Calculus. Min Yan Department of Mathematics Hong Kong University of Science and Technology

Transcription

1 Clculus Min Yn Deprtment of Mthemtics Hong Kong University of Science nd Technology My, 7

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3 Contents Limit 7 Limit of Sequence 7 Arithmetic Rule 9 Sndwich Rule 3 Some Bsic Limits 5 4 Order Rule 9 5 Subsequence Rigorous Definition of Sequence Limit 4 Rigorous Definition 6 The Art of Estimtion 8 3 Rigorous Proof of Limits 3 4 Rigorous Proof of Limit Properties 33 3 Criterion for Convergence 37 3 Monotone Sequence 38 3 Appliction of Monotone Sequence 4 33 Cuchy Criterion 45 4 Infinity 48 4 Divergence to Infinity 48 4 Arithmetic Rule for Infinity 5 43 Unbounded Monotone Sequence 5 5 Limit of Function 53 5 Properties of Function Limit 53 5 Limit of Composition Function One Sided Limit 6 54 Limit of Trigonometric Function 63 6 Rigorous Definition of Function Limit 66 6 Rigorous Proof of Bsic Limits 67 6 Rigorous Proof of Properties of Limit 7 63 Reltion to Sequence Limit More Properties of Function Limit 78 7 Continuity 8 7 Mening of Continuity 8 7 Intermedite Vlue Theorem 8 3

4 4 CONTENTS 73 Continuous Inverse Function Continuous Chnge of Vrible 88 Differentition 93 Liner Approximtion 93 Derivtive 94 Bsic Derivtive 95 3 Constnt Approximtion 98 4 One Sided Derivtive Property of Derivtive Arithmetic Combintion of Liner Approximtion Composition of Liner Approximtion 3 Implicit Liner Approximtion 9 3 Appliction of Liner Approximtion 3 3 Monotone Property nd Extrem 3 3 Detect the Monotone Property 5 33 Compre Functions 8 34 First Derivtive Test 35 Optimiztion Problem 4 Min Vlue Theorem 5 4 Men Vlue Theorem 5 4 Criterion for Constnt Function 7 43 L Hospitl s Rule 9 5 High Order Approximtion 33 5 Tylor Expnsion 36 5 High Order Approximtion by Substitution Combintion of High Order Approximtions Implicit High Order Differentition Two Theoreticl Exmples 5 6 Appliction of High Order Approximtion 5 6 High Derivtive Test 5 6 Convex Function Sketch of Grph 58 7 Numericl Appliction 63 7 Reminder Formul 64 7 Newton s Method 66 3 Integrtion 7 3 Are nd Definite Integrl 7 3 Are below Non-negtive Function 7 3 Definite Integrl of Continuous Function Property of Are nd Definite Integrl 78 3 Rigorous Definition of Integrl 8

5 CONTENTS 5 3 Wht is Are? 8 3 Drboux Sum Riemnn Sum Numericl Clcultion of Integrl Left nd Right Rule Midpoint Rule nd Trpezoidl Rule Simpson s Rule 9 34 Indefinite Integrl Fundmentl Theorem of Clculus Indefinite Integrl Properties of Integrtion 35 Liner Property 35 Integrtion by Prts Chnge of Vrible 6 36 Integrtion of Rtionl Function 3 36 Rtionl Function 3 n x + b 36 Rtionl Function of cx + d Rtionl Function of sin x nd cos x 4 37 Improper Integrl 4 37 Definition nd Property 4 37 Comprison Test Conditionl Convergence 5 38 Appliction to Geometry Length of Curve Are of Region Surfce of Revolution Solid of Revolution Cvlieri s Principle Polr Coordinte Curves in Polr Coordinte Geometry in Polr Coordinte 89 3 Appliction to Physics 93 3 Work nd Pressure 93 3 Center of Mss 96 4 Series 99 4 Series of Numbers 99 4 Sum of Series 3 4 Convergence of Series 3 4 Comprison Test 34 4 Integrl Test 35 4 Comprison Test 37

6 6 CONTENTS 43 Specil Comprison Test 3 43 Conditionl Convergence Test for Conditionl Convergence Absolute vs Conditionl Power Series 3 44 Convergence of Tylor Series 3 44 Rdius of Convergence Function Defined by Power Series Fourier Series Fourier Coefficient Complex Form of Fourier Series Derivtive nd Integrtion of Fourier Series Sum of Fourier Series Prsevl s Identity 345

7 Chpter Limit Limit of Sequence A sequence is n infinite list x, x,, x n, The n-th term of the sequence is x n, nd n is the index of the term In this course, we will lwys ssume tht ll the terms re rel numbers Here re some exmples x n = n:,, 3,, n, ; y n = :,,,,, ; z n = n :,,, n, ; u n = ( ) n :,,,, ( ) n, ; v n = sin n: sin, sin, sin 3,, sin n, Note tht the index does not hve to strt from For exmple, the sequence v n ctully strts from n = (or ny even integer) Moreover, sequence does not hve to be given by formul For exmple, the deciml expnsions of π give sequence w n : 3, 3, 34, 34, 345, 3459, 3459, If n is the number of digits fter the deciml point, then the sequence w n strts t n = Now we look t the trend of the exmples bove s n gets bigger We find tht x n gets bigger nd cn become s big s we wnt On the other hnd, y n remins constnt, z n gets smller nd cn become s smll s we wnt This mens tht y n pproches nd z n pproches Moreover, u n nd v n jump round nd do not pproch nything Finlly, w n is equl to π up to the n-th deciml plce, nd therefore pproches π 7

8 8 CHAPTER LIMIT x n w n y n v n z n u n n Figure : Sequences Definition (Intuitive) If x n pproches finite number l when n gets bigger nd bigger, then we sy tht the sequence x n converges to the limit l nd write lim x n = l n A sequence diverges if it does not pproch specific finite number when n gets bigger The sequences y n, z n, w n converge respectively to, nd π The sequences x n, u n, v n diverge Since the limit describes the behvior when n gets very big, we hve the following property Proposition If y n is obtined from x n by dding, deleting, or chnging finitely mny terms, then lim n x n = lim n y n The equlity in the proposition mens tht x n converges if nd only if y n converges Moreover, the two limits hve equl vlue when both converge Exmple The sequence n + is obtined from n by deleting the first two terms By lim n n = nd Proposition, we get lim n n = lim n n + = In generl, we hve lim n x n+k = lim n x n for ny integer k The exmple ssumes lim n n =, which is supposed to be intuitively obvious Although mthemtics is inspired by intuition, criticl feture of mthemtics is rigorous logic This mens tht we need to be cler wht bsic fcts re ssumed in ny rgument For the moment, we will lwys ssume tht we lredy know

9 LIMIT OF SEQUENCE 9 lim n c = c nd lim n n p = for p > After the two limits re rigorously estblished in Exmples nd 3, the conclusions bsed on the two limits become solid Arithmetic Rule Intuitively, if x is close to 3 nd y is close to 5, then the rithmetic combintions x+y nd xy re close to 3+5 = 8 nd 3 5 = 5 The intuition leds to the following property of limit Proposition 3 (Arithmetic Rule) Suppose lim n x n = l nd lim n y n = k Then lim (x n + y n ) = l + k, n lim cx n = cl, n lim x n y n = kl, n x n lim = l n y n k, where c is constnt nd k in the lst equlity The proposition sys lim n (x n + y n ) = lim n x n + lim n y n However, the equlity is of different nture from the equlity in Proposition, becuse the convergence of the limits on two sides re not equivlent: If the two limits on the right converge, then the limit on the left lso converges nd the two sides re equl However, for x n = ( ) n nd y n = ( ) n+, the limit lim n (x n + y n ) = on the left converges, but both limits on the right diverge Exercise Explin tht lim n x n = l if nd only if lim n (x n l) = Exercise Suppose x n nd y n converge Explin tht lim n x n y n = implies either lim n x n = or lim n y n = Moreover, explin tht the conclusion fils if x n nd y n re not ssumed to converge Exmple We hve lim n n + n n n + = lim n = = + n n + n = ( lim n + ) n ( lim n n + ) n lim n + lim n n lim n lim n n + lim n + + = n lim n n

