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Limit of a Sequence Advanced Level Pure Mathematics Advanced Level Pure Mathematics 7 Calculus I Chapter 7 Limit of a Sequence 7.1 Introduction 2 7.2 Sequences 2 7.3 Convergent Sequences 5 7.4 Divergent Sequences and Oscillating Sequences 6 7.5 Operations on Limits of Sequences 7 7.6 Sandwich Theorem for Sequences 13 7.7 Monotonic Sequences 15 7.8 The Number e 21 7.9 Some Worked Examples 22 Page 1 Limit of a Sequence 7.1 Advanced Level Pure Mathematics INTRODUCTION Some examples of sequences: 7.2 1 1 1 1 , , , , 1 , 2 2 4 8 1. 1, 2. 1, 1, 1, 1, , (1)r+1 , 3. cos x , cos2x , cos3x , , cos rx , SEQUENCES Definition 7.1 A sequence {xn} is a function on the set of real numbers, and is usually written as x1 , x2 , x3 , , xn , . 1. The term xn is called the general term of the sequence. Example {1, 2, 4, 8, ...} is a sequence of positive integers with general term xn 2 n1 . 2. If the sequence has infinite number of terms, it is called an infinite sequence. 3. If the sequence has finite number of terms, it is called a finite sequence. 4. Sn = x1 + x2 + x3 + + xn + is said to form a series. 5. Sn = x r = x1 + x2 + x3 + + xn is a finite series. n r 1 6. Sn = x r = x1 + x2 + x3 + + xn + is an infinite series. r 1 n How to find the series sum , S n x r ? By using r 1 (M1) Mathematical induction. (M2) Method of difference where xr f (r 1) f (r ) such that n n r 1 r 1 S n x r [ f (r 1) f (r )] f (n 1) f (1) (M3) Partial fractions and method of difference. Page 2 Limit of a Sequence (M4) Advanced Level Pure Mathematics Standard formulae: (i) A.P. : a , a+d , a+2d , , a+( n1)d , . Sn = 1 [2a (n 1)d ] . 2 (ii) G.P. : a , a r , a r2 , , a rn1 , . n Sn = a (1 r ) ; 1 r (iii) 1 + 2 + 3 + + n = S = n r= r 1 12 + 22 + 32 + + n2 = a where | r| < 1 . 1 r 1 n( n 1) 2 n r 2 = 1 n(n 1)( 2n 1) 6 3 = 1 2 n (n 1) 2 4 r 1 13 + 23 + 33 + + n3 = n r r 1 (M5) Arithmetic-Geometric Series (A.G.P.) Sequence : a , (a+d)R , (a+2d)R2 , , [a+( n1)d)Rn1 , . n n1 Sn = a [a (n 1)d ]R dRa (1 R2 ) . 1 R (1 R) In particular, 1 , 2x , 3x2 , , nxn1 , . n Sn = rx r 1 r 1 (M6) 1 (n 1) x n nx n 1 . (1 x) 2 Harmonic Series Hn = 1 1 1 1 ... 2 3 4 There is no simple formula for the sum of the first n terms. (M7) Difference equation / Recurrence relation Recurrent sequence : (1) a n+1 = Aan for n = 1 , 2 , 3 , ... . (2) an+2 = Aan+1 + Ban for n = 1 , 2 , 3 , ... . where the coefficients A and B are constants. For case (1) , we have an+1 = Aan = A(Aan1) = A2 an1 = A3 an2 == Ar a1 If a1 = Aa , then an+1 = An+1 a. Page 3 Limit of a Sequence Advanced Level Pure Mathematics For case (2) , we have an+2 Aan+1 Ban = 0 . Let , be the roots of the auxiliary equation t 2 At B = 0 . (i) If , then an = c1n + c2n. (ii) If = , then an = (nc1 + c2 )n. Example 1 A sequence of real numbers an is defined by a0 = 0 , a1 = 1 and an+2 = an+1 + an for all n = 0, 1, ... . 1 Show that for all non-negative integers n , an = (n n ), where , are roots of the 5 equation x2 + x 1 = 0 with > . Soln. an+2 = an+1 + an an+2 + an+1 an = 0 Consider the auxiliary equation t 2 + t 1 = 0 , we have t = = 1 5 . 