CHAPTER 1. LIMIT Example 1. 1.16. The sequence n (2n)! satisfies n→2n-1n∞2n(2n-1)=4=0.75 lim By the order rule, we have -<0.8 for sufficiently big n, say for n> N(in fact N=8 is enough). Then for n>N, we have INN<0.8-A N+ C.0.8.C=0.8-x By Example 1.1.11, we have limn-oo 08n=0. Since C is a constant, by the sandwich rule, we get lim, Exercises 1. 1.52 and 1.1.53 summarise the idea of the example Exercise 1.1.52. Prove that if _n <c for a constant c< 1, then n converges to 0 Exercise 1. 1.53. Prove that if limn-oo-=I and 1, then n converges to 0 Exercise 1. 1.54. Find a such that the sequence converges to 0, P, q>0 Vn (n!)p (n!)2 11 (3n)! 6. 1.1.5 Subsequence A subsequence is obtained by choosing infinitely many terms from a sequence. We denote a subsequence by where the indices satisfy The following are some examples a. 6. a T2,6,24,·,T
22 CHAPTER 1. LIMIT Example 1.1.16. The sequence xn = 3 n (n!)2 (2n)! satisfies limn→∞ xn xn−1 = limn→∞ 3n 2 2n(2n − 1) = 3 4 = 0.75. By the order rule, we have xn xn−1 < 0.8 for sufficiently big n, say for n > N (in fact, N = 8 is enough). Then for n > N, we have 0 < xn = xn xn−1 xn−1 xn−2 · · · xN+1 xN xN < 0.8 n−N xN = C · 0.8 n , C = 0.8 −N xN . By Example 1.1.11, we have limn→∞ 0.8 n = 0. Since C is a constant, by the sandwich rule, we get limn→∞ xn = 0. Exercises 1.1.52 and 1.1.53 summarise the idea of the example. Exercise 1.1.52. Prove that if xn xn−1 ≤ c for a constant c < 1, then xn converges to 0. Exercise 1.1.53. Prove that if limn→∞ xn xn−1 = l and |l| < 1, then xn converges to 0. Exercise 1.1.54. Find a such that the sequence converges to 0, p, q > 0. 1. (2n)! (n!)2 a n . 2. (n!)2 (3n)!a n . 3. (n!)3 (3n)!a n . 4. p (2n)! n! a n . 5. √ n!a n 2 . 6. a n 2 n! . 7. a n 2 √ n! . 8. (n!)pa n . 9. a n (n!)p . 10. n qa n (n!)p . 11. (n!)p ((2n)!)q a n . 12. n 5 (n!)p ((2n)!)q a n . 1.1.5 Subsequence A subsequence is obtained by choosing infinitely many terms from a sequence. We denote a subsequence by xnk : xn1 , xn2 , . . . , xnk , . . . , where the indices satisfy n1 < n2 < · · · < nk < · · · . The following are some examples x2k : x2, x4, x6, x8, . . . , x2k, . . . , x2k−1 : x1, x3, x5, x7, . . . , x2k−1, . . . , x2 k : x2, x4, x8, x16, . . . , x2 k , . . . , xk! : x1, x2, x6, x24, . . . , xk! , . . .
