its least upper boud is called the supremum of S ad deoted by sup(s). For a elemet x R we have: x = sup(s) S x ad for all r < x } there is some {{ s S with r < s }. S r A subset S R is bouded from below if ad oly if S := { s s S} is bouded from above, because S r r S for all r R. If this is the case, the S has a greatest lower boud, called the ifimum of S, deoted by if(s) ad we have if(s) = sup( S). Observe that the supremum of a bouded set S may be a member of S (e.g. if S = [0, ]) or may ot be a member of S (e.g. if S = [0, )). The existece of suprema of bouded sets is resposible for may costructios of real umbers. Most promietly, 2 is defied as sup{x Q x 2 < 2}.
2 Sequeces Further, we pick up a example from sectio. ow. Give a real umber r > 0 ad atural umbers, m, we have a good ituitio what r m is: r m is the m th root of r. However, what shall we thik of r 2? Or better: How is r 2 defied? We will aswer this ow with the aid of the axiom above. More geerally, let r, p be real umbers. If r we defie r p := sup{r x x Q ad x p}. We exted this defiitio for 0 < r < by r p = ( r ) p ad show that We have r p = if{r x x Q ad x p} : r p = sup{( r )x x Q ad x p} = = sup{r x x Q ad x p} = = sup{r x x Q ad x p} = see below = if{r x x Q ad x p}. The last equality eeds justificatio. Sice r < it is clear that for all x, y Q with x p y we have r x r y. Hece u := sup{r x x Q ad x p} if{r x x Q ad x p} =: v. We ow show that v u = (ad therefore v = u). We kow already v u ad we show that for every real umber δ > 0 we have v u < + δ. Choose N with + δ > r ad choose ratioal umbers x p x 2 with x 2 x <. By defiitio of u ad v we have v rx ad u r x2. Therefore v u rx r x2 Fially ( r ) < + δ sice r = ( r )x2 x by choice of < + δ as r > ad x2 x< < ( r ). from the Biomial Theorem ( + δ)..4.5. Defiitio. Let (a ) N be a sequece. (a) (a ) N is called strictly icreasig if a < a 2 < a 3 <... ad strictly decreasig if a > a 2 > a 3 >... (a ) N is called strictly mootoe if it is strictly icreasig or strictly decreasig. (b) (a ) N is called icreasig if a a 2 a 3... ad decreasig if a a 2 a 3... (a ) N is called mootoe if it is icreasig or decreasig. The ext theorem is a powerful tool to compute limits. We ll see examples shortly.
Bouded ad mootoe sequeces 3.4.6. Mootoe covergece theorem. Every mootoe ad bouded sequece has a limit. Proof. Let (a ) N be our mootoe sequece. We first assume that (a ) N is icreasig, i.e. a a 2 a 3... By assumptio, (a ) N is bouded from above, which meas that the value set A = {a N} is bouded from above. By the completeess axiom.4.3 of R, A has a supremum s ad we claim that s is the limit of (a ) N. To see this, take ε > 0. Sice s is the least upper boud of A, s ε is ot a upper boud of A, i.e. there is some N N with s ε < a N. Now if N, the as (a ) N is icreasig as s=sup A s ε < a N a s. I particular a s ε for all N as required. It remais to do the case whe (a ) N is decreasig. However, i this case the sequece ( a ) N is icreasig (ad agai bouded). By what we have show ( a ) N has a limit ad by the scalar multiplicatio rule, also (a ) N has a limit.
