Mathematical Analysis. Min Yan

Mthemticl Anlysis Min Yn Februry 4, 008

Contents 1 Limit nd Continuity 7 11 Limit of Sequence 8 111 Definition 8 11 Property 13 113 Infinity nd Infinitesiml 18 114 Additionl Exercise 0 1 Convergence of Sequence Limit 1 11 Necessry Condition 1 Supremum nd Infimum 4 13 Monotone Sequence 7 14 Convergent Subsequence 30 15 Convergence of Cuchy Sequence 35 16 Open Cover 36 17 Additionl Exercise 38 13 Limit of Function 39 131 Definition 39 13 Vrition 4 133 Property 45 134 Limit of Trignometric Function 49 135 Limit of Exponentil Function 51 136 More Property 55 137 Order of Infinity nd Infinitesiml 58 138 Additionl Exercise 60 14 Continuous Function 6 141 Definition 6 14 Uniformly Continuous Function 66 143 Mximum nd Minimum 68 144 Intermedite Vlue Theorem 69 145 Invertible Continuous Function 71 146 Inverse Trignometric nd Logrithmic Functions 74 147 Additionl Exercise 78 Differentition 81 1 Approximtion nd Differentition 8 11 Approximtion 8 1 Differentition 84 13 Derivtive 85 3

4 CONTENTS 14 Tngent Line nd Rte of Chnge 87 15 Rules of Computtion 90 16 Bsic Exmple 9 17 Derivtive of Inverse Function 96 18 Additionl Exercise 99 Appliction of Differentition 99 1 Mximum nd Minimum 100 Men Vlue Theorem 10 3 Monotone Function 105 4 L Ĥospitl s Rule 107 5 Additionl Exercise 11 3 High Order Approximtion 114 31 Qudrtic Approximtion 114 3 High Order Derivtive 116 33 Tylor Expnsion 11 34 Reminder 16 35 Mximum nd Minimum 18 36 Convex nd Concve 130 37 Additionl Exercise 134 3 Integrtion 139 31 Riemnn Integrtion 140 311 Riemnn Sum 140 31 Integrbility Criterion 143 313 Integrbility of Continuous nd Monotone Functions 146 314 Properties of Integrtion 148 315 Additionl Exercise 154 3 Antiderivtive 158 31 Fundmentl Theorem of Clculus 159 3 Antiderivtive 164 33 Integrtion by Prts 169 34 Chnge of Vrible 171 35 Additionl Exercise 174 33 Topics on Integrtion 178 331 Integrtion of Rtionl Functions 178 33 Improper Integrtion 183 333 Riemnn-Stieltjes Integrtion 187 334 Bounded Vrition Function 193 335 Additionl Exercise 199 4 Series 03 41 Series of Numbers 04 411 Sum of Series 04 41 Comprison Test 07 413 Conditionl Convergence 11 414 Additionl Exercise 15 4 Series of Functions 18

CONTENTS 5 41 Uniform Convergence 18 4 Properties of Uniform Convergence 3 43 Power Series 9 44 Fourier Series 34 45 Additionl Exercise 40 5 Multivrible Function 45 51 Limit nd Continuity 46 511 Limit in Eucliden Spce 46 51 Topology in Eucliden Spce 49 513 Multivrible Function 54 514 Continuous Function 56 515 Multivrible Mp 58 516 Exercise 6 5 Multivrible Algebr 64 51 Liner Trnsform 64 5 Biliner nd Qudrtic Form 68 53 Multiliner Mp nd Polynomil 75 6 Multivrible Differentition 85 61 Differentition 86 611 Differentibility nd Derivtive 86 61 Prtil Derivtive 87 613 Rules of Differentition 91 614 Directionl Derivtive 94 6 Inverse nd Implicit Function 98 61 Inverse Differentition 98 6 Implicit Differentition 30 63 Hypersurfce 304 64 Exercise 307 63 High Order Differentition 309 631 Qurtic Approximtion 309 63 High Order Prtil Derivtive 313 633 Tylor Expnsion 316 634 Mximum nd Minimum 319 635 Constrined Extreme 33 636 Exercise 330 7 Multivrible Integrtion 333 71 Riemnn Integrtion 334 711 Volume in Eucliden Spce 334 71 Riemnn Sum 339 713 Properties of Integrtion 343 714 Fubini Theorem 347 715 Volume in Vector Spce 35 716 Chnge of Vrible 356 717 Improper Integrtion 36

6 CONTENTS 718 Exercise 367 7 Integrtion on Hypersurfce 369 71 Rectifible Curve 369 7 Integrtion of Function on Curve 373 73 Integrtion of 1-Form on Curve 376 74 Surfce Are 379 75 Integrtion of Function on Surfce 38 76 Integrtion of -Form on Surfce 384 77 Volume nd Integrtion on Hypersurfce 389 78 Exercise 394 73 Stokes Theorem 396 731 Green Theorem 396 73 Independence of Integrl on Pth 401 733 Stokes Theorem 406 734 Guss Theorem 411 735 Exercise 416 8 Clculus on Mnifold 417 81 Differentible Mnifold 418 811 Mnifold 418 81 Tngent Spce 418 813 Differentil Form 418 814 Topics 418 8 Integrtion on Mnifold 418 81 Prtition of Unity 418 8 Integrtion 418 83 Stokes Theorem 418

Chpter 1 Limit nd Continuity 7

8 CHAPTER 1 LIMIT AND CONTINUITY 11 Limit of Sequence A sequence is n infinite list x 1, x, x 3,, x n, x n+1, The sequence cn be briefly denoted s {x n } The subscript n is clled the index nd does not hve to strt from 1 For exmple, x 5, x 6, x 7,, x n, x n+1,, is lso sequence, with the index strting from 5 In this chpter, the terms x n of sequence re ssumed to be rel numbers nd cn be plotted on the rel number line A sequence hs limit l if the following impliction hppens: n is big = x n is close to l Intuitively, this mens tht the sequence ccumultes round l However, to give rigorous definition, more considertion is required for the mening of big, close nd implies The bigness of number n is usully mesured by n > N for some big N (for exmple, we sy n is in the thousnds if N = 1, 000) The closeness between two numbers u nd v is usully mesured by (the smllness of) the size of u v The impliction mens tht the predetermined smllness of x n l my be chieved by the bigness of n 111 Definition Definition 111 A sequence {x n } of rel numbers hs limit l (or converges to l), nd denoted lim n x n = l, if for ny ɛ > 0, there is N, such tht n > N = x n l < ɛ (111) A sequence is convergent if it hs (finite) limit Otherwise, the sequence is divergent l ɛ ɛ x 1 x x 5 x N+1 x x x 3 N+3 N+ x N+4 x 4 Figure 11: for ny ɛ, there is N Note the logicl reltion between ɛ nd N The predetermined smllness ɛ for x n l is rbitrrily given, while the size N for n is to be found fter ɛ is given Thus the choice of N usully depends on ɛ nd is often expressed s function of ɛ

11 LIMIT OF SEQUENCE 9 l ɛ ɛ 1 5 N+1 N+4 Figure 1: nother plot of converging sequence Since the limit is bout the long term behvior of sequence getting closer to trget, only smll ɛ nd big N need to be considered in the rgument For exmple, the limit of sequence is not chnged if the first 100 terms re replced by other rbitrry numbers See Exercise 119 for more exmples Exmple 111 Intuitively, the bigger n is, the smller 1 n gets This suggests 1 lim n n = 0 Rigorously following the definition, for ny ɛ > 0, choose N = 1 ɛ Then n > N = 1 n 0 = 1 n < 1 N = ɛ { } 1 Thus the impliction (111) is estblished for the sequence n How did we find the suitble N? Our gol is to chieve 1 n 0 < ɛ This is the sme s n > 1 ɛ, which suggests us to tke N = 1 Note tht our choice of N ɛ my not be nturl number { Exmple 11 The sequence n n + ( 1) n 3, 3, 4 5, 5 4, 6 7, 7 6, n Plotting the sequence suggests lim n n + ( 1) n = 1 For the rigorous rgument, we observe tht n n + ( 1) n 1 = 1 n + ( 1) n 1 n 1 }, with index strting from n =, is In order for the left side to be less thn ɛ, it is sufficient to mke is the sme s n > 1 ɛ + 1 1 < ɛ, which n 1

