Notes #3 Sequences Limit Theorems Monotone and Subsequences Bolzano-WeierstraßTheorem Limsup & Liminf of Sequences Cauchy Sequences and Completeness
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1 Notes #3 Sequeces Limit Theorems Mootoe ad Subsequeces Bolzao-WeierstraßTheorem Limsup & Limif of Sequeces Cauchy Sequeces ad Completeess This sectio of otes focuses o some of the basics of sequeces of real umbers ad, to provide a geeralizatio, complex umbers, as well. We ll itroduce ad elaborate o the familiar otios of limits ad covergece. Historically, covergece of sequeces were first cosidered rigorously i the early 1800 s by Cauchy ad Bolzao (idepedetly). Cosider the followig properties of real ad complex umbers. Theorem 3.1: (Triagle Iequality). For ay u, v C, u + v u + v. Recall that for the complex umber z = a + bi we have z = z z = a 2 + b 2. I the case that u, v are purely real, the we are simply talkig about the absolute value. I this case, the proof of the Triagle Theorem is straightforward 0 u + v 2 =(u + v) 2 = u 2 +2uv + v 2 u 2 +2 u v + v 2 =( u + v ) 2. From this, the iequality follows immediately. There are a umber of useful corollaries to the theorem, icludig the followig two: 3.1a u v u w + w v. 3.1b u v u v. The fuctio provides a meas of providig a metric for R ad C. I geeral, a metric o a set S is a fuctio d : S S R that satisfies the followig p,q,r S: M1 d(p, q) 0 M2 d(p, q) =0 p = q M3 d(p, q) =d(q,p) M4 d(p, q) d(p, r) +d(r, q).
2 M4 is also kow as the triagle iequality. A metric, as its ame implies, is a way of measurig a distace betwee elemets of a set. There are typically may differet fuctios that could serve as metrics for a give set. For our purposes we will use the Euclidea metric defied o C (ad by iclusio, R) byd(u, v) = u v. For a discussio of covergece sequeces i a metric space X, we eed the otio of a eighborhood of a poit. Give a poit x, we are iterested i the set of poits that are close to x. The set N ɛ (x) ={y X d(x, y) <ɛ} is said to be a ɛ-eighborhood of x. Whe restricted to R, wehaven ɛ (x) =(x ɛ, x + ɛ). This is just the symmetric ope iterval of radius ɛ cetered o x. I the complex plae (or R 2 for that matter), N ɛ (z) is just the ope disk cetered at z. Sequeces ad Covergece A sequece of real umbers is a fuctio x : N R. To represet the sequece we geerally focus o the ordiate values {x() : N} ad usually write x() =x for the th term of the sequece ad {x } for the sequece itself. Similarly, for sequeces of complex umbers. Defiitio:We say that a sequece {x } (real or complex) coverges if there is a poit l such that for each ɛ>0 0 N such that x N ɛ (l) 0. I this case we write lim x = lim x = l or x l. (This is the same as sayig that x l <ɛfor sufficietly large values of.) Defiitio:If a sequece does ot coverge the it is said to diverge. For real sequeces we say that {x } diverges to if for each ɛ>0 0 N such that x >ɛ > 0. We write, loosely, lim x =. (This is the same as sayig that x is arbitrarily large for sufficietly large values of.) Defiitio: A sequece {x } is bouded if there is value M R such that x M, N. EXERCISES 1. Show that the limit of a coverget sequece is uique. 2. Show that every coverget sequece is bouded. 3. For each of the followig real sequeces, show that they coverge usig the defiitio. a) { 2 +3} b) { 2 +1} c) { 1 ( 1) } { ( d) )} 4. Show that each of the followig sequeces diverge. a) {( 1) + 1 } b) {si(π 2 )} 5. Let 0 a<1. Prove that lim a =0. 6. Let {x } be a sequece i R with lim x = l. Show that lim x 2 = l2. 7. Let {x } R + with lim x = l. Show that lim x = l.
3 8. Prove or give a couterexample: a) If x l, the x l. b) If x l, the x l. Limit Theorems Some importat properties of limits of sequeces are icluded i the followig: Theorem 3.2: Let {a } ad {b } be sequeces i R with lim a = A ad lim b = B. The, a) lim(a ± b )=A ± B. b) lim(a b )=A B. c) If b 0, ad B 0, the lim(b 1)=B 1. d) lim(a + c) =A + c, for c R. e) lim ca = ca, for c R. f) If b 0, ad B 0, the lim( a b )= A B. Proof: Note that d), e), ad f) are almost immediate cosequeces of a), b), ad c). We ll leave the proof of these ad parts a) ad c) for the exercises. For part b) cosider the sequece {a b }. To show that a b AB we eed to show that for each ɛ>0 there is a 0 N such that a b AB <ɛfor all 0. Sice {a } ad {b } coverge they are both bouded. I particular, let α be ay costat so that a α, for all. Note that, a b AB = a b a B + a B AB a b a B + a B AB = a b B + B a A α b B + B a A. Now, let ɛ>0ad sice a A ad b B we ca choose a positive iteger a so ɛ that a A < 2( B +1) wheever > a, ad b so that b B < ɛ 2α for all > b. If we choose 0 = max{ a, b }, we get (for > 0 ) a b AB α b B + B a A <α ( ɛ ) ɛ + 2α 2( B +1) < ɛ 2 + ɛ 2 = ɛ.