4 Sequeces.5. Null sequeces..5.. Defiitio. A ull sequece is a sequece (a ) N that coverges to 0. Hece, by defiitio of (a ) N coverges to 0, a sequece (a ) N is a ull sequece if ad oly if ( ) ε > 0 N N N a < ε..5.2. Observatio. Let (a ) N be a sequece. (i) For every real umber r, (a ) N coverges to r if ad oly if (a r) N is a ull sequece. This is true because the defiitio of (a ) N coverges to r is literally the same as coditio ( ) above, formulated for the sequece (a r) N. (ii) The sequece (a ) N is a ull sequece if ad oly if ( a ) N is a ull sequece, because coditio ( ) for the sequece ( a ) N reads as Now otice that a = a. ε > 0 N N N a < ε. Null sequeces are of particular iterest for the followig reaso: Suppose we are give a sequece (a ) N ad we have a guess that it is coverget with a specific limit r; the by.5.2(i) we may work with the sequece (a r) N ad try to show that this is a ull sequece. I geeral this is easier to verify tha lim a = r..5.3. Properties of ull sequeces. Let (a ) N be a ull sequece. (i) If c R the (c a ) N is a ull sequece. (ii) If (b ) N is aother ull sequece, the (a + b ) N is a ull sequece. (iii) (Sadwich rule for ull sequeces) If (b ) N is a sequece with b a for all but fiitely may N, the (b ) N is a ull sequece. (iv) If (b ) N is a bouded sequeces, the (a b ) N is a ull sequece. (v) If a 0 for all, ad p > 0 is a real umber, the (a p ) N is a ull sequece. Proof. (i) holds true by the scalar multiplicatio rule.3. ad (ii) holds true by the sum rule.3.2 (iii). Let ε > 0. We eed to fid some N N such that b < ε wheever N. As (a ) N is a ull sequece by assumptio, there is some N N with a < ε for all N. Sice b a for all but fiitely may N, there is some N 2 N with b a for all N 2. Hece if we take N to be greater or equal tha N ad N 2 the both coditios apply: For each N we have b as N N2 as N N a < ε. (iv). Sice (b ) N is bouded there is some B R with b B. By (i), (B a ) N is also a ull sequece. 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 sequece.
a p < a k < (ε k ) k = ε for all N. Null sequeces 5 (v). Let k N be such that k < p. Let ε > 0 ad w.l.o.g. assume ε <. Sice 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 coclude that.5.4. The stadard list of ull sequeces. () lim = 0 for every p R with p > 0. p Proof. By.2.3(c) we kow that lim.5.3(v). = 0, hece () holds true by (2) lim c = 0 for every c R with c >. Proof. By.5.2(ii) we may also assume that c 0, hece we may assume c >. The ( c ) N is strictly decreasig ad by.4.6, ( c ) N coverges to some r R. We apply the scalar multiplicatio rule ad obtai c r = c lim c = lim c c = lim As c > this meas r = 0. c see questio 5 = lim c = r. (3) lim p c = 0 for all p R ad every c R with c >. Proof. Agai we may assume that c >. Let k N be with p < k. Sice p < k (see the defiitio of p ) we may apply the sadwich rule for ull sequeces ad replace p by k, hece we may assume that p N. Claim. The sequece ( c ) N is bouded for all c >. To see this, write c = + x with x > 0 (so x = c ). By the Biomial Theorem, c = ( + x) = + ( ) x + ( ) x 2 +... + 2 ( ) x ( ) x 2 = 2 ( ) x 2. 2 Hece if 2 x 2, the c, i other words c. Now 2 x 2 is equivalet to + 2 x. This shows that the terms 2 c are bouded above by if + 2 x. From this we clearly get claim. 2 Claim 2. lim c = 0 for all c >. To see this, apply claim to c (which is also > ). Thus ( c ) N is bouded. Sice ( c ) N is a ull sequece (by (2) applied to c) we see that c = c c is a product of a bouded ad a ull sequece. By.5.3(iv), this shows claim 2. Now we ca prove (3) as follows: Recall from the begiig of the proof that we work with p N. We apply claim 2 for p c (which also is > ). Thus = 0. By.5.3(v) the also lim c p p lim c = lim ( ) p = 0 c p
6 Sequeces c (4) lim! = 0 for all c R. Proof. Agai we may assume that c > 0. We have c! = c c c... c 2 3.... We fix a atural umber K c. For N with > K we the have c! = ck K! ( K)-times {}}{ c c c... c (K + )... ck K! c. Sice ( ) N is a ull sequece ad ck K! c is a costat, we see that ck K! c 0 as. By the sadwich rule for ull sequeces, we get (4). Please read the assertios of.5.4 carefully. For example, item (2) says that for every d R with d <, the sequece (d ) N is a ull sequece. Now give a example of a sequece (a ) N that is ot a ull sequece with the property a < for all N.