10 CHAPTER LIMIT The rithmetic rule is used in the second nd third equlities The limits lim n c = c nd lim n = re used in the fourth equlity n Exercise 3 Find the limits n + n 3 4 n 3 + 4n n 3 n (n + )(n + ) (n + )(n + ) n + n 3 5 (n + )(n + ) n 8 ( n) 3 n 3 + 3n 3 n 3n + 3n 4n + 6 n (n + )(n + ) 9 (n + 3) 3 (n 3 ) Exercise 4 Find the limits n + n n + 4 n 3 n n ( n + )(n + ) (n + )( n + ) n + n 3 5 ( n + )( n + ) n 8 ( 3 n) 3 3 n + 3n 3 n 3n + 3 n 4n + 6 n ( n + )( n + ) 9 ( 3 n + 3) 3 ( n ) Exercise 5 Find the limits n + n + b 4 n + n + b n + c 7 n 3 + b (c n + d) 6 n + n + b 5 ( n + )( n + b) cn + d 8 ( 3 n + b) (c n + d) 3 3 n + n + bn + c 6 cn + d ( n + )( n + b) 9 ( n + b) (c 3 n + d) 3 Exercise 6 Show tht p n p + p n p + + n + lim n b q n q + b q n q = + + b n + b Exercise 7 Find the limits, if p < q, p, if p = q nd b q b q n n 5 5 (n + ) n ( n + ) n 5 Exercise 8 Find the limits

11 LIMIT OF SEQUENCE n n + n n 4 n + n + b n + c n + d 7 n 3 + n + b n3 + c n + d n n + n n 5 n + n + b n + c n + d 8 n + n 3 + b n + c n 3 + d 3 n n n + n 6 n + n + c n + b n + d 9 n + 3 n + b n + c 3 n + d Exercise 9 Find the limits n + n + n + b n + c n + c n + d 3 ( ) n + n + b ( ) n + c n + d n + n + n + b n + b n + c n + c n + d n + d 4 ( n ) + n + b ( n ) + c n + d Exercise Find the limits, p, q > n p + n q + b 3 n p + n q + b np + c n q + d n p + bn q + c n q + bn p + c 4 n p + n p + n q + b n q + b Sndwich Rule The following property reflects the intuition tht if x nd z re close to 3, then nything between x nd z should lso be close to 3 Proposition 4 (Sndwich Rule) Suppose x n y n z n for sufficiently big n If lim n x n = lim n z n = l, then lim n y n = l Note tht something holds for sufficiently big n is the sme s something fils for only finitely mny n Exmple 3 By n 3 > n for sufficiently big n (in fct, n > 3 is enough), we hve < < n 3 n Then by lim n = lim n n = nd the sndwich rule, we get lim n n 3 = On the other hnd, for sufficiently big n, we hve n + < n nd n > n, nd therefore < n + n < n n = n

12 CHAPTER LIMIT By lim n n = lim n n = (rithmetic rule used) nd the sndwich rule, we get lim n n + n = Exmple 4 By sin n, we hve n sin n n n, n + n + sin n n By lim n n =, lim n n = nd the sndwich rule, we get lim sin n n n = Moreover, by lim n = lim n = (see rgument in Exmple n + n ) nd the sndwich rule, we get lim n n + sin n = Exercise Prove tht lim n x n = implies lim n x n = Exercise Find the limits, > 3n 4 n + 3 4n 3 n + b 4 n + b cn + d Exercise 3 Find the limits cos n n ( ) n n sin n n cos n n n + ( ) n cos n n + ( ) n cos n n + ( ) n cos n n + sin n ( ) n n + ( ) n + ( ) n 3 3 n cos n sin n + ( ) n cos n n + ( ) n sin n + cos n n 3 n + n + ( ) n n sin n + cos n n n + sin n n + cos n n + sin n n cos n ( ) n (n + ) n + ( ) n+ ( ) n (n + ) ( ) n n 5 Exercise 4 Find the limits

13 LIMIT OF SEQUENCE 3 n + n + ( ) n b 4 n + ( ) n n + ( ) n b 7 cos n + n + b sin n 3 n + n + b 5 ( ) n (n + b) n + c( ) n+ n + d 8 cos n + n + b sin n 3 n + c + d 3 n + n + b 6 ( ) n (n + b) + c ( ) n n + d 9 n + b sin n cn + d sin n Exercise 5 Find the limits, p > sin n n p sin(n + ) n p + ( ) n 3 sin n + b n p 4 + c cos(sin n) n p b sin n Exmple 5 For >, the sequence n + n stisfies By lim n < n + n = ( n + n)( n + + n) n + + n = n + + n < n n = nd the sndwich rule, we get lim n ( n + n) = Similr rgument lso shows the limit for < Exmple 6 The sequence n + n stisfies n + < < n + = + n n n ( By lim n + ) n + = + = nd the sndwich rule, we get lim n = n n Exercise 6 Show tht lim n ( n + n) = for < Exercise 7 Use the ide of Exmple 5 to estimte lim n n + n n + n nd then find n + Exercise 8 Show tht lim n n + b > b nd < b = You my need seprte rgument for Exercise 9 Find the limits n + n + b n + n + c + n + d

14 4 CHAPTER LIMIT 3 n + + n + b n + c + n + d 4 n + n + b n + c n + d n + n + bn + c n + n + b n + cn + d 5 n + + b n + c + d 6 n( n + n + b) 7 n + c( n + n + b) 8 n + + n + b n + c n 9 n + n + n + n + b n + c n + d n n + n + n + n + bn + c n + n + b n + cn + d Exercise Find the limits n + sin n n + b cos n n + sin n n + b cos n n + ( ) n 3 n + ( ) n b 5 n + ( ) n ( n + n + b) 6 n + n + sin n n + bn + cos n 7 n + n + sin n n + bn + cos n 4 n + + sin n n + c + ( ) n 8 n + n + b n + ( ) n c Exercise Find the limits 3 n + 3 n + b 3 n + n + b 3 3 n ( 3 n + 3 n + b) 4 3 n( 3 n + 3 n + b) Exercise Find the limits ( ) n 5 n + ( ) n 54 n + 3 ( ) n 4 n + ( ) n + p n + b Exercise 3 Find the limits ( n + sin n n + b cos n ) p ( n ) p + n + b n + ( ) n 3 c ( ) n + p n + bn + c Exercise 4 Suppose lim n x n = Use the rithmetic rule nd the sndwich rule to prove tht, if x n, then lim n x p n = Of course we expect the condition x n to be unnecessry See Exmple