2 1 5 1 5 and = 2 2 Let an = c1 ( 1 5 )n + c ( 1 2 2 2 5 )n , where c1 , c2 R. Since a0 = 0 , we have 0 = c1 + c2 .................(1) Since a1 = 1 , we have 1 = c1 ( 1 5 ) + c ( 1 2 2 2 5 ) ...................(2) On solving (1) and (2) , we obtain c1 = Hence, an = 1 1 and c2 = . 5 5 1 n n ( ) , where , are roots of the equation x2 + x 1 = 0 5 with > . Page 4 Limit of a Sequence 7.3 Advanced Level Pure Mathematics CONVERGENT SEQUENCES A CONVERGENT SEQUENCE {xn} is a sequence whose terms will approach a Definition 7.2 finite value a as n tends to infinite. We say that xn a as n . Symbolically, lim x n a , which is called the limit of the convergent sequence. n 1 2 n Example If xn = ( ) , the sequence {xn} is convergent to 0 . Example The sequence { a } is convergent to 1. Example The sequence { 1 n Definition 7.3 sin r } is also convergent to 0. r A sequence {xn} is said to converge to a if and only if for any > 0 , there exists a positive integer N such that when n N , we have |xn a| < . a is called the limit value of {xn} and we write lim x n a . n Definition 7.4 A sequence {xn} not convergent is called DIVERGENT. Theorem 7.1 The limit value of a convergent sequence is UNIQUE. i.e. if xn a and xn b as n , then a = b. Theorem 7.2 All convergent sequences are BOUNDED. i.e. | xn | M for all n N . Important Facts: (1) If lim xn a , then lim | xn || a | . n n (2) If lim xn a , then lim xn p a for p = 1, 2, 3, ... n n (3) {xn } converges to a iff every subsequences of {xn} converges to a. i.e. odd sequence x1 , x3 , x5 , and even sequence x2 , x4 , x6 , both converges to a. (4) (5) lim xn 0 if | x | < 1. n lim n 1 0. n Page 5 Limit of a Sequence (6) (7) Advanced Level Pure Mathematics 1 lim x n 1 where x 0 . n lim 1 1n e . n n 0 if 0 1 (8) lim 1 if 0 n n if 0 7.4 DIVERGENT SEQUENCES AND OSCILLATING SEQUENCES Definition7.5 A sequence {xn} is said to diverge to positive infinity if for any positive real number M, there exists a positive integer N such that when n > N, xn > M. We write l i m x . A sequence {xn} is said to diverge to negative infinity if for any positive real number M , there exists a positive integer N such that when n > N , xn < M. We write l i m x . Theorem 7.3 Let {xn} be a sequence with xn 0 . 1 0. n x n Then lim xn if and only if lim n lim n lim N.B. n Definition 7.6 n 1 0 n Oscillating sequences are neither convergent nor diverging to infinity. Example {1, 1, 1, 1, 1, 1, ...} is an oscillating sequence. Example {0 , 4 , 0 , 8 , ... , n[1+(1)n ] , ...} is a infinitely oscillating sequence. Example (1) xn = sin n (2) xn = n cos n (3) xn = n (1)n (4) xn = 1 (1)n n Page 6 Limit of a Sequence 7.5 Advanced Level Pure Mathematics OPERATIONS ON LIMITS OF SEQUENCES Let {xn} and {yn} be two convergent sequences. Theorem 7.4 lim ( xn yn ) lim xn lim yn . (a) n n n lim ( xn yn ) lim xn lim yn . (b) n n n x lim xn lim n n , where lim yn 0 . n n y yn n nlim (c) lim (kxn ) k lim xn , where k is a constant. (d) n n (e) For any positive integer m, (i) lim ( xn ) ( lim xn )m , m n n 1 1 (ii) lim ( xn m ) ( lim xn ) m , n n (iii) lim xn m lim xn . n Example 2 n2 n 2 n 2n 2 2n 4 Find (a) lim n 2 4n (b) lim . n n 4n 1 a) HF p.60 (2.4a) b) HF p.61 (2.5) Example 3 Find. lim ( n 1 n ) n HF p.61 (2.6b) *N.B. We cannot use these rules of operations if {xn} or {yn} is NOT convergent. 