1. 1. LIMIT OF SEQUENCE If n starts at n= l, then mi>l, which further implies nk k for all k Proposition 1.1.6. If a sequence converges to L, then any subsequence converges to l. Conversely, if a sequence is the union of finitely many subsequences that all converge to the same limit l, then the whole sequence converges to l 1 1 Example1117. Since -o is a subsequence of,imn→x=0 implies limn→∞m5= 0. We also know limn-oVns0 implies limn-oo=Obut not vice versa Example 1. 1.18. The sequence n+(-1)3 is the union of the odd subsequence (2k-1)-32k-4 2k+3 (2k-1)+2-2k+1 and the even subsequence Both subsequences con- verge to 1, either by direct computation, or by regarding them also as subsequences n+3 which converge to 1. Then we conclude lim, n+(-1)3 n→ Example 1.1. 19. The sequence(-1)n has one subsequence(-1)2=1 converging to 1 and another subsequence(-1 1 converging to-1. Since the two limits are different, by Proposition 1. 1.6, the sequence(-1)" diverges Example 1. 1.20. The sequence sin na converges to 0 when a is an integer multiple of T. Now assume 0< a< T. For any natural number k, the interval k, (k+1T of length T contains the following interval of length a(both intervals have the same middle point) k+1)丌 For even k, we have sin r> sin2/cos>0 on [ak, bk]. For odd k,we have sin < - e <0 on Jak, bkl Since the arithmetic sequence a, 2a, 3a,... has increment a, which is the length of ak, bk], we must have nk a E [ ak, bk] for some natural number nk. Then sin nika is a subsequence of sin na satisfying sin n2k 2 cos>0, and sin n2k+1a is a subsequence satisfying sin m2k <-coS Therefore the two subsequences cannot converge to the same limit. As a result, the sequence sin na diverges Now for general a that is not an integer multiple of T, we have a= 2NT+b for an integer N and b satisfying 0<b<T. Then we have sin na=+sin nb. We have shown that sin mb diverges, so that sin na diverges We conclude that sin na converges if and only if a is an integer multiple of T
1.1. LIMIT OF SEQUENCE 23 If xn starts at n = 1, then n1 ≥ 1, which further implies nk ≥ k for all k. Proposition 1.1.6. If a sequence converges to l, then any subsequence converges to l. Conversely, if a sequence is the union of finitely many subsequences that all converge to the same limit l, then the whole sequence converges to l. Example 1.1.17. Since 1 n2 is a subsequence of 1 n , limn→∞ 1 n = 0 implies limn→∞ 1 n2 = 0. We also know limn→∞ 1 √ n = 0 implies limn→∞ 1 n = 0 but not vice versa. Example 1.1.18. The sequence n + (−1)n3 n − (−1)n2 is the union of the odd subsequence (2k − 1) − 3 (2k − 1) + 2 = 2k − 4 2k + 1 and the even subsequence 2k + 3 2k − 2 . Both subsequences converge to 1, either by direct computation, or by regarding them also as subsequences of n − 4 n + 1 and n + 3 n − 2 , which converge to 1. Then we conclude limn→∞ n + (−1)n3 n − (−1)n2 = 1. Example 1.1.19. The sequence (−1)n has one subsequence (−1)2k = 1 converging to 1 and another subsequence (−1)2k−1 = −1 converging to −1. Since the two limits are different, by Proposition 1.1.6, the sequence (−1)n diverges. Example 1.1.20. The sequence sin na converges to 0 when a is an integer multiple of π. Now assume 0 < a < π. For any natural number k, the interval [kπ,(k + 1)π] of length π contains the following interval of length a (both intervals have the same middle point) [ak, bk] = kπ + π − a 2 ,(k + 1)π − π − a 2 . For even k, we have sin x ≥ sin π − a 2 = cos a 2 > 0 on [ak, bk]. For odd k, we have sin x ≤ − cos a 2 < 0 on [ak, bk]. Since the arithmetic sequence a, 2a, 3a, . . . has increment a, which is the length of [ak, bk], we must have nka ∈ [ak, bk] for some natural number nk. Then sin n2ka is a subsequence of sin na satisfying sin n2k ≥ cos a 2 > 0, and sin n2k+1a is a subsequence satisfying sin n2k ≤ − cos a 2 . Therefore the two subsequences cannot converge to the same limit. As a result, the sequence sin na diverges. Now for general a that is not an integer multiple of π, we have a = 2Nπ ± b for an integer N and b satisfying 0 < b < π. Then we have sin na = ± sin nb. We have shown that sin nb diverges, so that sin na diverges. We conclude that sin na converges if and only if a is an integer multiple of π