15 LIMIT OF SEQUENCE 5 3 Some Bsic Limits Using lim n n Exmple 7 We show tht = nd the sndwich rule, we my estblish some bsic limits lim n n =, for > First ssume Then x n = n, nd This implies = ( + x n ) n = + nx n + n(n ) x n + + x n n > nx n x n < n By the sndwich rule nd lim n n =, we get lim n x n = Then by the rithmetic rule, this further implies lim n n = lim n (x n + ) = lim n x n + = For the cse <, let b = Then by the rithmetic rule, lim n n = lim n n = b Exmple 8 Exmple 7 cn be extended to lim n lim n n b = = n n = Let x n = n n Then we hve x n > for sufficiently lrge n (in fct, n is enough), nd This implies n = ( + x n ) n = + nx n + n(n ) x n + + x n n > x n < n n(n ) x n By lim n n = (see Exmple or 3) nd the sndwich rule, we get lim n x n = This further implies lim n n n = lim n x n + =

16 6 CHAPTER LIMIT Exmple 9 The following n-th root type limits cn be compred with the limits in Exmples 7 nd 8 < n n + < n n = n n n, < n n+ < n n, < (n n) n n < (n ) n n / = ( n n) 4 By Exmples 7, 8 nd the rithmetic rule, the sequences on the right converge to Then by the sndwich rule, we get lim n Exmple We hve n n + = lim n n+ = lim (n n) n n = n n 3 = n 3 n < n n + 3 n < n 3 n + 3 n = 3 n By Exmple 7, we hve lim n 3 n = 3 Then by the sndwich rule, we get lim n n n + 3 n = 3 For nother exmple, we hve Tking the n-th root, we get 3 n > 3 n n = 3 n + 3 n n > 3 n 3 > n 3 n n > 3 n 3 By lim n 3 n 3 = 3 nd the sndwich rule, we get lim n n 3 n n = 3 Exercise 5 Prove tht if x n b for some constnts, b > nd sufficiently big n, then lim n n x n = Exercise 6 Find the limits, > n n n n 3 n c n 4 (n + ) c n 5 (n + b) c n 6 (n + b) c n 7 ( n + ) c n 8 (n ) n+3 9 (n + b) c n+d (n + b) cn n +dn+e (n + b) c n+d (n + b) cn+d n +en+f Exercise 7 Find the limits, >

17 LIMIT OF SEQUENCE 7 n n + sin n n 3 n + ( ) n sin n n n + b sin n n 4 n + ( ) n b sin n 5 (n cos n) 6 (n + b sin n) n+sin n n n +c cos n Exercise 8 Find the limits, p, q > n n p + sin n n n p + n q 3 n+ n p + n q 4 n n p + n q Exercise 9 Find the limits n 5 n 4 n 4 n 5 n 3 4 n n 7 (5 n 4 n ) n+ n 5 n 3 4 n 5 n 4 n 5 n 8 (5 n 4 n ) n 3 n 5 n 3 4 n + n 6 n 4 n + ( ) n 5 n 9 (5 n 4 n ) n+ n + Exercise 3 Find the limits, > b > n n + b n 4 n n b n+ 7 ( n + b n ) n n n n b n 5 n+ n + b n 8 ( n b n ) n n 3 n n + ( ) n b n 6 n n b n 9 ( n ( ) n b n ) n n Exercise 3 For, b, c >, find lim n n n + b n + c n Exercise 3 For, prove lim n n = by using Exmple We show tht = ( n ) ( ( n ) n + ( n ) n + + n + ) lim n n =, for < First ssume < < nd write = Then b > nd + b < n = ( + b) n = < n(n ) + nb + b + + b n nb By lim n nb = nd the sndwich rule, we get lim n n = If < <, then < < nd lim n n = lim n n = By Exercise, we get lim n n = Exmple Exmple cn be extended to lim n nn =, for <

18 8 CHAPTER LIMIT This follows from < n n n = ( + b) n = n < n(n ) + nb + b + + b n the limit lim n = nd the sndwich rule (n )b Exercise 58 gives further extension of the limit n = n(n ) b (n )b, Exmple 3 We show tht n lim n n! for the specil cse = 4 For n > 4, we hve =, for ny < 4n n! = n n = 45 4! n 4 5 By lim n 4! n = nd the sndwich rule, we get lim 4 n n n! = Exercise 59 suggests how to show the limit in generl Exercise 33 Show tht lim n n n = for < in two wys The first is by using the ides from Exmples nd The second is by using lim n n n = for < Exercise 34 Show tht lim n n 54 n = for < Wht bout lim n n 54 n? Wht bout lim n n p n? Exercise 35 Show tht lim n n n! = for = 54 nd = 54 Exercise 36 Show tht lim n n n! = for = 54 nd = 54 Exercise 37 Show tht lim n n! n (n)! = for ny Exercise 38 Find the limits n + n 3 n n 5 n + n 3 n 7 n3 n ( + n ) n n 4 (n + ) n 6 n n + ( 3) n 4 n 8 5 n n6 n+ 3 n 3n+ Exercise 39 Find the limits Some convergence depends on nd p You my try some specil vlues of nd p first

19 LIMIT OF SEQUENCE 9 n p n n n p 4 5 n p n n p n! 7 8 n p n! n n p n! n p n 3 n! n! n (n)! 3 n p n 6 n p n n! 9 n p n n! n!n p n (n)! Exercise 4 Find the limits n + 3n + 5 n n! 3 n + n3 n + 5! n! 5 n + n! + (n )! 3 n n! + (n )! n + 3n + 5 n n! n + n Exercise 4 Prove n n! > Exercise 4 Prove n! n n lim n n! n n = lim n 4 Order Rule 4 n 3 n (n )! (n + )! n Then use this to prove lim n < (n!) nd n (n)! < n + (n!) (n)! = Wht bout lim n 6 n n! + 3 n (n )! 4 n (n )! 5 n n! n n! = for n > Then use this to prove (n!) k (kn)!? The following property reflects the intuition tht bigger sequence should hve bigger limit Proposition 5 (Order Rule) Suppose lim n x n = l nd lim n y n = k If x n y n for sufficiently big n, then l k If l < k, then x n < y n for sufficiently big n By tking y n = l, we get the following specil cses of the property for converging sequence x n If x n l for sufficiently big n, then lim n x n l If lim n x n < l, then x n < l for sufficiently big n Similr sttements with reversed inequlities lso hold (see Exercise 43) Note the non-strict inequlity in the first sttement of Proposition 5 nd the strict inequlity the second sttement For exmple, we hve x n = n < y n = n, but lim n x n lim n y n The exmple lso stisfies lim n x n lim n y n but x n y n, even for sufficiently big n

20 CHAPTER LIMIT Exercise 43 Explin how to get the following specil cses of the order rule If x n l for sufficiently big n, then lim n x n l If lim n x n > l, then x n > l for sufficiently big n n + n Exmple 4 By lim n = nd the order rule, we know < n n + n + n n n + n < 3 for sufficiently big n This implies < n n + n n + < n 3 for sufficiently big n By lim n n 3 = nd the sndwich rule, we get n n + n lim n n n + = Exmple 5 We showed lim n n 3 n n = 3 in Exmple Here we use different method, with the help of the order rule By n ( ) n ( ) n 3 n n = 3 n, we only need to find lim n n By 3 3 ( ) n ) Exmple, we hve lim n ( = By the order rule, therefore, 3 we hve ( ) n < < 3 for sufficiently big n This implies tht n < n ( ) n < n 3 for sufficiently big n Then by Exmple 7 nd the sndwich rule, we get ( ) n n lim n =, nd we conclude tht 3 lim n n 3n n n = 3 lim n ( ) n = 3 3 Exercise 44 Prove tht if lim n x n = l >, then lim n n x n = Moreover, find sequence stisfying lim n x n = nd lim n n x n = Cn we hve x n converging to nd n x n converging to 3? Exercise 45 Find the limits