The algebra of limits 7.6. The algebra of limits..6.. Sadwich rule for sequeces. Let (a ) N, (b ) N ad (c ) N be sequeces with the followig properties. (a) Both (a ) N ad (c ) N coverge to the same real umber r. (b) a b c for all N. The also (b ) N coverges to r. Proof. We show that (b a ) N is a ull sequece. The by the sum rule, lim b exists ad is equal to lim (a + b a ) = lim a + lim (b a ) = r + 0 = r, as desired. By (b) we have 0 b a c a ad by the sadwich rule for ull sequeces.5.3(iii) it suffices to show that (c a ) N is a ull sequece. However, by the scalar multiplicatio rule, ( a ) N coverges with lim ( a ) = r ad so by the sum rule agai lim (c a ) = lim (c + ( a )) = lim c + lim ( a ) = r r = 0, as desired..6.2. Compatibility of limits ad order. Let (a ) N ad (b ) N be coverget sequeces. (i) If a b for all but fiitely may, the lim a lim b. (ii) If lim a < lim b, the for all but fiitely may we have a < b. Proof. Let r = lim a ad s = lim b. We first show (ii). Let ε = s r 2 all N, a r < ε, i particular a < r + ε. There is some N 2 N such that for all N 2, b s < ε, i particular s ε < b. Thus for all N := max{n, N 2 } we have a < r + ε = s ε < b. Therefore, for all except possibly for =,..., N we have a < b. (i). Suppose by way of cotradictio that a b for all but fiitely may ad r > s. By (ii) with iterchaged role of (a ) N ad (b ) N we kow that for all, hece r + ε = r+s 2 = s ε. There is some N N such that for but fiitely may, a > b. But this cotradicts the assumptio a b for all but fiitely may.
8 Sequeces.6.3. The algebra of limits. Let (a ) N ad (b ) N be coverget sequeces. (i) (Sum rule) The sum (a + b ) N of the sequeces (a ) N ad (b ) N is coverget ad lim (a + b ) = lim a + lim b. (ii) (Muliplicatio rule) (a b ) N (called the product of the sequeces (a ) N ad (b ) N ) is coverget ad lim (a b ) = ( lim a ) ( lim b ). Observe that the scalar multiplicatio rule is a special istace of the multiplicatio rule: If c R, the the scalar multiplicatio rule is the multiplicatio rule applied to the costat sequece (c) N. (iii) (Divisio rule) If b 0 for all N, ad (b ) N is ot a ull sequece, the also ( a b ) N is coverget ad (iv) (Modulus rule) ( a ) N is coverget ad lim (a ) = lim a. b lim b lim a = lim a. (v) (Root rule) If a 0 for all N ad p N the also ( p a ) N is coverget ad lim p a = p lim a. Proof. (i) is the sum rule.3.2 ad is stated here agai for the sake of completeess. For the rest of the proof we write r = lim a ad s = lim b (ii). We must show that (a b ) N coverges to r s ad it suffices to show that (a b r s) N is a ull sequece. Our strategy here is the followig: We shall estimate the modulus of the terms a b r s by terms of aother ull sequece ad the apply the sadwich rule for ull sequeces.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 sequece with th term a b s + a r s is a ull sequece: (a) Sice a r ( ), ( a r ) N is a ull sequece ad by the scalar multiplicatio rule also ( a r s) N is a ull sequece.