10 CHAPTER 1 LIMIT AND CONTINUITY Bsed on the nlysis, for ny ɛ > 0, choose N = 1 + 1 Then ɛ n > N = n n + ( 1) n 1 = 1 n + ( 1) n 1 n 1 < 1 N 1 = ɛ Exmple 113 Consider the sequence 14, 141, 1414, 1414, 14141, 141413, 1414135, 14141356, of the finer nd finer deciml pproximtions of The intuition suggests tht lim n x n = The rigorous verifiction mens tht for ny ɛ > 0, we need to find N, such tht n > N = x n < ɛ Since the n-th term x n is the expnsion up to the n-th deciml point, it stisfies x n < 10 n Therefore it suffices to find N so tht the following impliction holds n > N = 10 n < ɛ Assume 1 > ɛ > 0 Then ɛ hs the deciml expnsion ɛ = 000 0E N E N+1 E N+, with E N is from {1,,, 9} In other words, N is the loction of the first nonzero digit in the deciml expnsion of ɛ Then for n > N we hve ɛ 000 0E N 000 01 = 10 N > 10 n The rgument bove ssumes 1 > ɛ > 0 This is not problem becuse if we cn chieve x n l < 05 for n > N, then we cn certinly chieve x n l < ɛ for ny ɛ 1 nd for the sme n > N In other words, we my dd the ssumption tht ɛ is less thn certin fixed number without hurting the overll rigorous rgument for the limit Exercise 111 Rigorously verify the limits 1 lim n n n = lim n 1 n = 0 3 lim n ( n + 1 n) = 0 1 4 lim n n /3 = 0 n1/ 5 lim n cos n n = 0 6 lim n n cos n n + sin n = 1 Exercise 11 Let positive rel number > 0 hve the deciml expnsion = XZ 1 Z Z n Z n+1, where X is non-negtive integer, nd Z n is single digit integer from {0, 1,,, 9} t the n-th deciml point Prove the sequence XZ 1, XZ 1 Z, XZ 1 Z Z 3, XZ 1 Z Z 3 Z 4, of finer nd finer deciml pproximtions converges to

11 LIMIT OF SEQUENCE 11 Exercise 113 Suppose x n l y n nd lim n (x n y n ) = 0 Prove lim n x n = lim n y n = l Exercise 114 Suppose x n l y n nd lim n y n = 0 Prove lim n x n = l Exercise 115 Suppose lim n x n = l Prove lim n x n = l Is the converse true? Exercise 116 Suppose lim n x n = l Prove lim n x n+3 = l Is the converse true? (See Proposition 11 for more generl sttement) Exercise 117 Prove tht the limit is not chnged if finitely mny terms re modified In other words, if there is N, such tht x n = y n for n > N, then lim n x n = l if nd only if lim n y n = l Exercise 118 Prove the uniqueness of the limit In other words, if lim n x n = l nd lim n x n = l, then l = l Exercise 119 Prove the following re equivlent definitions of lim n x n = l 1 For ny c > ɛ > 0, where c is some fixed number, there is N, such tht x n l < ɛ for ll n > N For ny ɛ > 0, there is nturl number N, such tht x n l < ɛ for ll n > N 3 For ny ɛ > 0, there is N, such tht x n l ɛ for ll n > N 4 For ny ɛ > 0, there is N, such tht x n l < ɛ for ll n N 5 For ny ɛ > 0, there is N, such tht x n l ɛ for ll n > N Exercise 1110 Which re equivlent to the definition of lim n x n = l? 1 For ɛ = 0001, we hve N = 1000, such tht x n l < ɛ for ll n > N For ny ɛ stisfying 0001 ɛ > 0, there is N, such tht x n l < ɛ for ll n > N 3 For ny ɛ > 0001, there is N, such tht x n l < ɛ for ll n N 4 For ny ɛ > 0, there is nturl number N, such tht x n l ɛ for ll n N 5 For ny ɛ > 0, there is N, such tht x n l < ɛ for ll n > N 6 For ny ɛ > 0, there is N, such tht x n l < ɛ + 1 for ll n > N 7 For ny ɛ > 0, we hve N = 1000, such tht x n l < ɛ for ll n > N 8 For ny ɛ > 0, there re infinitely mny n, such tht x n l < ɛ 9 For infinitely mny ɛ > 0, there is N, such tht x n l < ɛ for ll n > N 10 For ny ɛ > 0, there is N, such tht l ɛ < x n < l + ɛ for ll n > N 11 For ny nturl number K, there is N, such tht x n l < 1 K for ll n > N The subsequence exmples re the importnt bsic limits The nlysis leding to the suitble choice of N for the given ɛ will be given The rigorous rgument in line with the definition of limit is left to the reder

1 CHAPTER 1 LIMIT AND CONTINUITY Exmple 114 For ny > 0, we hve lim n 1 = 0 (11) n The inequlity 1 n < ɛ is the sme s 1 n < ɛ 1 Thus choosing N = ɛ 1 should mke the impliction (111) hold Exmple 115 If < 1, then we hve Let 1 = 1 + b, then b > 0 nd 1 n = (1 + b)n = 1 + nb + lim n n = 0 (113) n(n 1) b + > nb This implies n < 1 nb In order to get n < ɛ, therefore, it suffices to mke sure 1 nb < ɛ This suggests us to choose N = 1 bɛ Exmple 116 We hve Let x n = n n 1 Then x n > 0 nd lim n n = ( n n ) n = 1 + nxn + n n = 1 (114) n(n 1) x n + > n(n 1) x n This implies x n < n 1 In order to get n n 1 = x n < ɛ, therefore, it suffices to mke sure n 1 < ɛ Thus we my choose N = ɛ + 1 Exmple 117 For ny, we hve n lim = 0 (115) n n! Suppose M > is n integer Then for n > M, we hve n n! = M M! M + 1 M + n M M! n Thus in order to get n M n! < ɛ, we only need to mke sure M! {M, M+1 } n < ɛ This leds to the choice N = mx M!ɛ Exercise 1111 Prove n! n n < 1 (n!) nd n (n)! < 1 for n > 1 Then use this to n + 1 n! prove lim n n n = lim (n!) n (n)! = 0 Exercise 111 Use the binry expnsion of n = (1+1) n to prove n > n Then prove lim n n = 0 Exercise 1113 Prove lim n n 3 = 1 nd lim n n n + n + 3 = 1 Exercise 1114 Prove n n n! > Then use this to prove lim n n(n 1) 1 n n! = 0