4 Theorem 3.3: Suppose {a } coverges to 0, ad {b } is bouded. The, a b 0. Theorem 3.4: (Squeeze) Suppose {a }, {b }, ad {c } are sequeces with a b c,. If lim a = lim c = l, the b l. The followig theorem addresses the covergece of some bechmark sequeces. Theorem 3.5: a) If p>0, the lim 1 p =0. b) If p>0, the lim p =1. c) lim =1. d) If p>0 ad α R, the lim α p =0. e) If p < 1, the lim p =0. f) For all p R, lim p! =0. EXERCISES 1. Complete the proof of Theorem (a) If {a } ad {a + b } both coverge, does {b } coverge? (b) If {a } ad {a b } both coverge, does {b } coverge? 3. For each of the followig determie the limit of the give sequece. { 3 2 } } a) 5 2 b) { { } 1 c) 1+ 1 d) { e) 2 + } f) { ( + a )} 4. Let {x } be a real sequece with x 0 ad x 0 for all. Prove: lim x si( 1 )= Suppose {a } is a positive real sequece ad that a +1 l. Show that {a } coverges to 0 if l<1 ad diverges if l>1. What happes if l = a 1?.
5 Mootoe Sequeces ad Subsequeces Defiitio: A sequece {a } of real umbers is said to be: a) mootoe icreasig if a a +1 for all ; b) mootoe decreasig if a a +1 for all ; ad c) mootoe if it satisfies either a) or b). Theorem 3.6: If a bouded sequece of reals {x } is mootoe, the it coverges. Nested Itervals Property: A sequece of closed itervals I = {I } where I = [a,b ]isested if a a +1 b +1 b for all. IfI is ested the I =. Defiitio: Give a sequece {a } ad a sequece { k } of atural umbers, the the sequece {a k } is said to be a subsequece of {a }.Ifa k m, the m is said to be the subsequetial limit. l, the every subsequece also co- Theorem 3.7: For the sequece {a },ifa verges to l. Bolzao-Weierstrass Theorem First some essetial defiitios ad results: Defiitio: Let E X (where X is R or C) the l X is a limit poit of E if for each ɛ>0 there exists a poit p E Nɛ (l). Otherwise, l is a isolated poit. Theorem 3.8: Let E X, the: a) If E has a limit poit l, the for each ɛ>0,e N ɛ (l) is a ifiite set. b) If E has a limit poit l, the there is a sequece {a } Esuch that a l, ad a l for each. Oe cosequece of the above observatios is that a fiite set has o limit poits. We should ote that may ifiite sets have o limit poits either for example, Z. However, if we have a ifitie set that is also bouded, the set defiitely has a limit poit. This is stated i the followig theorem. Theorem 3.9: (Bolzao-Weierstrass) Every bouded ifiite set has a limit poit. The sequetial versio of this theorem (sometimes referred to as the B-W Thm. too) is: Theorem 3.9 (alterate BW): Every bouded sequece has a coverget subsequece.
6 EXERCISES 1. Show that 0 ad 2 are the oly subsequetial limits of the sequece {1 ( 1) }. 2. (a) Costruct a sequece whose subsequetial limits are exactly 0, 1, ad. (b) Costruct a sequece whose complete set of subsequetial limits is deumerable. 3. Fid all subsequetial limits of the followig sequeces i R: a) { si π } b) 1 ( 1) 2 4. Describe the set of limit poits ad the set of isolated poits for each of the followig sets: a) { ( 1) + 1 : N} b) { 1 ( 1) : N } c) (0, 1) {2} d) N e) R \ Q f) Q (0, 1) 5. Let A R with sup A = α. Show that if α A, the α is a limit poit of A. 6. Costruct sets i R ad C \ R such that each: a) has exactly two limit poits b) is ifiite with o limit poits c) is coutable with a coutable umber of limit poits d) is coutable with a ucoutable umber of limit poits 7. Prove that every sequece i R has a mootoe subsequece. Limsup ad Limif Defiitio: Let {a } be a sequece i R. The the limit superior of {a } is defied as lim sup a = if k The limit iferior of {a } is defied to be lim if a = sup k sup {a }. k if {a }. k It should be oted that lim if a diverges to ±. = lim sup a if ad oly if a coverges i R or
7 Cauchy Sequeces Defiitio: A sequece {p } is a Cauchy sequece if for each ɛ > 0 there is a positive iteger ɛ such that p m p <ɛfor all m, > ɛ. Note that if {x } is Cauchy, the lim x k+1 x k =0. k Theorem 3.10: (a) Every Cauchy sequece is bouded. (b) If {p } is Cauchy ad has a coverget subsequece, the {p } coverges. Theorem 3.11: A sequece of real umbers is Cauchy if ad oly if it coverges.
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