The algebra of limits 9 (b) Sice b s ( ), ( b s ) N is a ull sequece. Sice (a ) N is coverget, it is bouded (cf..4.2). Now by.5.3(iv), the product of a ull sequece ad a bouded sequece is itself a ull sequece. Thus ( a b s ) N is a ull sequece Applyig the sum rule to (a) ad (b) shows that the sequece ( a b s + a r s ) N is a ull sequece. We may therefore apply the sadwich rule for ull sequeces.5.3(iii) i combiatio with our computatio ( ) to obtai that also ( a b r s ) N is a ull sequece. This is what we wated to show. (iii). So here s = lim b 0 by assumptio. We show that ( b s ) N is a ull sequece, i other words ( b ) N is coverget ad lim ( b ) = lim b ; item (iii) the follows from the multiplicatio rule applied to (a ) N ad ( b ) N. By the defiitio of covergece of (b ) N ad the choice ε = s 2, there is some K N such that b s < s for all K. 2 We ow apply the derived triagle rule.2.(iv) ad get for K: b = s + (b s) derived triagle rule I particular if K, the b 2 s ad s b s as b s < s 2 b s = s b ( ) b s = s b s s b b s 2 s = 2 s 2 s b. Sice (b ) N coverges to s, s b 0 ( ). By the scalar multiplicatio rule (or by (ii)), also 2 s s b 2 0 ( ). So by the sadwich rule for ull sequeces.5.3(iii) applied to ( ), also ( b s ) N is a ull sequece. (iv). We must show that lim a = r. By the derived triagle iequality we kow for each N: a r a r. Sice the right had side here teds to 0 as we ca apply the sadwich rule for ull sequeces.5.3(iii) ad get that a r 0 as. This is what we had to show. (v). If (a ) N is a ull sequece, the we ca apply.5.3(v) ad see that also ( p a ) N is a ull sequece, which cofirms (v). Therefore we may assume that (a ) N is ot a ull sequece, i other words For each N we have r = lim a 0. (+) ( p a p r) ( p a p + p a p 2 p r +... + p a p r p 2 + p r p ) = a r. Sice a, r 0 we ca estimate (++) p r p p a p + p a p 2 p r +... + p a p r p 2 + p r p. By combiig (+) ad (++) we obtai p a p r p r p a r for all N. s 2.
20 Sequeces Sice r 0 this ca be writte as p a p r p r p a r for all N. Hece by the sadwich rules for ull sequeces.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 sequece. This shows that ( p a ) N coverges to p r..6.4. Example. Let a = 23 2 2+ 2 +. We wat to decide whether (a 3 ) N is coverget ad if it is coverget to fid its limit. We caot apply the divisio rule directly, because either the sequece ( 2 3 2 ) N or the sequece (2 + 2 + 3 ) N is coverget. The idea is to apply the followig strategy: We idetify the domiat term i the expressio 23 2 2+ 2 + 3, 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 2 2 + 3 + Now we ca apply the algebra of limits together with the stadard list of ull a = 23 2 2 + 2 + 3 = ( 2 3 2 ) 3 (2 + 2 + 3 ) = 3 sequeces to see that the ew umerator 2 3 coverges to 2 (covice yourself that this is true) ad the ew deomiator 2 + 3 + coverges to. Hece by the divisio rule, the quotiet (which is our origial a ) coverges to 2 = 2. Let us do aother example of this type. Let a = 3!+4 5 2 +6!. The domiat term is! (this ca be see from the stadard list of ull sequeces, see.5.4 (4), (3)). Therefore 3! + 4 a = 5 2 + 6! = 3 + 4! 5 2! + 6 From the stadard list of ull sequeces, we kow that 4! ad 2! coverge to 0. Hece by the algebra of limits, a coverges to 3 6 = 2. Be aware that the recipe of.6.4 is ot always applicable i a obvious way. For example oe ca use this strategy to compute the limit of the sequece ( + ) N by writig ( + ) ( + + ) + = + + ad the applyig the recipe of.6.4. This will be carried out i the examples classes. O the other had the recipe of.6.4 is ot always successful, for example look at the sequeces (! ) N. The it is ot clear what the domiat term is. It is ot difficult to see that (! ) N is a ull sequece (this is treated i the examples classes). Or, look at the sequece ( c) N where c > 0. We claim that ( c) N coverges to. 3
The algebra of limits 2 To see this, we first assume that c. the ( c) N is bouded below by ad bouded above by c. Moreover ( c) N is decreasig ad so by the mootoe covergece theorem, it coverges to some r R. By the compatibility of limits ad order (cf..6.2), r. We must show that r =. We have r = lim c by the root rule = lim c = lim 2 c Now for every coverget sequece (a ) N, also the sequece (a 2 ) N is coverget ad lim a = lim a 2. Applyig this to our sequece ( c) N gives lim 2 c = lim c = r. So r = r ad r, which implies r =. This shows the claim if c. If c <, the c > ad from the previous case we get lim c =. Now we apply the divisio rule to obtai lim c = lim = lim = lim c c lim c = =. Other sequeces eve eed more hard work. For example look at the sequece ( ) N. Is it coverget? We tackle this questio ow. First a lemma..6.5. 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! {}}{..., sice the last factor k + is < for k >. Thus ( k We ca ow show that.6.6. ( ) N is coverget with lim =. Proof. For all 3 we have by.6.5: ( ) ( + ) = + < k k= < + k k = + = + +, k= k= ) < k k!. k a strict iequality occurs whe k = thus ( + ) +, i other words + + for all 3.