11 LIMIT OF SEQUENCE 13 11 Property A sequence is bounded if there is constnt B, such tht x n B for ny n This is equivlent to the existence of constnts B 1 nd B, such tht B 1 x n B for ny n The constnts B, B 1, B re respectively clled bound, lower bound nd n upper bound Proposition 11 Convergent sequences re bounded Proof Suppose lim n x n = l For ɛ = 1 > 0, there is N, such tht n > N = x n l < 1 Moreover, by tking bigger nturl number if necessry, N my be further ssumed to be nturl number Then x N+1, x N+,, hve upper bound l + 1 nd lower bound l 1, nd the whole sequence hs mx{x 1, x,, x N, l + 1}, min{x 1, x,, x N, l 1} s upper nd lower bounds Exercise 1115 Prove tht if x n < B for n > N, then the whole sequence {x n } is bounded This implies tht the boundedness is not chnged by modifying finitely mny terms in sequence Exercise 1116 Suppose lim n x n = 0 nd y n is bounded Prove lim n x n y n = 0 Proposition 113 (Arithmetic Rule) Suppose Then lim x n = l, n lim (x n + y n ) = l + k, n lim y n = k n lim x n y n = lk, n where y n 0 nd k 0 re ssumed in the third equlity Proof For ny ɛ > 0, there re N 1 nd N, such tht Then x n lim = l n y n k, n > N 1 = x n l < ɛ, n > N = y n k < ɛ n > mx{n 1, N } = (x n + y n ) (l + k) x n l + y n k < ɛ + ɛ = ɛ This completes the proof tht lim n (x n + y n ) = l + k By Proposition 11, we hve y n < B for fixed number B nd ll n For ny ɛ > 0, there re N 1 nd N, such tht n > N 1 = x n l < ɛ B, n > N = y n k < ɛ l

14 CHAPTER 1 LIMIT AND CONTINUITY Then n > N = mx{n 1, N } = x n y n lk = (x n y n ly n ) + (ly n lk) x n l y n + l y n k < ɛ B B + l ɛ l = ɛ This completes the proof tht lim n x n y n = lk 1 Assume y n 0 nd k 0 We will prove lim n = 1 By the product y n k property of the limit, this implies x n 1 lim = lim x n lim = l 1 n y n n n y n k = l k { ɛ k For ny ɛ > 0, we hve ɛ = min, k } > 0 Then there is N, such tht n > N = y n k < ɛ y n k < ɛ k, y n k < k = y n k < ɛ k, y n > k ɛ k = 1 1 y n k = y n k < y n k This completes the proof tht lim n 1 y n = 1 k k k = ɛ Exmple 118 By the limit (11) nd the rithmetic rule, we hve lim n n n = lim n 1 1 n Here is more complicted exmple = lim n 1 lim n 1 lim n n = 1 0 = lim n n 3 + n + 1 + 1 n 3 + 10n + 1 = lim n + 1 n 3 n + 10 1 n + 1 n 3 1 + = ( 1 lim n n ( 1 + 10 lim n n ) ( 1 + lim n n ) ( 1 + lim n n = 1 + 0 + 0 3 + 10 0 + 0 3 = 1 The ide cn be generlized to obtin p n p + p 1 n p 1 + + 1 n + 0 0 if p < q, b q 0 lim n b q n q + b q 1 n q 1 = p + + b 1 n + b 0 if p = q, b q 0 (116) b q ) 3 ) 3

11 LIMIT OF SEQUENCE 15 Exercise 1117 Suppose lim n x n = l, lim n y n = k Prove lim n mx{x n, y n } = mx{l, k}, lim n min{x n, y n } = min{l, k} You my use the formul mx{x, y} = 1 (x + y + x y ) nd the similr one for min{x, y} Proposition 114 (Order Rule) Suppose both {x n } nd {y n } converge 1 If x n y n for big n, then lim n x n lim n y n If lim n x n > lim n y n, then x n > y n for big n A specil cse of the property is tht lim n x n < l implies x n < l for sufficiently big n, nd x n l implies lim n x n l Proof We prove the second sttement first Suppose lim n x n > lim n y n Then by Proposition 113, lim n (x n y n ) = lim n x n lim n y n > 0 For ɛ = lim n (x n y n ) > 0, there is N, such tht n > N = (x n y n ) ɛ < ɛ = x n y n > ɛ ɛ = 0 x n > y n By exchnging x n nd y n in the second sttement, we find tht lim x n < lim y n = x n < y n for big n n n This further implies tht we cnnot hve x n y n for big n The combined impliction lim x n < lim y n = opposite of (x n y n for big n) n n is equivlent to the first sttement In the second prt of the proof bove, we used the logicl fct tht A = B is the sme s (not B) = (not A) Moreover, we note tht the following two sttements re not opposite of ech other 1 There is N, such tht x n < y n for n > N There is N, such tht x n y n for n > N Proposition 115 (Sndwich Rule) Suppose Then x n y n z n, lim x n = lim z n = l n n lim y n = l n Proof For ny ɛ > 0, there re N 1 nd N, such tht Then n > N 1 = x n l < ɛ, n > N = z n l < ɛ n > mx{n 1, N } = ɛ < x n l y n l z n l < ɛ = y n l < ɛ

16 CHAPTER 1 LIMIT AND CONTINUITY { cos n } Exmple 119 To find the limit of the sequence, we compre it with { } n 1 the sequence in Exmple 111 Since 1 n n cos n 1 n n nd lim 1 n n = lim n 1 n = 0, we get lim cos n n n = 0 sin n By similr reson, we get lim n n = 0 nd lim ( 1) n n = 0 Then by n the rithmetic rule, lim n n cos n n + ( 1) n sin n = lim 1 n n ( = lim n 1 cos n n 1 + sin n ( 1) n n n ( ) cos n ) 1 1 lim n ( ) n n sin n ( 1) 1 + lim n (lim n ) n n n = 0 1 0 1 + 0 0 = 0 Exmple 1110 Suppose {x n } is sequence stisfying x n l < 1 Then we n hve l 1 n < x n < l + 1 ( n Since lim n l 1 ) ( = lim n l + 1 ) = l, by the n n sndwich rule, we get lim n x n = l Exmple 1111 For ny > 1 nd n >, we hve 1 < n < n n Thus by the limit (114) nd the sndwich rule, we hve lim n n = 1 On the other hnd, for 0 < < 1, we hve b = 1 > 1 nd lim n n = lim n 1 n b = 1 lim n n b = 1 Combining ll the cses, we get lim n n = 1 for ny > 0 Furthermore, we hve 1 < (n + ) 1 n+b < (n) n, 1 < (n + n + b) 1 n+c < (n ) n for sufficiently big n By ( lim (n) n = lim n ( lim n (n ) n = lim n n n lim n n ) ( lim n n n ) = (1 1) = 1, n n ) 4 = 1 1 4 = 1, nd the sndwich rule, we get lim n (n + ) 1 n+b = limn (n + n + b) 1 n+c = 1 The sme ide leds to the limit lim ( pn p + p 1 n p 1 + + 1 n + 0 ) 1 n+ = 1 (117) n n p Exmple 111 Consider lim n for > 1 nd ny p The specil cse n p = 0 is the limit (113), nd the specil cse p = 1, = is Exercise 111 Let