21 LIMIT OF SEQUENCE n 5 n n4 n 3 n 5 n n4 n 5 (5 n n4 n ) n n + n+ 5 n n4 n 4 n 5 n ( ) n n4 n 6 (5 n ( ) n n4 n ) n+( ) n n + Exercise 46 Find the limits 3 n n 5n n4 n n n 5n ( ) n n4 n n+ n 5n n4 n sin n 4 (n 4 n 5 n ) n n n n n n + 4 n+ + 3n n n 3n + 3n n n 3n + n n + 3n (n 3n + 3n n (n 3n + 3n n ) n n ) n Exercise 47 Find the limits, > b n n+ + b n n n+ + ( ) n b n 3 n n+ + ( ) n b n 4 n 4 n 5b n 5 n 4 n + 5b n+ 6 7 n + b n n n + b n 8 n n n + (n + )b n 9 n+ n n + nb n+ ( n + b n ) n+ n + ((n + ) n + b n ) n n ( n + b n ) n 3 ( n + ( ) n b n ) ( ) n n Exercise 48 Find the limits,, b, c > n n n + nb n + c n 3 n (n + sin n) n + b n + n c n n n (b n + ) + nc n 4 n n (n + b n ( + nc n )) Exercise 49 Suppose polynomil p(n) = p n p + n n p + + n+ hs leding coefficient p > Prove tht p(n) > for sufficiently big n Exercise 5 Suppose, b, c >, nd p, q, r re polynomils with positive leding coefficients Find the limit of n p(n) n + q(n)b n + r(n)c n Exercise 5 Find the limits,, b, p, q > n n p + b sin n 3 n+ n p + bn q 5 n n p + bn q n n p + bn q 4 n n p + bn q 6 (n p + bn q ) n

22 CHAPTER LIMIT Exmple 6 The sequence x n = 3n (n!) lim n By the order rule, we hve x n (n)! 3n = lim x n n N = 8 is enough) Then for n > N, we hve stisfies n(n ) = 3 4 = 75 x n < 8 for sufficiently big n, sy for n > N (in fct, x n < x n = x n x n x n x n xn+ x N x N < 8 n N x N = C 8 n, C = 8 N x N By Exmple, we hve lim n 8 n = Since C is constnt, by the sndwich rule, we get lim n x n = Exercises 5 nd 53 summrise the ide of the exmple Exercise 5 Prove tht if x n x n Exercise 53 Prove tht if lim n c for constnt c <, then x n converges to x n x n = l nd l <, then x n converges to Exercise 54 Find such tht the sequence converges to, p, q > (n)! (n!) n (n)! 4 n n 7 n! n! (n!) (3n)! n 5 n! n 8 (n!) p n n q n (n!) p (n!) p ((n)!) q n 3 (n!) 3 (3n)! n 6 n n! 9 n (n!) p n 5 (n!) p ((n)!) q n 5 Subsequence A subsequence is obtined by choosing infinitely mny terms from sequence We denote subsequence by where the indices stisfy The following re some exmples x nk : x n, x n,, x nk,, n < n < < n k < x k : x, x 4, x 6, x 8,, x k,, x k : x, x 3, x 5, x 7,, x k,, x k : x, x 4, x 8, x 6,, x k,, x k! : x, x, x 6, x 4,, x k!,

23 LIMIT OF SEQUENCE 3 If x n strts t n =, then n, which further implies n k k for ll k Proposition 6 If sequence converges to l, then ny subsequence converges to l Conversely, if sequence is the union of finitely mny subsequences tht ll converge to the sme limit l, then the whole sequence converges to l Exmple 7 Since n is subsequence of n, lim n n = implies lim n n = We lso know lim n n = implies lim n = but not vice vers n Exmple 8 The sequence n + ( )n 3 is the union of the odd subsequence n ( ) n (k ) 3 (k ) + = k 4 k + 3 nd the even subsequence Both subsequences converge to, either by direct computtion, or by regrding them lso s subsequences k + k of n 4 n + nd n + 3 n, which converge to Then we conclude lim n + ( ) n 3 n n ( ) n = Exmple 9 The sequence ( ) n hs one subsequence ( ) k = converging to nd nother subsequence ( ) k = converging to Since the two limits re different, by Proposition 6, the sequence ( ) n diverges Exmple The sequence sin n converges to when is n integer multiple of π Now ssume < < π For ny nturl number k, the intervl [kπ, (k + )π] of length π contins the following intervl of length (both intervls hve the sme middle point) [ [ k, b k ] = kπ + π, (k + )π π ] ( ) π For even k, we hve sin x sin = cos > on [ k, b k ] For odd k, we hve sin x cos < on [ k, b k ] Since the rithmetic sequence,, 3, hs increment, which is the length of [ k, b k ], we must hve n k [ k, b k ] for some nturl number n k Then sin n k is subsequence of sin n stisfying sin n k cos >, nd sin n k+ is subsequence stisfying sin n k cos Therefore the two subsequences cnnot converge to the sme limit As result, the sequence sin n diverges Now for generl tht is not n integer multiple of π, we hve = Nπ ± b for n integer N nd b stisfying < b < π Then we hve sin n = ± sin nb We hve shown tht sin nb diverges, so tht sin n diverges We conclude tht sin n converges if nd only if is n integer multiple of π

24 4 CHAPTER LIMIT Exercise 55 Find the limit n! + n! (n!) n! 3 ((n + )!) n! 4 ((n + ( ) n )!) n! 5 (n!) (n+)! 6 ( n + 3 n ) n Exercise 56 Explin convergence or divergence ( )n 6 ( n n + ( ) n ) n sin nπ cos nπ 3 n ( )n n 3 n ( )n 4 5 ( ) n n + 3 n ( ) n ( ) n n n 3 7 ( ( )nn + 3 n ) n 8 ( n + 3 ( )nn ) n 9 tn nπ 3 ( ) n sin nπ 3 Exercise 57 Find ll such tht the sequence cos n converges 3 n sin nπ 3 n cos nπ + n sin nπ 3 n + cos nπ Exmple We prove tht lim n x n = implies lim n x p n = Exercises 58 nd 59 extend the result The sequence x n is the union of two subsequences x k nd x k (short for x m k nd x nk ) stisfying ll x k nd ll x k By Proposition 6, the ssumption lim n x n = implies tht lim k x k = lim k x k = Pick integers M nd N stisfying M < p < N Then x k implies x km x kp x kn By the rithmetic rule, we hve lim k x km = (lim k x k )M = M = nd similrly lim k x kn = Then by the sndwich rule, we get lim k x kp = Similr proof shows tht lim k x kp = Since the sequence x p n is the union of two subsequences x p k nd x kp, by Proposition 6 gin, we get lim n x p n = Exercise 58 Suppose lim n x n = nd y n is bounded Prove tht lim n x yn n = Exercise 59 Suppose lim n x n = l > By pplying Exmple to the sequence x n l, prove tht lim n x p n = l p Rigorous Definition of Sequence Limit The sttement lim n n = mens tht gets smller nd smller s n gets n bigger nd bigger To mke the sttement rigorous, we need to be more specific bout smller nd bigger Is big? The nswer depends on the context A villge of people is big, nd city of people is smll (even tiny) Similrly, rope of dimeter less