22 Sequeces By the mootoe covergece theorem ( ) N coverges to some r ad we have to show that r =. We have r = lim 2 = lim 2 = lim ( 2 2 2 ) = lim ( 2 2 ) = = ( lim 2 2) ( lim ). Now lim 2 2 = (see the computatio before.6.5) ad by the root rule of.6.3, lim = lim. We obtai r = = r. Hece r = r ad r = as desired. lim
The Bolzao-Weierstraß Theorem ad Cauchy sequeces 23.7. The Bolzao-Weierstraß Theorem ad Cauchy sequeces. This sectio cosist of two further fudametal statemets (.7. ad.7.2) at the begiig of real aalysis. They are ot examiable ad icluded for the iterested reader. Surprisigly the followig is true..7.. Theorem. (Bolzao-Weierstraß) Every bouded sequece has a coverget subsequece. Proof. Let (a ) N be our bouded sequece, 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. Sice A k A for all k ad A is bouded by assumptio, also A k is bouded. So by the completeess axiom for R (.4.3), A k has ideed a supremum i R, which justifies the defiitio of s k. As A k [ B, B], also s k [ B, B]. Sice A A 2 A 3... we have s s 2... Hece (s k ) k N is a mootoe ad bouded sequece. So by.4.6, (s k ) k N coverges to some real umber r. We ow defie a subsequece (a i ) i N of (a ) N by defiig < 2 <... as follows: Choice of. Sice s k r (k ) there is some K N such that r s k for all k K. Sice s K is the supremum of A K, there is some {K, K +,...} with s K < a s K. I particular a s K. We ca ow estimate the distace of a to r by a r = a s K + s K r a s K + s K r + = 2. Choice of i, provided we have already chose i. Sice s k r (k ) there is some K N such that r s k i for all k K. We may certaily choose K > i if ecessary. Sice s K is the supremum of A K, there is some i {K, K +,...} with s K i < a i s K. I particular s K a i i. Agai we estimate the distace 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 = 2 i. Thus we have defied < 2 < 3 <... ad we kow that for all i N, a i r 2 i. By the sadwich rule for ull sequeces.5.3(iii), this meas that a i r 0 (i ), i.e. (a i ) i N coverges to r..7.2. Defiitio. (Augusti-Louis Cauchy, 789-857) A sequece (a ) N is called a Cauchy sequece (proouced like co she ), if for every ε > 0 there is some N N such that for all, k N we have a a k < ε.