11 LIMIT OF SEQUENCE 17 = 1 + b Since > 1, we hve b > 0 Fix nturl number P p Then for n > P, n n(n 1) = 1 + nb + b + + n(n 1) (n P ) > b P +1 (P + 1)! n(n 1) (n P ) b P +1 + (P + 1)! Thus 0 < n p n np n < (P + 1)! n P n(n 1) (n P ) b P +1 = 1 n n n (n 1) (n ) n (n P ) (P + 1)! b P +1 1 By lim n n = 0, lim n n = 1, the fct tht P nd b re fixed constnts, n k nd the rithmetic rule, the right side hs limit 0 s n By the sndwich rule, we conclude tht lim n n p = 0 for > 1 nd ny p (118) n Exercise 1118 Redo Exercise 114 by using the sndwich rule Exercise 1119 Let > 0 be constnt Then 1 n the limit (114) to prove lim n n = 1 Exercise 110 Compute the limits < < n for big n Use this nd 1 lim n n 7/4 3n 3/ (3n 3/4 n 1/ + 1)(n + ) lim n ( n + n n) 3 lim n 3 n 1 8 7 n + ( 1) n+1 8 n 1 + (n + 1)( 5) n 4 lim n ( 1) n n + 1 n 3 + ( 1) n 5 lim n n! + 10 n n 10 + n n 6 lim n (n + n + 3)(n! n + 5 n ) (n + )! n + 5 n 7 lim n n n + cos n + sin n 8 lim n n 11 n + 5 n + 5 11 n 9 lim n n 11 n n 5 n (n + 1) n 10 lim n 1 + n + n 3 + n n 1 3 5 (n 1) 11 lim n 4 5 n Exercise 111 Compute the limits n 1 lim n n, where 1 + 1 lim n (n + ) n n +bn+c 3 lim n n n + b n + c n, where, b, c > 0 ( ) 1 4 lim n + 1 + + 1 1 + n + n n + n

18 CHAPTER 1 LIMIT AND CONTINUITY 113 Infinity nd Infinitesiml A chnging numericl quntity is n infinity if it tends to get rbitrrily big For sequences, this mens the following Definition 116 A sequence {x n } diverges to infinity, denoted lim n x n =, if for ny b, there is N, such tht n > N = x n > b (119) It diverges to positive infinity, denoted lim n x n = +, if for ny b, there is N, such tht n > N = x n > b (1110) It diverges to negtive infinity, denoted lim n x n =, if for ny b, there is N, such tht n > N = x n < b (1111) Exmple 1113 We rigorously verify lim n n choose N = b Then = + For ny b > 0, n + ( 1) n n > N = n n + ( 1) n n n + 1 > n n > N = b Exercise 11 Rigorously verify the divergence to infinity 1 lim n (100 + 10n n ) = lim n ( 1) n n + sin n = 3 lim n n + (n 1) + + + 1 = + Exercise 113 Infinities must be unbounded Is the converse true? Exercise 114 Suppose lim n x n = +, lim n y n = + Prove lim n (x n + y n ) = +, lim n x n y n = + Exercise 115 Suppose lim n x n = nd x n x n+1 < c for some constnt c Prove tht either lim n x n = + or lim n x n = Furthermore, if lim n x n = + nd x > x 1, then prove x < x n < x + c for some n A chnging numericl quntity is n infinitesiml if it tends to get rbitrrily smll For sequences, this mens tht for ny ɛ > 0, there is N, such tht n > N = x n < ɛ (111) This is simply lim n x n = 0 Note tht the implictions (119) nd (111) re equivlent by tking ɛ = 1 Therefore we hve b { } 1 {x n } is n infinity is n infinitesiml x n

11 LIMIT OF SEQUENCE 19 For exmple, the infinitesimls (11),{ (113), } (115), { }(118) tell us tht n! n {n } (for > 0), { n } (for > 1),, nd (for > 1) re infinities Moreover, the first cse in the limit (116) tells us p n p + p 1 n p 1 + + 1 n + 0 lim = if p > q, n b q n q + b q 1 n q 1 p 0 + + b 1 n + b 0 On the other hnd, since lim n x n = l is equivlent to lim n (x n l) = 0, we hve {x n } converges to l {x n l} is n infinitesiml For exmple, the limit (114) tells us tht { n n 1} is n infinitesiml Exercise 116 { } How to chrcterize positive infinity {x n } in terms of the infinitesiml 1? x n Exercise 117 Explin the infinities n! 1 lim n = for ny 0 n n! lim n n = if + b 0 + bn 3 lim n 1 n n 1 = + 4 lim n 1 n n n n = Some properties of finite limits cn be extended to infinities nd infinitesimls For exmple, if lim n x n = + nd lim n y n = +, then lim n (x n + y n ) = + The property cn be denoted s the rithmetic rule (+ ) + (+ ) = + Moreover, if lim n x n = 1, lim n y n = 0 nd y n < 0 for big n, then lim n x n y n = Thus we hve nother rithmetic rule 1 = Common sense suggests more rithmetic rules such s 0 c+ =, c = (for c 0), =, n n p c 0 = (for c 0), c = 0, where c is finite number nd represents sequence convergent to c We must be creful in pplying rithmetic rules involving infinities nd infinitesimls For exmple, we hve lim n lim n n 1 = 0, n 1 n 1 = 1, lim n 1 = 0, n lim n so tht 0 hs no definite vlue 0 n 1 = +, n lim lim n = 0, n n n = 0, n 1

0 CHAPTER 1 LIMIT AND CONTINUITY Exmple 1114 By the rithmetic rule, we hve lim n (n + 3)( ) n = lim n (n + 3) lim n ( )n = = ( lim n + 1 ) = lim n n n + lim 1 = (+ ) + 0 = + n n n lim n n n n n 1 = lim n n n lim n ( n n 1) = lim n Exercise 118 Prove the properties of infinities 1 0 + = + 1 (bounded)+ = : If {x n } is bounded nd lim n y n =, then lim n (x n + y n ) = min{+, + } = + : If lim n x n = lim n y n = +, then lim n min{x n, y n } = + 3 Sndwich rule: If x n y n nd lim n y n = +, then lim n x n = + 4 (> c > 0) (+ ) = + : If x n > c for some constnt c > 0 nd lim n y n = +, then lim n x n y n = + Exercise 119 Show tht it is not necessrily true tht + = by constructing exmples of sequences {x n } nd {y n } tht diverge but one of the following holds 1 lim n (x n + y n ) = lim n (x n + y n ) = + 3 {x n + y n } is bounded nd divergent Exercise 1130 Show tht one cnnot mke definite conclusion on 0 by constructing exmples of sequences {x n } nd {y n }, such tht lim n x n = 0 nd lim n y n = but one of the following holds 1 lim n x n y n = lim n x n y n = 0 3 lim n x n y n = 4 {x n y n } is bounded nd divergent Exercise 1131 Provide counterexmples to the wrong rithmetic rules + = 1, (+ ) (+ ) = 0, 0 = 0, 0 =, 0 = 1 + 114 Additionl Exercise Rtio Rule Exercise 113 Suppose x n+1 x n y n+1 y n 1 Prove tht x n c y n for some constnt c

1 CONVERGENCE OF SEQUENCE LIMIT 1 Prove tht lim n y n = 0 implies lim n x n = 0 3 Prove tht lim n x n = implies lim n y n = Exercise 1133 Suppose lim n x n+1 by looking t the vlue of l? x n = l Wht cn you sy bout lim n x n (n!) n Exercise 1134 Use the rtio rule to study the limits (118) nd lim n (n)! Power Rule Exercise 1135 Suppose lim n x n = l > 0 Prove lim n x α n = l α by the following steps 1 Assume x n 1 nd l = 1 By using the sndwich rule nd the fct tht lim n x n = 1 for ny integer, prove tht lim n x α n = 1 for ny number α The sme rgument pplies to the cse x n 1 Use min{x n, 1} x n mx{x n, 1}, Exercise 1117 nd the sndwich rule to remove the ssumption x n 1 in the first prt 3 Use the rithmetic rule to prove the limit for generl l Averge Rule Exercise 1136 Suppose lim n x n = l Let y n = x 1 + x + + x n n 1 Prove tht if x n l < ɛ for n > N, where N is nturl number, then n > N = y n l < x 1 + x + + x N + N l n Use the first prt nd Proposition 11 to prove lim n y n = l 3 Wht hppens if l = + or? Exercise 1137 Find suitble condition on sequence { n } of positive numbers, such tht lim n x n = l implies lim n 1 x 1 + x + + n x n 1 + + + n = l 1 Convergence of Sequence Limit The discussion of convergent sequences in Section 11 is bsed on the explicit vlue of the limit However, there re mny cses tht sequence must be convergent, but the vlue of the limit is not known The limit of the world record in 100 meter dsh is one such exmple In such cses, the existence of the limit cnnot be estblished by using the definition lone A more fundmentl theory is needed + ɛ