1 1 Example lim n lim n lim n n n n n ( lim n , i.e. {n} diverges ) n Page 7 Limit of a Sequence Advanced Level Pure Mathematics 1 2 n Find lim 2 2 2 . n n n n Wrong proof : Example 4 k = 0 for k n. n n2 Since lim 2 n 1 lim 2 2 2 = 0 + 0 + = 0 n n n It is invalid since the sum of an infinite number of terms, each term tends to So n zero, may not be zero. HF p.62 (2.7) Example 5 Soln. 1 2 3 4 n Show that the limit value lim + (1) n 1 does not exist. n n n n n n When n is even, let n = 2k , where k is an integer. Then 1 2 3 4 n + ( 1) n 1 n n n n n 1 2 3 4 2k 1 2k + = 2k 2k 2k 2k 2k 2k 1 [(1 2) (3 4) (2k 1 2k )] = 2k 1 1 (k ) = = 2k 2 1 n 1 2 3 4 lim + (1) n 1 = n n 2 n n n n HF p.63 (2.8) Theorem 7.5 Let {xn} and {yn} be two sequences. (a) If lim x n 0 and | yn| M for all n , then lim x n y n 0 . n n Page 8 Limit of a Sequence Advanced Level Pure Mathematics (b) If lim x n and | yn| M for all n , then lim ( x n y n ) . n n (c) If lim x n and yn 0 for all n > N and yn does not converge to 0 , then n lim x n y n n Example 6 Soln. sin n . n n Evaluate lim Since |sinn| 1 for all n and lim Example 7 n (1) n n n Evaluate (a) lim 1 sin n = 0 , lim = 0. n n n (b) lim (n cos n) n (c) N.B. For Theorem 2.5 , lim yn does not necessarily exist. n Example 8 1 1 1 Evaluate lim n (1) n . n 1 3 3 5 (2 n 1)(2 n 1) HF p.67 (2.12) Example A sequence xn is defined by Page 9 lim n sin n n Limit of a Sequence Advanced Level Pure Mathematics x1 1 , x2 2 and 2 xn 2 xn 1 xn 0 Show that for all positive integers n , 1 xn 1 2 Hence evaluate lim xn . n Solution HF p.63 (2.9) Page 10 n2 ( n 1,2,3, ) Limit of a Sequence Example 9 Advanced Level Pure Mathematics A sequence xn is defined by x1 1 and xn 1 xn Find lim x2 n . n Solution HF p.65 (2.10) Page 11 1 ( n 1,2,3, ) 3n Limit of a Sequence 7.6 Advanced Level Pure Mathematics SANDWICH THEOREM FOR SEQUENCES Theorem If xn yn , then lim xn lim yn . Theorem 7.6 SANDWICH THEOREM FOR SEQUENCES n n Let {xn} , {yn} and {zn} be three sequences such that xn yn zn lim xn lim zn a and n lim yn a . Then Example10 Soln. n n sin n 0. n n Prove that lim Since 1 sin n 1 So Since 1 n lim 1 n sin n n lim 1 n 1 for n 1 n 0 Hence, by SANDWICH THEOREM for sequence, lim n Example 11 sin n 0. n 1 1 1 Find lim 2 . 2 n (2n)2 n (n 1) since 1 1 1 for all k = 0 , 1, 2, . . ., n ( 2n ) 2 (n k ) 2 n 2 HF p.72 (2.14) Page 12 Limit of a Sequence Advanced Level Pure Mathematics N.B. It is wrong to say 1 1 1 lim 2 2 n (2n)2 n (n 1) = 1 1 1 lim lim 2 2 n n n ( n 1) n (2 n) 2 = = 0 + 0 + ... 0 lim (totally infinitely many 0) 0 is an indeterminate form. 0 Other indeterminate forms : , , 0 , 1 , , etc. 0 because Example 12 1 1 1 Evaluate lim . 