25 RIGOROUS DEFINITION OF SEQUENCE LIMIT 5 thn one millimeter is considered thin But the hir is considered thin only if the dimeter is less thn 5 millimeter So big or smll mkes sense only when compred with some reference quntity We sy n is in the thousnds if n > nd in the millions if n > The reference quntities nd give sense of the scle of bigness In this spirit, the sttement lim n = mens the following list of infinitely mny n implictions n > = n > = n > = n > = n <, n <, n <, n <, n For nother exmple, lim n n! n > = n > = = mens the following implictions n n! < 3, n n! < 5, So the generl shpe of the implictions is n > N = x n l < ɛ Note tht the reltion between N (mesuring the bigness of n) nd ɛ (mesuring the smllness of x n l ) my be different for different limits The problem with infinitely mny implictions is tht our lnguge is finite In prctice, we cnnot verify ll the implictions one by one Even if we hve verified the truth of the first one million implictions, there is no gurntee tht the one million nd the first impliction is true To mthemticlly estblish the truth of ll implictions, we hve to formulte one finite sttement tht includes the considertion for ll N nd ll ɛ

26 6 CHAPTER LIMIT Rigorous Definition Definition (Rigorous) A sequence x n converges to finite number l, nd denoted lim n x n = l, if for ny ɛ >, there is N, such tht n > N implies x n l < ɛ In cse N is nturl number (which cn lwys be rrnged if needed), the definition mens tht, for ny given horizontl ɛ-bnd round l, we cn find N, such tht ll the terms x N+, x N+, x N+3, fter N lie in the shded re in Figure x x x 3 x 4 x N+ x N+ x N+3 x n xn+ l + ɛ l l ɛ N n Figure : n > N implies x n l < ɛ Exmple For ny ɛ >, choose N = ɛ Then n > N = n = n < N = ɛ This verifies the rigorous definition of lim n n = By pplying the rigorous definition to ɛ =,,, we recover the infinitely mny implictions we wish to chieve This justifies the rigorous definition of limit Exmple For the constnt sequence x n = c, we rigorously prove For ny ɛ >, choose N = Then lim c = c n n > = x n c = c c = < ɛ In fct, the right side is lwys true, regrdless of the left side Exmple 3 We rigorously prove lim n =, for p > np

27 RIGOROUS DEFINITION OF SEQUENCE LIMIT 7 For ny ɛ >, choose N = Then ɛ p n > N = n p = n < p N = ɛ p We need to be more specific on the logicl foundtion for the rguments We will ssume the bsic knowledge of rel numbers, which re the four rithmetic opertions x + y, x y, xy, x y, the exponentil opertion xy (for x > ), the order x < y (or y > x, nd x y mens x < y or x = y), nd the properties for these opertions For exmple, we ssume tht we lredy know x > y > implies x < y nd x p > y p for p > These properties re used in the exmple bove More importnt bout the knowledge ssumed bove is the knowledge tht re not ssumed nd therefore cnnot be used until fter the knowledge is estblished In prticulr, we do not ssume ny knowledge bout the logrithm The logrithm nd its properties will be rigorously estblished in Exmple 75 s the inverse of exponentil n Exmple 4 To rigorously prove lim n =, for ny ɛ >, we hve n + n > N = ɛ = n n + = n + < N + = ( ) = ɛ ɛ + Therefore the sequence converges to How did we choose N = this is equivlent to n + N = should work ɛ? We wnt to chieve ɛ < ɛ, which we cn solve to get n > n n + < ɛ Since, choosing ɛ Exmple 5 To rigorously prove the limit in Exmple 5, we estimte the difference between the sequence nd the expected limit n + n = (n + ) n n + + n < n This shows tht for ny ɛ >, it is sufficient to hve words, we should choose N = 4 ɛ < ɛ, or n > 4 In other n ɛ

28 8 CHAPTER LIMIT The discussion bove is the nlysis of the problem, which you my write on your scrtch pper The forml rigorous rgument you re supposed to present is the following: For ny ɛ >, choose N = 4 ɛ Then n > N = n + n = < < = ɛ n + + n n N The Art of Estimtion In Exmples 4, the formul for N is obtined by solving n n + < ɛ in exct wy However, this my not be so esy in generl For exmple, for the limit in Exmple, we need to solve n + n n n + = 3n n n + < ɛ While the exct solution cn be found, the formul for N is rther complicted For more complicted exmple, it my not even be possible to find the formul for the exct solution We note tht finding the exct solution of x n l < ɛ is the sme s finding N = N(ɛ), such tht n > N x n l < ɛ However, in order to rigorously prove the limit, only = direction is needed The weker gol cn often be chieved in much simpler wy Exmple 6 Consider the limit in Exmple For n >, we hve n + n n n + = 3n n n + < 3n n n = 3 n 3 Since n < ɛ implies n + n n n + < ɛ, nd 3 < ɛ is equivlent to n n > 3 ɛ +, we find tht choosing N = 3 + is sufficient ɛ n > N = 3 ɛ + = n + n n n + = 3n n n + < 3 n < 3 N = ɛ Exercise Show tht n n + < n nd then rigorously prove lim n n n + = The key for the rigorous proof of limits is to find simple nd good enough estimtion We emphsize tht there is no need to find the best estimtion Any estimtion tht cn fulfill the rigorous definition of limit is good enough

29 RIGOROUS DEFINITION OF SEQUENCE LIMIT 9 Everydy life is full of good enough estimtions Mstering the rt of such estimtions is very useful for not just lerning clculus, but lso for mking smrt judgement in rel life Exmple 7 If bottle is % bigger in size thn nother bottle, how much bigger is in volume? The exct formul is the cube of the comprison in size ( + ) 3 = Since 3 = 6, 3 =, nd 3 is much smller thn, the bottle is little more thn 7% bigger in volume Exmple 8 The 3 GDP per cpit is 9,8USD for Chin nd 53,USD for the United Sttes, in terms of PPP (purchsing power prity) The percentge of the nnul GDP growth for the three yers up to 3 re 93, 77, 77 for Chin nd 8, 8, 9 for the United Sttes Wht do we expect the number of yers for Chin to ctch up to the United Sttes? First we need to estimte how much fster is the Chinese GDP growing compred to the United Sttes The comprison for 3 is (77 9) = + 58 Similrly, we get the (pproximte) comprisons + 75 nd + 49 for the other two yers Among the three comprisons, we my choose more conservtive + 5 This mens tht we ssume Chinese GDP per cpit grows 5% fster thn the United Sttes for the next mny yers Bsed on the ssumption of 5%, the number of yers n for Chin to ctch up to the United Sttes is obtined exctly by solving ( + 5) n = + n 5 + n(n ) n = 53, 9, 8 55 If we use + n 5 to pproximte ( + 5) n, then we get n 55 = 9 5 n(n ) However, 9 yers is too pessimistic becuse for n = 9, the third term 5 is quite sizble, so tht + n 5 is not good pproximtion of ( + 5) n An n exercise for the rt of estimtion, we try to void using clcultor in 53, getting better estimtion By 9, 8 3, we my solve n = m, ( + 5) m = + m 5 + m(m ) m 3