24 Sequeces Hece i a Cauchy sequece, the terms get closer ad closer to each other. However, oe has to uderstad the expressio get closer ad closer to each other i the sese of the defiitio above, because there are sequeces (a ) N with a + a 0 as (so here also i some sese the terms get closer ad closer to each other), such that (a ) N is ot Cauchy. We will see a example of this type later, whe we are dealig with series. For the impatiet reader: choose a = + 2 +... +. Every coverget sequece (a ) N (with limit r) is Cauchy: Let ε > 0 ad choose N N with a r < ε 2 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 2 + ε 2 = ε..7.3. Theorem. Every Cauchy sequece coverges. Proof. Let (a ) N be our Cauchy sequece. 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 coverget subsequece (a i ) i N. Let R be the limit of that subsequece. Fially we will show that (a ) N coverges to r. Applyig the defiitio of a Cauchy sequece 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 cotaied 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 coverget subsequece (a i ) i N, which has a limit r. We show that (a ) N coverges to r. Let ε > 0. Sice lim i a i = r there is some I N such that for all i I we have a i r < ε 2. Sice (a ) N is a Cauchy sequece there is some N N such that for all, m N we have a a m < ε 2. 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. Sice i, N we have a a i < ε 2. Sice i I we have a i r < ε 2. Hece from ( ) we get a r < ε 2 + ε 2 = ε. Hece a sequece coverges if ad oly if it is Cauchy. The advatage of the otio of a Cauchy-sequece is that we do ot eed to kow the limit of the sequece if we wat to cofirm covergece. We will see i the ext sectio that some limits are very hard to compute, whereas it is ofte easy to see why a particular sequece is coverget.
Defiitio ad first examples 25 2. Series 2.. Defiitio ad first examples. Give a sequece (a ) N we ca form a ew sequece (s ) N defied by s := a s 2 := a + a 2 s 3 := a + a 2 + a 3. s := a + a 2 +... + a. The umber s is called the -th partial sum of the sequece (a ) N. As usual, a + a 2 +... + a is deoted by k= a k 2... Mai Defiitio for series. If the sequece (s ) N of partial sums of a sequece (a ) N coverges, the we say the series a coverges. = Moreover if lim s = r we write a = r. = If (s ) N diverges, the we say the series = a diverges. Please follow this defiitio carefully ad do ot iterpret other thigs ito the expressio = a. By defiitio, = a oly exists if the sequece of partial sums of (a ) N is coverget. For example it does ot make sese to talk about the object = ( ) apart from askig whether it is a coverget series. The situatio is quite similar to the oe for limits: The expressio lim a does ot make sese if (a ) N does ot coverge. Eve if a series coverges oe has to carefully iterpret this covergece. See the warig after 2..7 below. 2..2. Example. (Geometric series) For every x R ad each N we have ( x) ( + x + x 2 +... + x ) = + x + x 2 +... + x x x 2 x 3... x + = x + Hece if x <, the x + 0 as (from the stadard list of ull sequeces) ad s := + x + x 2 +... + x = x+ x x as.
26 Series Therefore the so called geometric series k=0 xk coverges with k=0 x k = x for x <. 2..3. Propositio. If = a coverges, the (a ) N is a ull sequece. Proof. Let r = = a ad for each N let s = k= a k be the th partial sum. The a + = s + s r r as, by the algebra of limits. Hece a 0 as. The coverse of 2..3 is ot true, as the ext example shows. 2..4. Example. (Harmoic series) The harmoic series k= k is diverget. Proof. Let (s ) N be the sequece of partial sums of ( ) N. It suffices to show by iductio o that s 2 + 2 for each N. For =, we have s 2 = a + a 2 = + 2. For the iductio step we have s 2 + s 2 = 2 + k=2 + + 2 k > k=2 + By the iductio hypothesis we have s 2 + 2, hece 2 + = 2 2 + = 2. s 2 + = s 2 + (s 2 + s 2 ) + 2 + 2 = + +. 2 2..5. Covergece of series with positive terms. If (a ) N is a sequece with a 0 for all N, the the followig are equivalet: (i) = a is coverget. (ii) The sequece of partial sums of (a ) N is bouded. Proof. Let s be the th partial sum of (a ) N. (i) (ii). By (i), (s ) N is coverget. Hece by.4.2, (ii) holds. (ii) (i). As a 0 for all, (s ) N obviously is icreasig. By (ii), (s ) N is bouded. Hece by the mootoe covergece theorem, (s ) N is coverget. This shows (i). 2..6. Example. The series = 2 is coverget.