2 n n2 2 n2 n n 1 HF p.75 EX 2B (1c) Example 13 Let a be a real number greater than 1. Prove that lim n n 0. an Since a > 1, let a = 1+h where h is a positive real number. Then an = (1+h)n = HF p.72 (2.5) Page 13 Limit of a Sequence Advanced Level Pure Mathematics Example 14 Let A be a positive real number and {an} be a sequence of real numbers such that a1 A and an+1 = 1 A2 an for n 1. 2 an (a) Show that an A for all positive integers n. Hence, show that an A 1 (an1 A ). 2 (b) Find lim an by using sandwich theroem . n HF p.74 (2.18) Theorem 2.7 (a) If lim xn and there exists a positive integer N such that xn yn as n > N , n then lim y n . n (b) If lim xn and there exists a positive integer N such that xn yn as n > N , then n lim yn . n Example 15 Since 1 1 1 Find lim . n n 1 n 2 2 n [compare with Example14 and 15] 1 1 1 1 1 1 n 1 n2 2n nn nn 2n HF p.75 EX 2B (1d) Page 14 Limit of a Sequence 7.7 Advanced Level Pure Mathematics MONOTONIC SEQUENCES (1) A sequence {xn} is said to be monotonic increasing if and only if xn xn+1 for n = 1, 2, 3, ... . Definition 7.7 (2) A sequence {xn} is said to be monotonic decreasing if and only if xn xn+1 for n = 1, 2, 3, ... . (3) A sequence {xn} is said to be monotonic if and only if it is either increasing or decreasing. (4) A sequence {xn} is said to be strictly increasing or strictly decreasing if and only if xn < xn+1 or xn > xn+1 for all n. Example 16 Show that the sequence { Soln. 1 } is strictly decreasing. n 1 1 > for all n N n n 1 1 So the sequence { } is strictly decreasing. n Since n Example 17 Let a sequence {xn} be defined by xn = r 1 monotonic increasing. HF p.77 (2.19b) Page 1 r . Show that the sequence is 15 Limit of a Sequence Definition 7.8 Advanced Level Pure Mathematics (a) A sequence {xn} is said to be BOUNDED ABOVE if and only if there exists a constant M such that xn M for n = 1, 2, 3, ... . (b) A sequence {xn} is said to be BOUNDED BELOW if and only if there exists a constant M such that xn M for n = 1, 2, 3, ... . Theorem 7.8 (a) If a monotonic increasing sequence is bounded above, then the sequence is convergent and limit of {xn} exists. (b) If a monotonic decreasing sequence is bounded below, then the sequence is convergent and limit of {xn} exists. decreasing and bounded below increasing and bounded above n 1 for n 1 . r 1 3 1 Example 18 Let the sequence {xn} be defined by xn = r (a) Show that {xn} is an increasing sequence and bounded above. (b) Hence, show that {xn} is convergent. HF p.78 (2.20) Page 16 Limit of a Sequence N.B. Advanced Level Pure Mathematics (1) {xn} converges and xn < L , then lim xn L. n (2) {xn} converges and xn > L , then lim xn L. n Example 19 Let a and b be two positive real numbers. A sequence {xn} is defined by x n 1 ab 2 x n a 1 2 for n 1. It is given that 0 < x1 < b. (a) Show that xn b for all positive integer n. (b) Show that {xn} is monotonic increasing. Hence, show that {xn} is convergent. (c) Find lim xn . n HF p.82 (2.24) Page 17 Limit of a Sequence Advanced Level Pure Mathematics Example 20 Let a and b be two real numbers such that a > b > 0 . Two sequences {an} and {bn} are defined by an a n 1 bn 1 ab , bn a n 1bn 1 for n > 1 and a1 , b1 ab . 