CHAPTER 1. LIMIT Exercise 1.1.55. Find the limit 2.(ml!)m 4.(7+(-1 +3)n Exercise 1. 1.56. Explain convergence or divergence. n7n7 7.(2-1)n+3y)2 n sll +2 (-1)2n+3 n-(-1)2 9. tan 1)n2 n+2 cos Exercise 1.1.57. Find all a such that the sequence cos na converges. Example 1. 1.21. We prove that limn-yoo n= l implies limn-oo P= 1. Exercises 1.1.58 and 1.1. 59 extend the result The sequence an is the union of two subsequences xk and ak(short for m, and Ink)satisfying all k 2 1 and all ak 1. By Proposition 1.1.6, the assumption limn→xn=1 implies that limk→oxk=limk→xk=1 Pick integers M and N satisfying M<P<N. Then k> l implies k< cpp<sk. By the arithmetic rule, we have limk-oo K=(limk-+00 k)M and similarly limk 1. Then by the sandwich rule, we get limk-oo apP=1 Similar proof shows that limk-yoo pp=1. Since the sequence ap is the union of two subsequences xkp and kp, by Proposition 1.1.6 again, we get limn-oo P=1 Exercise 1. 1.58. Suppose limn-oo In= l and yn is bounded. Prove that limn-ooIm=1 Exercise 1. 1.59. Suppose limn+oo In=l>0. By applying Example 1. 1.21 to the sequence prove that limn→oxh= 1.2 Rigorous Definition of Sequence Limit 1 The statement limn→∞n 0 means that-gets smaller and smaller as n gets bigger and bigger. To make the statement rigorous, we need to be more specif about smaller and bigger Is 1000 big? The answer depends on the context. A village of 1000 people is big, and a city of 1000 people is small(even tiny). Similarly, a rope of diameter less
24 CHAPTER 1. LIMIT Exercise 1.1.55. Find the limit. 1. √ n! + 1 − √ n! − 1. 2. (n!) 1 n! . 3. ((n + 1)!) 1 n! . 4. ((n + (−1)n )!) 1 n! . 5. (n!) 1 (n+1)! . 6. (2n 2−1 + 3n 2 ) 1 n2 . Exercise 1.1.56. Explain convergence or divergence. 1. 2 (−1)n . 2. n (−1)n n . 3. n (−1)n . 4. (−1)nn + 3 n − (−1)n2 . 5. (−1)nn 2 n3 − 1 . 6. √ n p n + (−1)n − √ n . 7. (2(−1)nn + 3n ) 1 n . 8. (2n + 3(−1)nn ) 1 n . 9. tan nπ 3 . 10. (−1)n sin nπ 3 . 11. sin nπ 2 cos nπ 3 . 12. n sin nπ 3 n cos nπ 2 + 2 . 13. n − sin nπ 3 n + 2 cos nπ 2 . Exercise 1.1.57. Find all a such that the sequence cos na converges. Example 1.1.21. We prove that limn→∞ xn = 1 implies limn→∞ x p n = 1. Exercises 1.1.58 and 1.1.59 extend the result. The sequence xn is the union of two subsequences x 0 k and x 00 k (short for xmk and xnk ) satisfying all x 0 k ≥ 1 and all x 00 k ≤ 1. By Proposition 1.1.6, the assumption limn→∞ xn = 1 implies that limk→∞ x 0 k = limk→∞ x 00 k = 1. Pick integers M and N satisfying M < p < N. Then x 0 k ≥ 1 implies x 0 k M ≤ x 0 k p ≤ x 0 k N . By the arithmetic rule, we have limk→∞ x 0 k M = (limk→∞ x 0 k ) M = 1M = 1 and similarly limk→∞ x 0 k N = 1. Then by the sandwich rule, we get limk→∞ x 0 k p = 1. Similar proof shows that limk→∞ x 00 k p = 1. Since the sequence x p n is the union of two subsequences x 0 k p and x 00 k p , by Proposition 1.1.6 again, we get limn→∞ x p n = 1. Exercise 1.1.58. Suppose limn→∞ xn = 1 and yn is bounded. Prove that limn→∞ x yn n = 1. Exercise 1.1.59. Suppose limn→∞ xn = l > 0. By applying Example 1.1.21 to the sequence xn l , prove that limn→∞ x p n = l p . 1.2 Rigorous Definition of Sequence Limit The statement limn→∞ 1 n = 0 means that 1 n gets smaller and smaller as n gets bigger and bigger. To make the statement rigorous, we need to be more specific about smaller and bigger. Is 1000 big? The answer depends on the context. A village of 1000 people is big, and a city of 1000 people is small (even tiny). Similarly, a rope of diameter less
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT than one millimeter is considered thin. But the hair is considered thin only if the diameter is less than 0. 05 millimeter So big or small makes sense only when compared with some reference quantity We say n is in the thousands if n> 1000 and in the millions if n >10000000 The reference quantities 1000 and 1000000 give a sense of the scale of bigness. In this spirit, the statement limn-oo -=0 means the following list of infinitely many plications 0<1, 0<0 n>100→|--0<0.01, n n>1000000=→ 0|<0.000001 For another example, limn-yoo ni=0 means the following implications n>10→→ 0<0.000 nI n>20=→ 0|<0.000000005 So the general shape of the implications is Note that the relation between N(measuring the bigness of n) and E(measuring the smallness of lin-lD) may be different for different limit The problem with infinitely many implications is that our language is finite In practice, we cannot verify all the implications one by verified the truth of the first one million implications, there is no guarantee that the one million and the first implication is true. To mathematically establish the truth of all implications, we have to formulate one finite statement that includes the consideration for alln and all e