Defiitio ad first examples 27 Proof. Let (s ) N be the sequece of partial sums of ( 2 ) N. The s = + 2 2 + 3 3 + 4 4 +... + + 2 + 2 3 + 3 4 +... + ( ) = = + ( 2 ) + ( 2 3 ) + ( 3 4 ) +... + ( ) = + 2 + 2 3 + 3 4 +... + = + < 2 Hece by 2..5, k= k 2 is coverget. 2..7. Example. (Alteratig harmoic series) The alteratig harmoic series ( ) + ( = 2 + 3 4 +...) is coverget. k= = Proof. Let (s ) N be the sequece of partial sums of ( ( )+ ) N. For N we the have s 2 = ( 2 ) + ( 3 4 ) +... + ( 2 2 ) = (because k k = (k ) k ) = (2 ) 2 + (4 ) 4 +... + (2 ) 2 = ( ) = (2k ) 2k 2 (2k ) 2 k 2 (see 2..6) k2 k= Equality ( ) shows that (s 2 ) N is icreasig ad so (s 2 ) N is mootoe ad bouded (by k= k ). By the mootoe covergece theorem, s 2 2 coverges to some real umber r. O the other had s 2+ = s 2 + a 2+ = s 2 + 2+ which coverges to r + 0 = r by the sum rule. It follows that also (s ) N coverges to r (exercise!) Sice the alteratig harmoic series coverges, oe might be tempted to iterpret ( ) + = as the sum of all elemets i the set {, 2, 3, 4,...}. However this is false! The reaso is that the summatio order plays a crucial role: Amazigly, for every real umber r, oe ca rearrage the sequece, 2, 3, 4,... to obtai a ew sequece (b ) N such that k= b is coverget with limit r. This statemet blatatly cotradicts our aive ituitio; more precisely it tells us that we have to be very careful whe attachig ituitio to coverget series. k= k=
28 Series For those who are iterested, we give a proof (ad a precise statemet) of the Warig above. This is ot examiable. 2..8. Riema s Rearragemet Theorem. Let (a ) N be a sequece such that = a is coverget, but = a is diverget; such series are sometimes called coditioally coverget (the choice a = ( ) is a example here). The for every real umber r, there is a bijectio f : N N (such maps are called permutatios of N ad this is the precise formulatio of rearragig ) such that = a f() is coverget with a f() = r. = Proof. The proof makes use of some results i sectio 2.2 without metioig this explicitly ad you might wat to read this sectio first. For example the assumptio that = a is diverget says that = a is ot absolutely coverget. We may assume that a 0, otherwise we replace a by a ad r by r (otice that if = a f() = r, the also = a f() = r). First we defie two ew sequeces. For N let { a if a 0 b = 0 if a 0 { 0 if a 0 c = a if a 0 Obviously a = b + c ad a = b c. Claim. = b ad = c are diverget ad for every x R, there is some N such that = b k > x > = c k. Proof of claim. Suppose = b coverges. The also = c = = a = b coverges ad so = a = = b = c coverges, i cotradictio to our assumptio. Thus = b is diverget. Sice b 0 for all, this is equivalet to sayig that for every x R, there is some N such that = b k > x. Similarly, the assumptio that = c coverges leads to a cotradictio. Sice c < 0 for all, for every x R, there is some N such that x > = c k. This shows claim. We ow defie f : N N by defiig f() iductively. We choose f() =. Assumig f(),..., f() has already bee defied, we defie f( + ) by { mi{k N \ {f(),..., f()} a k 0} if a f() +... + a f() r f( + ) = mi{k N \ {f(),..., f()} a k < 0} if a f() +... + a f() > r Note that this makes sese, sice by claim, there are ifiitely may k such that a k 0 ad ifiitely may k such that a k < 0. Claim 2. f is bijective. Proof of claim 2. By the iductive defiitio of f, we kow f( + ) f(i) for ay i <. Hece f is ijective. I order to show that f is surjective, we show by iductio o l N that l is i the image of f. For l = this holds true as f() =. Suppose,..., l are