30 3 CHAPTER LIMIT We get n 3 m(m ) = 5 Since 5 is still sizble for m = 6 (but 5 giving much better pproximtion thn n = 9), the ctul n should be somewht smller thn 5 We try n = 4 nd estimte ( + 5) n by the first three terms ( + 5) So it looks like somewhere between 4 nd 45 is good estimtion We conclude tht, if Chinese GDP per cpit growth is 5% ( very optimistic ssumption) fster thn the United Sttes in the next 5 yers, then Chin will ctch up to the United Sttes in 4 some yers Exercise I wish to pint wll mesuring 3 meters tll nd 6 meters wide, give or tke % in ech direction If the cost of pint is $35 per squre meters, how much should I py for the pint? Exercise 3 In supermrket, I bought four items t $595, $635, $55, $7 The sles tx is 8% The finl bill is round $38 Is the bill correct? Exercise 4 In 9, Argentin nd Cnd hd the sme GDP per cpit In, the GDP per cpit is 9,3USD for Argentin nd 4,USD for Cnd On verge, how much fster is Cndin GDP growing nnully compred with Argentin in the th century? Next we leve rel life estimtions nd try some exmples in clculus Exmple 9 If x is close to 3 nd y is close to 5, then x 3y is close to = 9 We wish to be more precise bout the sttement, sy, we wnt to find tolernce for x nd y, such tht x 3y is within ± of 9 We hve (x 3y) ( 9) = (x 3) 3(y 5) x y 5 For the difference to be within ±, we only need to mke sure x y 5 < This cn be esily chieved by x 3 < = 4 nd y 5 < 4 +3 Exmple Agin we ssume x nd y re close to 3 nd 5 Now we wnt to find the percentge of tolernce, such tht x 3y is within ± of 9 We cn certinly use the nswer in Exmple 9 nd find the percentge % for x nd 4 8% for y This implies tht, if both x nd y re within 5 8% of 3 nd 5, then x 3y is within ± of 9 The better (or more honest) wy is to directly solve the problem Let δ nd δ be the percentge of tolernce for x nd y Then x = 3( + δ ) nd y = 5( + δ ), nd (x 3y) ( 9) = (x 3) 3(y 5) 3δ 3 5δ δ, δ = mx{ δ, δ }

31 RIGOROUS DEFINITION OF SEQUENCE LIMIT 3 To get δ to be within our trget of, we my tke our tolernce δ = 9% < 95 Exmple Assume x nd y re close to 3 nd 5 We wnt to find the tolernce for x nd y, such tht xy is within ± of 3 5 = 5 This mens finding δ >, such tht x 3 < δ, y 5 < δ = xy 5 < Under the ssumptions x 3 < δ nd y 5 < δ, we hve xy 5 xy 3y + 3y 5 x 3 y + 3 y 5 ( y + 3)δ We lso note tht, if we postulte δ, then y 5 < δ implies 4 < y < 6, so tht xy 5 ( y + 3)δ (6 + 3)δ = 9δ To get 9 δ to be within our trget of, we my tke our tolernce δ = < 9 Since this indeed stisfies δ, we conclude tht we cn tke δ = If the trgeted error ± is chnged to some other mount ±ɛ, then the sme rgument shows tht we cn tke the tolernce to be δ = ɛ Strictly speking, since we lso use δ in the rgument bove, we should tke δ = min{ ɛ, } Exercise 5 Find tolernce for x, y, z ner, 3, 5, such tht 5x 3y + 4z is within ±ɛ of Exercise 6 Find tolernce for x nd y ner nd, such tht xy is within ±ɛ of 4 Exercise 7 Find tolernce for x ner, such tht x is within ±ɛ of 4 Exercise 8 Find percentge of tolernce for x ner, such tht x 5 is within ± of 3 Rigorous Proof of Limits We revisit the limits derived before nd mke the rgument rigorous Exmple In Exmple 5, we rgued tht lim n ( n + n) = To mke the rgument rigorous, we use the estimtion in the erlier exmple In fct, regrdless of the sign of, we lwys hve n + n = ( n + n)( n + + n) n + + n = < n + + n n For the right side < ɛ, it is sufficient to hve n > Then we cn esily get n ɛ n > ɛ = n + n < ɛ

32 3 CHAPTER LIMIT This gives the rigorous proof of lim n ( n + n) = n + Exmple 3 In Exmple 6, we rgued tht lim n = To mke n the rgument rigorous, we use the estimtion in the erlier exmple The estimtion suggested tht it is sufficient to hve < ɛ Thus we get the following rigorous n rgument for the limit n < ɛ = < n + n < n + n = n < ɛ = n + n < ɛ Exercise 9 Rigorously prove the limits n + n 3 5 n + n 3 sin n n 3 4 n n + 3 n + n + b n 3n + 3n 4n n + n + b n n + n n n + n + b n + c n + d n + b cos n + n + b sin n n + n + b n + 3 n + b 4 3 n + 3 n Exercise Rigorously prove the limits, p > n p + b n p + n p + b 3 sin n + b n p + c Exmple 4 The estimtion in Exmple 7 tells us tht n < n for > This suggests tht for ny ɛ >, we my choose N = ɛ Then n > N = n < n < N = ɛ This rigorously proves tht lim n n = in cse Exmple 5 We try to rigorously prove lim n n n = for < Using the ide of Exmple, we write = Then < implies + b

33 RIGOROUS DEFINITION OF SEQUENCE LIMIT 33 b >, nd for n 3, we hve n n = n n = n ( + b) n n = n(n ) n(n )(n ) + nb + b! + b 3! b n n 3!n < = n(n )(n ) (n )(n )b < 3!n 3! b 3! 3 3 n n = b3 nb 3 Since 3! 3! < ɛ is the sme s n >, we hve nb3 b 3 ɛ n > 3! b 3 ɛ nd n 3 = n n < 3! nb < ɛ 3 { } 3! This shows tht we my choose N = mx b 3 ɛ, 3 It is cler from the proof tht we generlly hve lim n np n =, for ny p nd < n Exmple 6 We rigorously prove lim n = in Exmple 3 n! Choose nturl number M stisfying < M Then for n > M, we hve n n! < M n = M M M n! M M M + M M + M n M M M M! n = M M+ M! n Therefore for ny ɛ >, we hve { } M M+ n > mx M!ɛ, M = n n! < M M+ M! n < M M+ M! M M+ M!ɛ = ɛ Exercise Rigorously prove the limits n n n 54 n! 3 4 n p n! n 54 3 n n! 5 6 n p n n! n! n n 7 n p n, < 4 Rigorous Proof of Limit Properties The rigorous definition of limit llows us to rigorously prove some limit properties

34 34 CHAPTER LIMIT Exmple 7 Suppose lim n x n = l > We prove tht lim n xn = l First we clrify the problem The limit lim n x n = l mens the impliction For ny ɛ >, there is N, such tht n > N = x n l < ɛ The limit lim n xn = l mens the impliction For ny ɛ >, there is N, such tht n > N = x n l < ɛ We need to rgue is tht the first impliction implies the second impliction We hve x n l = ( x n l)( x n + l) xn + l = x n l xn + l x n l l Therefore for ny given ɛ >, the second impliction will hold s long s x n l l < ɛ, or x n l < lɛ The inequlity x n l < lɛ cn be chieved from the first impliction, provided we pply the first impliction to lɛ in plce of ɛ The nlysis bove leds to the following forml proof Let ɛ > By pplying the definition of lim n x n = l to lɛ >, there is N, such tht Then n > N = x n l < lɛ n > N = x n l < lɛ = x n l = ( x n l)( x n + l) xn + l = x n l xn + l x n l l < ɛ In the rgument, we tke dvntge of the fct tht the definition of limit cn be pplied to ny positive number, lɛ for exmple, insted of the given positive number ɛ Exmple 8 We prove the rithmetic rule lim n (x n + y n ) = lim n x n + lim n y n in Proposition 3 The concrete Exmple 9 provides ide of the proof Let lim n x n = l nd lim n y n = k Then for ny ɛ >, ɛ >, there re N, N, such tht n > N = x n l < ɛ, n > N = y n k < ɛ We expect to choose ɛ, ɛ s some modifiction of ɛ, s demonstrted in Exmple 7