CHAPTER 1 LIMIT AND CONTINUITY 11 Necessry Condition A subsequence of sequence {x n } is obtined by selecting some terms from the sequence The indices of the selected terms cn be rrnged s strictly incresing sequence n 1 < n < < n k <, nd the subsequence cn be denoted s {x nk } The following re two exmples of subsequences {x 3k }: x 3, x 6, x 9, x 1, x 10, x 15, {x k}: x, x 4, x 8, x 16, x 3, x 64, Note tht if {x n } strts from n = 1, then n k k Thus by reindexing the terms if necessry, we will lwys ssume n k k in subsequent proofs Proposition 11 Suppose sequence converges to l Then ll its subsequences converge to l Proof Suppose lim n x n = l For ny ɛ > 0, there is N, such tht n > N implies x n l < ɛ Then k > N = n k k > N = x nk l < ɛ Exmple 11 By lim n n n = 1, we hve lim k k k = 1 Tking the squre of the limit, we get lim n n n = (lim k k k) = 1 Exmple 1 The sequence {( 1) n } hs subsequences {( 1) k } = {1} nd {( 1) k+1 } = { 1} Since the two subsequences hve different limits, the originl sequence {( 1) n } diverges Exercise 11 Prove the sequences diverge 1 ( 1) n n + 1 n + ( 1) n n(n + 1) ( n + ) 3 3 ( n n + ( 1) n ) n ( 1) n 4 n sin nπ 3 n cos nπ + 5 ( 1) n n n + 1 6 x n = 1 n, x n+1 = n n 7 n n + 3 ( 1)nn 8 1 + n sin nπ 9 cos n nπ 3 Exercise 1 Prove tht lim n x n = l if nd only if lim k x k = l nd lim k x k+1 = l Exercise 13 Wht is wrong with the following ppliction of Propositions 113 nd 11: The sequence x n = ( 1) n stisfies x n+1 = x n Therefore lim n x n = lim n x n+1 = lim n x n, nd we get lim n x n = 0 Exercise 14 Prove tht if sequence diverges to infinity, then ll its subsequences diverge to infinity

1 CONVERGENCE OF SEQUENCE LIMIT 3 Theorem 1 (Cuchy 1 Criterion) Suppose sequence {x n } converges Then for ny ɛ > 0, there is N, such tht m, n > N = x m x n < ɛ Proof Suppose lim n x n = l For ny ɛ > 0, there is N, such tht n > N implies x n l < ɛ Then m, n > N implies x m x n = (x m l) (x n l) x m l + x n l < ɛ + ɛ = ɛ Cuchy criterion plys criticl role in nlysis Therefore the criterion is clled theorem insted of just proposition Moreover, the property described in the criterion is given specil nme Definition 13 A sequence {x n } is clled Cuchy sequence if for ny ɛ > 0, there is N, such tht m, n > N = x m x n < ɛ (11) Theorem 1 sys tht convergent sequences must be Cuchy sequences The converse tht ny Cuchy sequence is convergent is lso true nd is one of the most fundmentl results in nlysis Exmple 13 Consider the sequence {( 1) n } For ɛ = 1 > 0 nd ny N, we cn find n even n > N Then m = n+1 > N is odd nd x m x n = > ɛ Therefore the Cuchy criterion fils nd the sequence diverges Exmple 14 (Oresme ) The hrmonic sequence stisfies x n = 1 + 1 + 1 3 + + 1 n x n x n = 1 n + 1 + 1 n + + + 1 n 1 n + 1 n + + 1 n = n n = 1 Thus for ɛ = 1 nd ny N, we hve x m x n > 1 by tking ny nturl number n > N nd m = n Therefore the Cuchy criterion fils nd the hrmonic sequence diverges Exmple 15 We show the sequence {sin n} diverges The rel line is divided by kπ ± π 4 into intervls of length π > 1 Therefore for ny integer k, there re integers m nd n stisfying kπ + π 4 < m < kπ + 3π 4, kπ π 4 > n > kπ 3π 4 1 Augustin Louis Cuchy, born 1789 in Pris (Frnce), died 1857 in Sceux (Frnce) His contributions to mthemtics cn be seem by the numerous mthemticl terms bering his nme, including Cuchy integrl theorem (complex functions), Cuchy-Kovlevsky theorem (differentil equtions), Cuchy-Riemnn equtions, Cuchy sequences He produced 789 mthemtics ppers nd his collected works were published in 7 volumes Nicole Oresme, born 133 in Allemgne (Frnce), died 138 in Lisieux (Frnce)

4 CHAPTER 1 LIMIT AND CONTINUITY Moreover, by tking k to be big positive number, m nd n cn be s big s we wish Then sin m > 1, sin n < 1, nd we hve sin m sin n > Thus the sequence {sin n} is not Cuchy nd must diverge Exercise 15 Prove the sequences diverge 1 x n = 1 + 1 + 1 3 + + 1 n x n = 1 + 5 + 3 10 + + n n + 1 Exercise 16 Prove tht if lim n x n = + nd x n x n+1 < c for some constnt c < π, then {sin x n } diverges 1 Supremum nd Infimum To discuss the converse of Theorem 1, we hve to consider the difference between rtionl nd rel numbers Specificlly, consider the converse of Cuchy criterion stted for the rel nd the rtionl number systems: 1 Rel number Cuchy sequences lwys hve rel number limits Rtionl number Cuchy sequences lwys hve rtionl number limits The key distinction here is tht sequence of rtionl numbers my hve n irrtionl number s the limit For exmple, the rtionl number sequence of the deciml pproximtions of in Exmple 113 is Cuchy sequence but hs no rtionl number limit This shows tht the second sttement is wrong Therefore the truthfulness of the first sttement is closely relted to the fundmentl question of the definition of rel numbers In other words, the estblishment of the first property must lso point to the key difference between the rtionl nd the rel number systems One solution to the fundmentl question is to simply use the converse of Cuchy criterion s the wy of constructing rel numbers from rtionl numbers (by requiring tht ll Cuchy sequences converge) This is the topologicl pproch nd cn be delt with in the lrger context of the completion of metric spces Alterntively, rel numbers cn be constructed by considering the order mong the numbers The subsequent discussion will be bsed on this more intuitive pproch, which is clled the Dedekind 3 cut The order reltion between rel numbers enbles us to introduce the following concept Definition 14 Let X be nonempty set of numbers An upper bound of X is number B such tht x B for ny x X The supremum of X is the lest upper bound of the set nd is denoted sup X The supremum λ = sup X is chrcterized by the following properties 3 Julius Wilhelm Richrd Dedekind, born 1831 nd died 1916 in Brunschweig (Germny) Dedekind cme up with the ide of the cut on November 4 of 1858 while thinking how to tech clculus He mde importnt contributions to lgebric number theory His work introduced new style of mthemtics tht influenced genertions of mthemticins