2 2 (a) Prove that {an} is monotonic decreasing. Hence deduce that {bn} is monotonic increasing. (b) Prove that {an} and {bn} converge to the same limit. HF p.83 (2.25) Page 18 Limit of a Sequence Example21 Let x1 > x2 > 0 and x n 1 Advanced Level Pure Mathematics 1 ( x n x n 1 ) . 2 (a) Show that {x2n1} is a decreasing sequence and {x2n} is an increasing sequence . (b) Show that x2n1 > x2m for all positive integers n and m . (c) Hence, show that {x2n1} and {x2n} have a common limit and find it . HF p.86 EX 2C (14) Page 19 Limit of a Sequence Advanced Level Pure Mathematics Example 22 Given two positive numbers a and b where a > b and {an}, {bn} are two sequences defined by a1 2a n 1bn 1 a bn 1 ab ab , b1 and a n n 1 , bn for all n 2, 2 ab a n 1 bn 1 2 (a) Show that bn bn+1 an+1 an . (b) Show that {an} and {bn} have the same limit. Find this common limit. BR p.326 (8.4) Page 20 Limit of a Sequence 7.8 Advanced Level Pure Mathematics THE NUMBER e Consider the sequence of numbers defined by: 1 (1 ) n , n 1, 2, 3, n The following table give the value of the sequence corresponding to different values of n. n (1 1 2 3 10 100 1000 10000 100000 2 2.25 2.3704 2.5937 2.7048 2.71692 2.71875 2.71827 1000000 1 ) n 2.718281 1 As the value of n increases without bounds, the value of (1 ) increases steadily, but it seems to n increase slower and slower. We can see that it would stop somewhere around 2.7182…. This number plays an important role in advanced mathematics and is denoted by e. n Definition e = lim 1 1 , where n takes positive integral values. n n Or lim ( 1 n 1 1 1 ) e 1! 2! n! e is an important irrational number in calculus. N.B. Theorem 7.9 2e3. Example 23 1 Find lim 1 . n 3n n 1 1 3n 3n 1 1 3 1 3 1 = = = e3 lim 1 lim 1 1 3lim n n 3 n 3n n 3n n Soln. Example 24 Express the following limits in terms of e. n (a) 1 lim 1 n n 1 (d) 3 2 lim 1 2 n n n 3 (b) lim 1 n 2n n HF p.88 EX2D (2.26, 1b, c, f) Page 21 n 1 (c) lim 1 n 2n n 1 Limit of a Sequence Advanced Level Pure Mathematics 7.9 SOME WORKED EXAMPLES 2n 2n 1 3 Example 35 Let a1 = 2 , b1 = and an = an1 , bn = bn1 for n 2 . 2n 1 2n 2 (a) Prove that an > bn and an bn = 2n + 1 for n 1 . (b) Using (a), or otherwise, show that an2 > 2n + 1 for n 1 . 1 . n a n Hence find lim [HKAL98] (7 marks) Example 26 (a) Let x > 1 and define a sequence {an} by a n 1 1 a1 = x and an = for n 2 . 2 a n 1 2 (i) Show that an > 1 and an > an+1 for all n . (ii) Show that lim an = 1 . (8 marks) n (b) Let f : [1,] R be a continuous function satisfying x2 1 for all x 1 . f(x) = f 2 x Using (a), show that f(x) = f(1) for all x 1. Example 27 [HKAL96] (7 marks) For any > 0 , define a sequence of real numbers as follows : a1 = + 1 , an = an1 + for n > 1 . a n 1 (a) Prove that (i) an2 an12 + 2 for n 2 ; (ii) an2 2 + 2n + 1 for n 1 . (2 marks) (b) Using (a), show that for n 2 , an2 2 + 2n + 1 + 2 n 1 k 1 2 2k 1 . (3 marks) (c) Prove that for k 1 , 2 2k 1 2 k k 1 1 dx . 2 x 1 (2 marks) 2 2 a (d) Using the above results, show that lim n exists and find the limit. n n State with reasons whether lim an n Page 2 n 22 exists. [HKAL95] (8 marks)