35 RIGOROUS DEFINITION OF SEQUENCE LIMIT 35 Let N = mx{n, N } Then n > N = n > N, n > N = x n l < ɛ, y n k < ɛ = (x n + y n ) (l + k) x n l + y n k < ɛ + ɛ If ɛ + ɛ ɛ, then this rigorously proves lim n (x n + y n ) = l + k Of course this mens tht we my choose ɛ = ɛ = ɛ t the beginning of the rgument The nlysis bove leds to the following forml proof For ny ɛ >, pply the definition of lim n x n = l nd lim n y n = k to ɛ > We find N nd N, such tht n > N = x n l < ɛ, n > N = y n k < ɛ Then n > mx{n, N } = x n l < ɛ, y n k < ɛ = (x n + y n ) (l + k) x n l + y n k < ɛ + ɛ = ɛ Exmple 9 The rithmetic rule lim n x n y n = lim n x n lim n y n in Proposition 3 mens tht, if we know the pproximte vlues of the width nd height of rectngle, then multiplying the width nd height pproximtes the re of the rectngle The rigorous proof requires us to estimte how the pproximtion of the re is ffected by the pproximtions of the width nd height Exmple gives the key ide for such estimtion l k re= x l y y re= l y k x Figure : The error in product

36 36 CHAPTER LIMIT Let lim n x n = l nd lim n y n = k Then for ny ɛ >, ɛ >, there re N, N, such tht n > N = x n l < ɛ, n > N = y n k < ɛ Then for n > N = mx{n, N }, we hve (see Figure ) x n y n lk = (x n l)y n + l(y n k) x n l y n + l y n k < ɛ ( k + ɛ ) + l ɛ, where we use y n k < ɛ implying y n < k + ɛ The proof of lim n x n y n = lk will be complete if, for ny ɛ >, we cn choose ɛ > nd ɛ >, such tht ɛ ( k + ɛ ) + l ɛ ɛ This cn be chieved by choosing ɛ, ɛ stisfying ɛ, ɛ ( k + ) ɛ, l ɛ ɛ In other words, if we choose ɛ = ɛ ( k + ), { } ɛ ɛ = min, l t the very beginning of the proof, then we get rigorous proof of the rithmetic rule The forml writing of the proof is left to the reder Exmple The sndwich rule in Proposition 4 reflects the intuition tht, if x nd z re within ɛ of 5, then ny number y between x nd z is lso within ɛ of 5 x 5 < ɛ, z 5 < ɛ, x y z = y 5 < ɛ Geometriclly, this mens tht if x nd z lies inside n intervl, sy (5 ɛ, 5 + ɛ), then ny number y between x nd z lso lies in the intervl Suppose x n y n z n nd lim n x n = lim n z n = l For ny ɛ >, there re N nd N, such tht Then n > N = x n l < ɛ, n > N = z n l < ɛ n > N = mx{n, N } = x n l < ɛ, z n l < ɛ = l ɛ < x n, z n < l + ɛ = l ɛ < x n y n z n < l + ɛ y n l < ɛ

37 3 CRITERION FOR CONVERGENCE 37 Exmple The order rule in Proposition 5 reflects the intuition tht, if x is very close to 3 nd y is very close to 5, then x must be less thn y More specificlly, we know x < y when x nd y re within ± of 3 nd 5 Here is hlf of the distnce between 3 nd 5 Suppose x n y n, lim n x n = l, lim n y n = k For ny ɛ >, there is N, such tht (you should know from erlier exmples how to find this N) Picking ny n > N, we get n > N = x n l < ɛ, y n k < ɛ l ɛ < x n y n < k + ɛ Therefore we proved tht l ɛ < k + ɛ for ny ɛ > It is esy to see tht the property is the sme s l k Conversely, we ssume lim n x n = l, lim n y n = k, nd l < k For ny ɛ >, there is N, such tht n > N implies x n l < ɛ nd y n k < ɛ Then n > N = x n < l + ɛ, y n > k ɛ = y n x n > (k ɛ) (l + ɛ) = k l ɛ By choosing ɛ = k l y n > x n for n > N > t the beginning of the rgument, we conclude tht Exercise Prove tht if lim n x n = l, then lim n x n = l Exercise 3 Prove tht lim n x n l = if nd only if lim n x n = l Exercise 4 Prove tht if lim n x n = l, then lim n cx n = cl Exercise 5 Prove tht sequence x n converges if nd only if the subsequences x n nd x n+ converge to the sme limit This is specil cse of Proposition 6 Exercise 6 Suppose x n for sufficiently big n nd lim n x n = Prove tht lim n x p n = for ny p > Exercise 7 Suppose x n for sufficiently big n nd lim n x n = Suppose y n c for sufficiently big n nd some constnt c > Prove tht lim n x yn n = 3 Criterion for Convergence Any number close to 3 must be between nd 4, nd in prticulr hve the bsolute vlue no more thn 4 The intuition leds to the following result Theorem 3 If x n converges, then x n B for constnt B nd ll n

38 38 CHAPTER LIMIT The theorem bsiclly sys tht ny convergent sequence is bounded The number B is bound for the sequence If x n B for ll n, then we sy x n is bounded bove, nd B is n upper bound If x n B for ll n, then we sy x n is bounded below, nd B is lower bound A sequence is bounded if nd only if it is bounded bove nd bounded below n + ( ) n The sequences n, diverge becuse they re not bounded On the n + other hnd, the sequence,,,, is bounded but diverges Therefore the converse of Theorem 3 is not true in generl Exercise 3 Prove tht if x n is bounded for sufficiently big n, ie, x n B for n N, then x n is still bounded Exercise 3 Suppose x n is the union of two subsequences x k bounded if nd only if both x k nd x k re bounded nd x k Prove tht x n is 3 Monotone Sequence The converse of Theorem 3 holds under some dditionl ssumption A sequence x n is incresing if It is strictly incresing if x x x 3 x n x n+ x < x < x 3 < < x n < x n+ < The concepts of decresing nd strictly decresing cn be similrly defined Moreover, sequence is monotone if it is either incresing or decresing The sequences n, n, n re (strictly) decresing The sequences n, n re incresing Theorem 3 A monotone sequence converges if nd only if it is bounded An incresing sequence x n is lwys bounded below by its first term x Therefore x n is bounded if nd only if it is bounded bove Similrly, decresing sequence is bounded if nd only if it is bounded below The world record for meter dsh is decresing sequence bounded below by The proposition reflects the intuition tht there is limit on how fst humn being cn run We note tht the proposition does not tells us the exct vlue of the limit, just like we do not know the exct limit of the humn bility Exmple 3 Consider the sequence x n = n

39 3 CRITERION FOR CONVERGENCE 39 The sequence is clerly incresing Moreover, the sequence is bounded bove by x n (n )n ( = + ) ( + ) = n < Therefore the sequence converges The limit of the sequence is the sum of the infinite series n= We will see tht the sum is ctully π 6 ( n ) n n = n + Exercise 33 Show the convergence of sequences x n = n 3 x n = n 4 3 x n = (n )(n + ) 4 x n =! +! + + n! Exmple 3 The number is the limit of the sequence x n inductively given by x =, x n+ = + x n After trying first couple of terms, we expect the sequence to be incresing This cn be verified by induction We hve x = + > x = Moreover, if we ssume x n > x n, then x n+ = + x n > + x n = x n This proves inductively tht x n is indeed incresing Next we clim tht x n is bounded bove For n incresing sequence, we expect its limit to be the upper bound So we find the hypotheticl limit first Tking the limit on both sides of the equlity x n+ = + x n nd pplying the rithmetic rule, we get l = + l The solution is l = or Since x n >, by the order rule, we must hve l Therefore we conclude tht l =