Defiitio ad first examples 29 i the image of f, but l + is ot i the image of f.,..., l {f(),..., f( 0 )}. Case. a l+ 0. The ( ) l + = mi{k N \ {f(),..., f()} a k 0}, Take 0 N such that sice a l+ 0 ad by assumptio l + is ot i the image of f; as {,..., l} {f(),..., f( 0 )} {f(),..., f()}, o k < l + is i N \ {f(),..., f()}. Sice l + is ot i the image of f, equatio ( ) ad the defiitio of f( + ) show that for all 0 we have a f() +... + a f() > r. Cosequetly for every 0, f( + ) is the least idex k {f(),..., f()} for which a k < 0. Hece f( 0 + ), f( 0 + 2), f( 0 + 3),... is a eumeratio of all idices k > {f(),..., f( 0 )} with the property that a k < 0. These terms a k all are terms i the sequece (c ) N. However, by claim, for every x R the partial sums k= c k is smaller tha x for all but fiitely may, which cotradicts that for all 0 we have a f() +... + a f() > r. This fiishes the proof of claim 2 i case. Case 2. a l+ < 0. This is aalogue to the proof i case. It remais to show that = a f() = r. Let ε > 0. We have to fid some N N with r ε < a f() +... + a f() < r + ε for all N. Sice (a ) N is a ull sequece, there is some K N such that a k < ε for all k K. Claim 3. There is some N N with {,..., K} {f(),..., f(n)} ad r ε < a f() +... + a f(n) < r + ε Proof of claim 3. By claim 2, f is surjective ad therefore there is some N 0 N with {,..., K} {f(),..., f(n 0 )}. It follows that for every N N 0, f(n + ) {f(),..., f(n)} {,..., K}, thus f(n + ) K. Case. a f() +... + a f(n0) r. Sice ( k= b k) N exceeds every real umber (by claim ), there must be some N N 0 such that ( ) ( ) a f() +... + a f(n) r but a f() +... + a f(n+) > r As f(n + ) K we have a f(n+) < ε. By ( ), the defiitio of f(n + ) says a f(n+) 0, thus 0 a f(n+) < ε. But ow ( ) ad ( ) imply r ε < a f() +... + a f(n) < r + ε. Case 2. a f() +... + a f(n0) > r. Sice ( k= c k) N gets smaller tha every real umber (by claim ), there must be some N N 0 such that ( ) ( ) a f() +... + a f(n) > r but a f() +... + a f(n+) r As f(n + ) K we have a f(n+) < ε. By ( ), the defiitio of f(n + ) says a f(n+) < 0, thus ε < a f(n+) < 0. But ow ( ) ad ( ) imply r ε < a f() +... + a f(n) < r + ε.
30 Series This shows claim 3. Claim 4. For all N, if {,..., K} {f(),..., f()} ad r ε < a f() +...+a f() < r + ε, the also r ε < a f() +... + a f(+) < r + ε. Proof of claim 4. As f( + ) {f(),..., f()} {,..., K} we kow that f( + ) > K ad therefore a f(+) < ε Case. a f() +... + a f() r. By defiitio of f( + ) we the have a f(+) 0, thus 0 a f(+) < ε. By assumptio r ε < a f() +... + a f() r. Thus r ε < a f() +... + a f(+) < r + ε as desired. Case 2. a f() +... + a f() > r. By defiitio of f( + ) we the have a f(+) < 0, thus ε < a f(+) < 0. By assumptio r < a f() +... + a f() < r + ε. Thus r ε < a f() +... + a f(+) < r + ε as desired. This fiishes the proof of claim 4. We ca ow fiish the proof of the rearragemet theorem. Recall that we were workig with a real umber ε > 0 ad we have to ame some N N with r ε < a f() +... + a f() < r + ε for all N. We take N as i claim 3. It is the clear that for all N with N we have {,..., K} {f(),..., f()}. Hece we ca ru a iductio o startig at = N usig claim 4 to obtai r ε < a f() +... + a f() < r + ε for all N.