1.2. RIGOROUS DEFINITION OF SEQUENCE LIMIT 25 than one millimeter is considered thin. But the hair is considered thin only if the diameter is less than 0.05 millimeter. So big or small makes sense only when compared with some reference quantity. We say n is in the thousands if n > 1000 and in the millions if n > 10000000. The reference quantities 1000 and 1000000 give a sense of the scale of bigness. In this spirit, the statement limn→∞ 1 n = 0 means the following list of infinitely many implications n > 1 =⇒ 1 n − 0 < 1, n > 10 =⇒ 1 n − 0 < 0.1, n > 100 =⇒ 1 n − 0 < 0.01, . . . n > 1000000 =⇒ 1 n − 0 < 0.000001, . . . For another example, limn→∞ 2 n n! = 0 means the following implications n > 10 =⇒ 2 n n! − 0 < 0.0003, n > 20 =⇒ 2 n n! − 0 < 0.0000000000005, . . . So the general shape of the implications is n > N =⇒ |xn − l| < . Note that the relation between N (measuring the bigness of n) and (measuring the smallness of |xn − l|) may be different for different limits. The problem with infinitely many implications is that our language is finite. In practice, we cannot verify all the implications one by one. Even if we have verified the truth of the first one million implications, there is no guarantee that the one million and the first implication is true. To mathematically establish the truth of all implications, we have to formulate one finite statement that includes the consideration for all N and all
CHAPTER 1. LIMIT 1.2.1 Rigorous Definition Definition 1.2.1( Rigorous). A sequence In converges to a finite number L, and denoted limn-oo In = L, if for any E>0, there is N, such that n >N implies n-ll In case N is a natural number(which can always be arranged if needed ), the definition means that, for any given horizontal E-band around l, we can find N, such that all the terms N+1, N+2, TN+3,.. after N lie in the shaded area in Figure 1.2.1 l+∈ +2xN+3 N Figure 1.2.1: n> N implies In-I<e Example 1.2.1. For any E>0, choose N The N This verifies the rigorous definition of limn+.? By applying the rigorous definition to E=0. 1, 0.01,.. we recover the infinitely many implications we wish to achieve. This justifies the rigorous definition of limit Example 1.2.2. For the constant sequence an=c, we rigorously prove lim For any E>0, choose N=0. Then ln -c=c-c=0 In fact, the right side is always true, regardless of the left side Example 1.2.3. We rigorously pro lin 0. fo
26 CHAPTER 1. LIMIT 1.2.1 Rigorous Definition Definition 1.2.1 (Rigorous). A sequence xn converges to a finite number l, and denoted limn→∞ xn = l, if for any > 0, there is N, such that n > N implies |xn − l| < . In case N is a natural number (which can always be arranged if needed), the definition means that, for any given horizontal -band around l, we can find N, such that all the terms xN+1, xN+2, xN+3, . . . after N lie in the shaded area in Figure 1.2.1. l l + l − N n x1 x2 x3 x4 xN+1 xN+2 xN+3 xn xn+1 Figure 1.2.1: n > N implies |xn − l| < . Example 1.2.1. For any > 0, choose N = 1 . Then n > N =⇒ 1 n − 0 = 1 n < 1 N = . This verifies the rigorous definition of limn→∞ 1 n = 0. By applying the rigorous definition to = 0.1, 0.01, . . . , we recover the infinitely many implications we wish to achieve. This justifies the rigorous definition of limit. Example 1.2.2. For the constant sequence xn = c, we rigorously prove limn→∞ c = c. For any > 0, choose N = 0. Then n > 0 =⇒ |xn − c| = |c − c| = 0 < . In fact, the right side is always true, regardless of the left side. Example 1.2.3. We rigorously prove limn→∞ 1 np = 0, for p > 0