1 CONVERGENCE OF SEQUENCE LIMIT 5 1 λ is n upper bound: For ny x X, we hve x λ Any number smller thn λ is not n upper bound: For ny ɛ > 0, there is x X, such tht x > λ ɛ The lower bound nd the infimum inf X cn be similrly defined nd chrcterized Exmple 16 Both the set {1, } nd the intervl [0, ] hve s the supremum In generl, the mximum of set X is number ξ X such tht ξ x for ny x X, nd the mximum (if exists) is lwys the supremum On the other hnd, the intervl (0, ) hs no mximum but still hs s the supremum The similr discussion on minimum cn be mde Exmple 17 is the supremum of the set {14, 141, 1414, 1414, 14141, 141413, 1414135, 14141356, } of its deciml expnsions It is lso the supremum of the set { m n : m nd n re nturl numbers stisfying m < n } of positive rtionl numbers whose squres re less thn Exmple 18 Let L n be the length of n edge of the inscribed regulr n-gon in circle of rdius 1 Then π is the supremum of the set {3L 3, 4L 4, 5L 5, } of the circumferences of the n-gons Exercise 17 Find the suprem nd the infim 1 { + b:, b re rtionl < 3, b + 1 < 5} { } n : n is nturl number n + 1 3 { ( 1) n } n : n is nturl number n + 1 4 { m } n : m nd n re nturl numbers stisfying m > 3n 5 { 1 m + 1 } : m nd n re nturl numbers 3n 6 { } nrn : n 3 is nturl number, where R n is the length of n edge of the circumscribed regulr n-gon in circle of rdius 1 Exercise 18 Prove the supremum is unique Exercise 19 Suppose X is nonempty bounded set of numbers Prove tht λ = sup X is chrcterized by the following two properties 1 λ is n upper bound: For ny x X, we hve x λ λ is the limit of sequence in X: There re x n X, such tht λ = lim n x n The following re some properties of the supremum nd infimum

6 CHAPTER 1 LIMIT AND CONTINUITY Proposition 15 Suppose X nd Y re nonempty bounded sets of numbers 1 If x y for ny x X nd y Y, then sup X inf Y If x y c for ny x X nd y Y, then sup X sup Y c nd sup X inf Y c 3 If X + Y = {x + y : x X, y Y }, then sup(x + Y ) = sup X + sup Y nd inf(x + Y ) = inf X + inf Y 4 If cx = {cx: x X}, then sup(cx) = c sup X when c > 0 nd sup(cx) = c inf X when c < 0 5 If XY = {xy : x X, y Y } nd ll numbers in X, Y re positive, then sup(xy ) = sup X sup Y nd inf(xy ) = inf X inf Y 6 If X 1 = {x 1 : x X} nd ll numbers in X re positive, then sup X 1 = (inf X) 1 Proof In the first property, fix ny y Y Then y is n upper bound of X Therefore sup X y Since sup X y for ny y Y, sup X is lower bound of Y Therefore sup X inf Y Now consider the second property For ny ɛ > 0, there re x X nd y Y, such tht sup X ɛ < x sup X nd sup Y ɛ < y sup Y Then (sup X ɛ) sup Y < x y < sup X (sup Y ɛ), which mens exctly (x y) (sup X sup Y ) < ɛ This further implies sup X sup Y < x y + ɛ c + ɛ Since this holds for ny ɛ > 0, we conclude tht sup X sup Y c The inequlity sup X inf Y c cn be similrly proved Next we prove the third property For ny x X nd y Y, we hve x+ y sup X +sup Y, so tht sup X +sup Y is n upper bound of X +Y On the other hnd, for ny ɛ > 0, there re x, y X, Y, such tht x > sup X ɛ nd y > sup Y ɛ Then x+y X +Y stisfies x+y > sup X +sup Y ɛ Since ɛ > 0 is rbitrry, this shows tht ny number smller thn sup X + sup Y is not n upper bound of X + Y Therefore sup X + sup Y is the supremum of X + Y The proof of the rest re left s exercises Exercise 110 Finish the proof of Proposition 15 Exercise 111 Suppose X i re nonempty sets of numbers Let X = i X i nd l i = sup X i Prove tht sup X = sup i l i The existence of the supremum is wht distinguishes the rel numbers from the rtionl numbers

1 CONVERGENCE OF SEQUENCE LIMIT 7 Definition 16 Rel numbers is set with the usul rithmetic opertions nd the order stisfying the usul properties, nd the dditionl property tht ny bounded set of rel numbers hs the supremum The rithmetic opertions re ddition, subtrction, multipliction nd division An order on set S is reltion x < y defined for pirs x, y S, stisfying the following properties: trnsitivity: x < y nd y < z = x < z exclusivity: If x < y, then y < x does not hold The following re some (but not ll) of the usul rithmetic nd order properties commuttivity: + b = b +, b = b distributivity: (b + c) = b + c unit: There is specil number 1 such tht 1 = order comptible with ddition: < b = + c < b + c order comptible with multipliction: < b, 0 < c = c < bc Becuse of these properties, the rel numbers form n ordered field Since the rtionl numbers lso hs the rithmetic opertions nd the order stisfying these usul properties, the rtionl numbers lso form n ordered field Thus the key distinction between the rel nd rtionl numbers is the existence of the supremum A bounded set of rtionl numbers my not hve rtionl number supremum A bounded set of rel numbers lwys hs rel number supremum Due to the existence of supremum, the rel numbers form complete ordered field 13 Monotone Sequence A sequence {x n } is incresing if x n+1 x n It is strictly incresing if x n+1 > x n The concepts of decresing nd strictly decresing sequences cn be similrly defined A sequence is monotone if it is either incresing or decresing Proposition 17 Bounded monotone sequences of rel numbers re convergent Unbounded monotone sequences of rel numbers diverge to infinity Since n incresing sequence {x n } stisfies x n x 1, the sequence hs x 1 s lower bound Therefore it is bounded if nd only if it hs n upper bound, nd the proposition sys tht n incresing sequence with upper bound must be convergent Similr remrks cn be mde for decresing sequences

8 CHAPTER 1 LIMIT AND CONTINUITY Proof Let {x n } be bounded incresing sequence The sequence hs rel number supremum l = sup{x n } For ny ɛ > 0, by the second property chrcterizing the supremum, there is N, such tht x N > l ɛ Then becuse the sequence is incresing, n > N implies x n x N > l ɛ We lso hve x n l becuse l is n upper bound Thus we conclude tht n > N = l ɛ < x n l = x n l < ɛ This proves tht the sequence converges to l Let {x n } be n unbounded incresing sequence Then it hs no upper bound In other words, for ny b, there is N, such tht x N > b Since the sequence is incresing, we hve n > N = x n > x N > b Thus we conclude tht {x n } diverges to + The proof for the decresing sequences is similr Exmple 19 The sequence in Exmple 14 is clerly incresing Since it is divergent, the sequence hs no upper bound In fct, the proof of Proposition 17 tells us ( lim 1 + 1 n + 1 3 + + 1 ) = + n On the other hnd, the incresing sequence x n = 1 + 1 + 1 3 + + 1 n stisfies x n < 1 + 1 1 + 1 3 + + 1 (n 1)n ( 1 = 1 + 1 1 ) ( 1 + 1 ) ( 1 + + 3 n 1 1 ) n = 1 + 1 1 1 n <, nd must be convergent Much lter on, we will see tht the sequence converges if nd only if p > 1 x n = 1 + 1 p + 1 3 p + + 1 n p Exmple 110 Suppose sequence is given inductively by x 1 = 1, x n+1 = + xn We clim tht the sequence is incresing It is esy to see tht x n+1 > x n is equivlent to x n x n = (x n )(x n + 1) < 0 Since the sequence is clerly positive, the problem becomes x n < First x 1 < Second if x n <, then x n+1 < + = The inequlity x n < is therefore proved by induction Since the sequence is incresing nd hs upper bound, it is convergent The limit l stisfies l = lim n x n+1 = + lim x n = + l n Solving the eqution for l 0, we conclude tht the limit is l =