40 4 CHAPTER LIMIT The hypotheticl limit vlue suggests tht x n < for ll n Agin we verify this by induction We lredy hve x = < If we ssume x n <, then x n+ = + x n < + = This proves inductively tht x n < for ll n We conclude tht x n is incresing nd bounded bove By Theorem 3, the sequence converges, nd the hypotheticl limit vlue is the rel limit vlue Figure 3 suggests tht our conclusion ctully depends only on the generl shpe of the grph of the function, nd hs little to do with the exct formul + x y = x f(x ) f(x 3 ) y = f(x) f(x ) x x x 3 x 4 l Figure 3: Limit of inductively defined sequence Exercise 34 Suppose sequence x n stisfies x n+ = + x n Prove tht if < x <, then x n is incresing nd converges to Prove tht if x >, then x n is decresing nd converges to Exercise 35 For the three functions f(x) in Figure 3, study the convergence of the sequences x n defined by x n+ = f(x n ) Your nswer depends on the initil vlue x Exercise 36 Suppose sequence x n stisfies x n+ = (x n + x n ) Prove the following sttements If x >, then the sequence is incresing nd diverges If < x <, then the sequence is decresing nd converges to 3 If < x <, then the sequence is incresing nd converges to 4 If < x <, then the sequence is decresing for n nd converges to

41 3 CRITERION FOR CONVERGENCE 4 Figure 3: Three functions 5 If x <, then the sequence is incresing for n nd diverges Exercise 37 Determine the convergence of inductively defined sequences Your nswer my depend on the initil vlue x x n+ = x n x n+ = x n + 3 x n+ = x n 4 x n+ = x n 5 x n+ = + x n 6 x n+ = x n Exercise 38 Determine the convergence of inductively defined sequences, > some cses, the sequence my not be defined fter certin number of terms In x n+ = + x n x n+ = x n 3 x n+ = x n 4 x n+ = 3 + x n 5 x n+ = 3 x n 6 x n+ = 3 x n Exercise 39 Explin the continued frction expnsion = Wht if on the right side is chnged to some other positive number? Exercise 3 For ny, b >, define sequence by x =, x = b, x n = x n + x n Prove tht the sequence converges Exercise 3 The rithmetic nd the geometric mens of, b > re + b By repeting the process, we get two sequences defined by x =, y = b, x n+ = x n + y n, y n+ = x n y n nd b

42 4 CHAPTER LIMIT Prove tht x n x n+ y n+ y n for n, nd the two sequences converge to the sme limit Exercise 3 The Fiboncci sequence,,, 3, 5, 8, 3,, 34, is defined by x = x = nd x n+ = x n + x n Consider the sequence y n = x n+ x n Find the reltion between y n+ nd y n Assume y n converges, find the limit l 3 Use the reltion between y n+ nd y n to prove tht l is the upper bound of y k nd the lower bound of y k+ 4 Prove tht the subsequence y k is incresing nd the subsequence y k+ is decresing 5 Prove tht the sequence y n converges to l Exercise 33 To find for >, we strt with guess x > of the vlue of Noting tht x nd re on the two sides of, it is resonble to choose the verge x x = ) (x + x s the next guess This leds to the inductive formul x n+ = (x n + xn ) s wy of numericlly computing better nd better pproximte vlues of Prove tht lim n x n = We my lso use weighted verge x n+ = ) (x n + xn s the next guess Do we 3 still hve lim n x n = for the weighted verge? 3 Compre the two methods for specific vlues of nd b (sy = 4, b = ) Which wy is fster? 4 Cn you come up with similr scheme for numericlly computing 3? Wht choice of the weight gives you the fstest method? 3 Appliction of Monotone Sequence We use Theorem 3 to prove some limits nd define specil number e Exmple 33 We give nother rgument for lim n n = in Exmple

43 3 CRITERION FOR CONVERGENCE 43 First ssume < < Then the sequence n is decresing nd stisfies < n < Therefore the sequence converges to limit l By the remrk in Exmple, we lso hve lim n n = l Then by the rithmetic rule, we hve l = lim n n = lim n n = lim n n = l Since, we get l = For the cse < <, we my consider the even nd odd subsequences of n nd pply Proposition 6 Another wy is to pply the sndwich rule to n n n Exmple 34 We give nother rgument tht the sequence x n = 3n (n!) in Exmple 6 converges to By lim n = 75 < nd the order rule, we hve (n)! x n x n x n < for sufficiently big n Since x n is lwys positive, we hve x n < x n for x n sufficiently big n Therefore fter finitely mny terms, the sequence is decresing Moreover, is the lower bound of the sequence, so tht the sequence converges Let lim n x n = l Then we lso hve lim n x n = l If l, then lim n x n x n = lim n x n lim n x n = l l = But the limit is ctully 75 The contrdiction shows tht l = Exercise 34 Extend Exmple 33 to proof of lim n n n = for < x n Exercise 35 Extend Exmple 34 to prove tht, if lim n x n lim n x n = = l <, then Exmple 35 For the sequence ( + n) n, we compre two consecutive terms by

44 44 CHAPTER LIMIT their binomil expnsions ( + ) n = + n n n ( + ) n+ = + n + n(n ) n(n ) ! n n! n n = +! + ( ) +! n + ( ) ( ) ( n ), n! n n n + n! +! +! ( (n + )! ( ) + n + ) ( n + ( n + ) ( n + ) ( n + n ) n + ) ( n n + A close exmintion shows tht the sequence is incresing Moreover, by the computtion in Exmple 3, the first expnsion gives ( + ) n < + n! +! + + n! < (n )n < 3 By Theorem 3, the sequence converges We denote the limit by e Exercise 36 Find the limit ( ) n + n+ n Exercise 37 Let x n = ( e = lim + n = n n) ( n) n 3 ( + n) n+ ( + ) n 4 n ) ( ) n + n n Use induction to prove ( + x) n + nx for x > nd ny nturl number n Use the first prt to prove x n x n > This shows tht x n is decresing 3 Prove tht lim n x n = e ( 4 Prove tht n) n is incresing nd converges to e

45 3 CRITERION FOR CONVERGENCE 45 Exercise 38 Prove tht for n > k, we hve ( + ) n + n! + (! n Then use Proposition 5 to show tht ) + + k! e +! +! + + k! ( + k ) k ( ) ( ) ( k ) n n n Finlly, prove ( lim + n! +! + + ) = e n! 33 Cuchy Criterion Theorem 3 gives specil cse tht we know the convergence of sequence without knowing the ctul limit vlue Note tht the definition of limit mkes explicit use of the limit vlue nd therefore cnnot be used to derive the convergence here The following provides the criterion for the convergence in generl, gin without referring to the ctul limit vlue Theorem 33 (Cuchy Criterion) A sequence x n converges if nd only if for ny ɛ >, there is N, such tht m, n > N = x m x n < ɛ Sequences stisfying the property in the theorem re clled Cuchy sequences The theorem sys tht sequence converges if nd only if it is Cuchy sequence The necessity is esy to see If lim n x n = l, then for big m, n, both x m nd x n re very close to l (sy within ɛ ) This implies tht x m nd x n re very close (within ɛ + ɛ = ɛ) The proof of sufficiency is much more difficult nd relies on the following deep result tht touches the essentil difference between the rel nd rtionl numbers Theorem 34 (Bolzno-Weierstrss) Any bounded sequence hs convergent subsequence Using the theorem, the converse my be proved by the following steps A Cuchy sequence is bounded By Bolzno-Weierstrss Theorem, the sequence hs convergent subsequence 3 If Cuchy sequence hs subsequence converging to l, then the whole sequence converges to l