1 CONVERGENCE OF SEQUENCE LIMIT 9 Exmple 111 Let x n = ( 1 + 1 n) n The binomil expnsion tells us ( ) ( ) 1 n(n 1) 1 x n = 1 + n + + n! n ( ) n(n 1) 1 1 n + + n! n ( ) ( = 1 + 1 1! + 1! + + 1 n! 1 1 n ( 1 1 n + 1 3! ) ( 1 n ) n(n 1)(n ) 3! 1 1 ) ( 1 ) n n ( 1 n 1 ) n ( ) 1 3 n By compring the similr formul for x n+1, we find the sequence is strictly incresing The formul lso tells us (see Exmple 19) x n < 1 + 1 + 1! + 1 3! + + 1 n! < 1 + 1 + 1 1 + 1 3 + + 1 (n 1)n < 3, so tht the sequence hs n upper bound Therefore the sequence converges The limit hs specil nottion ( e = lim 1 + 1 n = 718818845904 n n) nd is fundmentl constnt of the nture Exercise 11 Prove the sequences converge 1 x n = 1 + 1 3 + 1 3 3 + + 1 n 3 x n = 1 + 1! + 1 3! + + 1 n! 3 x n = 1 + 1 + 1 3 3 + + 1 n n 4 x n = 1 + 1 + 1 + + 1 n Exercise 113 Consider the sequence x n = 1 + 1 + 1 3 + + 1 n n + α 1 Prove tht if α 1, then the sequence is strictly decresing Prove tht if α > 1, then the sequence is strictly incresing for n > 1 16α 8 3 Prove the sequence converges to the sme limit for ll α Exercise 114 For ny > 0, let x n = + + + +, where the squre root ppers n times Prove x n < x n+1 < 1 + 4 + 1 nd find the limit of the sequence (The result cn be extended to sequence defined by x 1 b, x n+1 = f(x n ), where f(x) be function stisfying b f(x) x for x b See Figure 13) Exercise 115 Prove the inductively defined sequences converge nd find the limits

30 CHAPTER 1 LIMIT AND CONTINUITY x 1 x x 3 x 4 x 5 b Figure 13: recursively defined convergent sequence 1 x 1 = 1, x n+1 = 1 + x n x 1 = 1, x n+1 = x n + x n Exercise 116 Discuss the convergence of the sequence defined by x 1 = α, x n+1 = β + γx n Do the sme to the sequence defined by x 1 = α, x n+1 = β + γ x n (This generlizes the first problem in Exercise 115) Exercise 117 Let, b > 0 Define sequences 1 =, b 1 = b, n = n 1 + b n 1, b n = n 1b n 1 Use + b b to prove the sequences converge Moreover, n 1 + b n 1 + b find the limits Exercise 118 Let x n = ( 1 + n) 1 n ( nd y n = 1 + n) 1 n+1 1 Use induction to prove (1+x) n 1+nx for x > 1 nd ny nturl number n By showing x n+1 decresing x n > 1 nd y n 1 y n > 1, prove {x n } is incresing nd {y n } is 3 Prove {x n } nd {y n } converge to the sme limit e 4 Prove e x n < e n 5 Prove lim n ( 1 1 n) n = e Exercise 119 Prove tht e 1 + 1 1! + 1! + + 1 n! for ny n 14 Convergent Subsequence By Proposition 11, ny convergent sequence is bounded While bounded sequences my not converge (see Exmple 1), the following still holds

1 CONVERGENCE OF SEQUENCE LIMIT 31 Theorem 18 (Bolzno 4 -Weierstrss 5 Theorem) A bounded sequence of rel numbers hs convergent subsequence Recll tht by Proposition 11, ny subsequence of convergent sequence is convergent Theorem 18 shows tht if the originl sequence is only ssumed to be bounded, then ny subsequence should be chnged to some subsequence Proof Let {x n } be bounded sequence Then ll x n lie in bounded intervl I = [, b] [ Divide I into two equl hlves I =, + b ] [ ] + b nd I =, b Then either I or I must contin infinitely mny x n We denote this intervl by I 1 = [ 1, b 1 ] nd find x n1 I 1 [ Further divide I 1 into two equl hlves I 1 = 1, ] 1 + b 1 nd I 1 = [ 1 + b 1, b 1 ] Then either I 1 or I 1 must contin infinitely mny x n We denote this intervl by I = [, b ] Becuse I contins infinitely mny x n, we cn find x n I with n > n 1 Keep going, we get sequence of intervls I = [, b] I 1 = [ 1, b 1 ] I = [, b ] I k = [ k, b k ] with the length of I k being b k k = b k Moreover, we hve subsequence {x nk } stisfying x nk I k The inclusion reltion between the intervls tells us 1 k b k b b 1 b Thus { k } nd {b k } re bounded nd monotone sequences By Proposition 17, both sequences converge Moreover, the length of I k nd the limit (113) tell us lim k (b k k ) = 0 Therefore the two sequences hve the sme limit Denote l = lim k k = lim k b k The property x nk I k mens k x nk b k By the sndwich rule, we get lim k x nk = l Thus we find convergent subsequence {x nk } 4 Bernrd Plcidus Johnn Nepomuk Bolzno, born 1781 nd died 1848 in Prgue (Bohemi, now Czech) Bolzno insisted tht mny results which were thought obvious required rigorous proof nd mde fundmentl contributions to the foundtion of mthemtics He understood the need to redefine nd enrich the concept of number itself nd define the Cuchy sequence four yers before Cuchy s work ppered 5 Krl Theodor Wilhelm Weierstrss, born 1815 in Ostenfelde, Westphli (now Germny), died 1848 in Berlin (Germny) In 1864, he found continuous but nowhere differentible function His lectures on nlytic functions, elliptic functions, belin functions nd clculus of vritions influenced mny genertions of mthemticins, nd his pproch still domintes the teching of nlysis tody

3 CHAPTER 1 LIMIT AND CONTINUITY Exmple 11 Let {x n }, {y n }, {z n } be sequences converging to l 1, l, l 3, respectively Then for the sequence x 1, y 1, z 1, x, y, z, x 3, y 3, z 3,, x n, y n, z n, the limits of convergent subsequences re l 1, l, l 3 We need to explin tht ny l l 1, l, l 3 is not the limit of ny subsequence There is ɛ > 0 such tht (tke ɛ = 1 min{ l l 1, l l, l l 3 }, for exmple) Then there re N 1, N, N 3, such tht l l 1 ɛ, l l ɛ, l l 3 ɛ n > N 1 = x n l 1 < ɛ, n > N = y n l < ɛ, n > N 3 = z n l 3 < ɛ Since l l 1 ɛ nd x n l 1 < ɛ imply x n l > ɛ, we hve n > mx{n 1, N, N 3 } = x n l > ɛ, y n l > ɛ, z n l > ɛ This implies tht l cnnot be the limit of ny convergent subsequence A more direct rgument bout the limits of convergent subsequences is the following Let l be the limit of convergent subsequence {w m } of the combined sequence The subsequence {w m } must contin infinitely mny terms from t lest one of the three sequences If {w m } contins infinitely mny terms from {x n }, then it contins subsequence {x nk } of {x n } By Proposition 11, we get l = lim w m = lim x nk = lim x n = l 1 The second equlity is due to the fct tht {x nk } is subsequence of {w m } The third equlity is due to {x nk } being subsequence of {x n } Exercise 10 For sequences in Exercise 11, find ll the limits of convergent subsequences Exercise 11 Any rel number is the limit of sequence of the form n 1 10, n 100, n 3 1000,, where n k re integers Bsed on this observtion, construct sequence so tht the limits of convergent subsequences re ll the numbers between 0 nd 1 Exercise 1 Prove tht number is the limit of convergent subsequence of {x n } if nd only if it is the limit of convergent subsequence of {x n n n} Exercise 13 Suppose {x n } nd {y n } re two bounded sequences Prove tht there re n k, such tht both subsequences {x nk } nd {y nk } converge The following technicl result provides criterion for number to be the limit of subsequence Proposition 19 l is the limit of convergent subsequence of {x n } if nd only if for ny ɛ > 0 nd N, there is n > N, such tht x n l < ɛ

1 CONVERGENCE OF SEQUENCE LIMIT 33 Proof Let l be the limit of convergent subsequence {x nk } Let ɛ > 0 nd N be given Then there is K, such tht k > K implies x nk l < ɛ It is esy to find k > K, such tht n k > N (tke k = mx{k, N} + 1, for exmple) Then for n = n k, we hve n > N nd x n l < ɛ Conversely, suppose for ny ɛ > 0 nd N, there is n > N, such tht x n l < ɛ Then for ɛ = 1, there is n 1 such tht x n1 l < 1 Next, for ɛ = 1 nd N = n 1, there is n > n 1 such tht x n l < 1 Keep going, by tking ɛ = 1 k + 1 nd N = n k to find x nk+1, we construct subsequence {x nk } stisfying x nk l < 1 k The inequlity implies lim k x nk = l Here is remrk tht is very useful for the discussion of subsequences Suppose P is property bout terms in sequence (x n > l or x n > x n+1, for exmples) Then the following sttements re equivlent: 1 For ny N, there is n > N, such tht x n hs property P There re infinitely mny x n with property P 3 There is subsequence {x nk } such tht ech term x nk hs property P In prticulr, the criterion for l to be the limit of convergent subsequence of {x n } is for ny ɛ > 0, there re infinitely mny x n stisfying x n l < ɛ Exercise 14 Let {x n } be sequence Suppose lim k l k = l nd for ech k, l k is the limit of convergent subsequence of {x n } Prove tht l is lso the limit of convergent subsequence Let {x n } be bounded sequence Then the set LIM{x n } of ll the limits of convergent subsequences of {x n } is lso bounded The supremum of LIM{x n } is clled the upper limit nd denoted lim n x n The infimum of LIM{x n } is clled the lower limit nd denoted lim n x n For exmple, for the sequence in Exmple 1114, the upper limit is mx{l 1, l, l 3 } nd the lower limit is min{l 1, l, l 3 } The following chrcterizes the upper limit The lower limit cn be similrly chrcterized Proposition 110 Suppose {x n } is bounded sequence nd l is number 1 If l < lim n x n, then there re infinitely mny x n > l If l > lim n x n, then there re only finitely mny x n > l Proof If l < lim n x n, then by the definition of the upper limit, some l > l is the limit of subsequence By Proposition 114, ll the terms in the subsequence except finitely mny will be bigger thn l Thus we find infinitely mny x n > l The second sttement is the sme s the following: If there re infinitely mny x n > l, then l lim n x n We will prove this equivlent sttement

34 CHAPTER 1 LIMIT AND CONTINUITY Since there re infinitely mny x n > l, there is subsequence {x nk } stisfying x nk > l By Theorem 18, the subsequence hs further convergent subsequence {x nkp } Then x nk > l implies lim x nkp l This gives number in LIM{x n } tht is no less thn l, so tht lim n x n = sup LIM{x n } l Proposition 111 The upper nd lower limits re limits of convergent subsequences Moreover, the sequence converges if nd only if the upper nd lower limits re equl The first conclusion is lim n x n, lim n x n LIM{x n } In the second conclusion, the equlity lim n x n = lim n x n = l mens LIM{x n } = {l}, which bsiclly sys tht ll convergent subsequences hve the sme limit Proof Denote l = lim x n For ny ɛ > 0, we hve l + ɛ > lim x n nd l ɛ < lim x n Applying Proposition 110, we know there re infinitely mny x n > l ɛ nd only finitely mny x n > l + ɛ Therefore there re infinitely mny x n stisfying l + ɛ x n > l ɛ Thus we hve proved tht for ny ɛ > 0, there re infinitely mny x n stisfying x n l ɛ By Proposition 19 (nd the remrk fter the proof), this shows tht l is the limit of convergent subsequence For the second prt, Proposition 11 sys tht if {x n } converges to l, then LIM{x n } = {l}, so tht lim x n = lim x n = l Conversely, suppose lim x n = lim x n = l Then for ny ɛ > 0, pplying the second prt of Proposition 110 to l + ɛ > lim x n, we find only finitely mny x n > l + ɛ Applying the similr property for the lower limit to l ɛ < lim x n, we lso find only finitely mny x n < l ɛ Thus x n l ɛ holds for ll but finitely mny x n If N is the biggest index for those x n tht do not stisfy x n l ɛ, then we get x n l ɛ for ll n > N This proves tht {x n } converges to l Exercise 15 Find ll the upper nd lower limits of bounded sequences in Exercise 11 Exercise 16 Prove the properties of upper nd lower limits 1 lim n ( x n ) = lim n x n lim n x n + lim n y n lim n (x n + y n ) lim n x n + lim n y n 3 If x n > 0, then lim n 1 x n = 1 lim n x n 4 If x n 0 nd y n 0, then lim n x n lim n y n lim n (x n y n ) lim n x n lim n y n x n+1 Exercise 17 Prove tht if lim n x n < 1, then lim n x n = 0 Prove x n+1 tht if lim n x n > 1, then lim n x n = Exercise 18 Prove tht the upper limit l of bounded sequence {x n } is chrcterized by the following two properties

1 CONVERGENCE OF SEQUENCE LIMIT 35 1 l is the limit of convergent subsequence For ny ɛ > 0, there is N, such tht x n < l + ɛ for ny n > N The chrcteriztion my be compred with the one for the supremum in Exercise 19 Exercise 19 Let {x n } be bounded sequence Let y n = sup{x n, x n+1, x n+, } Then {y n } is bounded decresing sequence Prove tht lim n y n = lim n x n Find the similr formul for lim n x n 15 Convergence of Cuchy Sequence Now we re redy to prove the converse of Theorem 1 Theorem 11 Any Cuchy sequence of rel numbers is convergent Proof Let {x n } be Cuchy sequence We clim the sequence is bounded For ɛ = 1 > 0, there is N, such tht m, n > N implies x m x n < 1 Tking m = N + 1, we find n > N implies x N+1 1 < x n < x N+1 + 1 Therefore mx{x 1, x,, x N, x N+1 + 1} is n upper bound for the sequence, nd min{x 1, x,, x N, x N+1 1} is lower bound By Theorem 18, there is subsequence {x nk } converging to limit l Thus for ny ɛ > 0, there is K, such tht k > K = x nk l < ɛ On the other hnd, since {x n } is Cuchy sequence, there is N, such tht m, n > N = x m x n < ɛ Now for ny n > N, we cn esily find some k > K, such tht n k > N (k = mx{k, N} + 1, for exmple) Then we hve both x nk l < ɛ nd x nk x n < ɛ The inequlities imply x n l < ɛ Thus we estblished the impliction n > N = x n l < ɛ The proof of Theorem 11 does not use the upper nd lower limits Alterntively, we observe tht for ny ɛ > 0, the Cuchy condition implies tht ny two subsequences will be within ɛ of ech other fter finitely mny terms This implies tht the difference between the limits of ny two convergent subsequences cnnot be more thn ɛ Since ɛ cn be rbitrrily smll, this implies the upper nd the lower limits must be the sme Exmple 113 For the sequence x n = 1 1 + 1 3 + ( 1)n n,