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Chapter 4 Sequences and Limit of Sequences 4.1 Sequences: Basic De…nitions De…nition 278 (sequence) A sequence is a function whose domain is a subset of the form fn 2 Z : n n0 for some n0 2 Zg The elements or the terms of the sequence, usually denoted xn will be of the form: xn = f (n). If the terms of a sequence are denoted xn , then the sequence is denoted (xn ) or fxn g. We can think of a sequence as a list of numbers. In this case, a sequence will look like: fxn g = fx1 ; x2 ; x3 ; :::g. The starting point, n0 is usually 1 but it does not have to be. However, it is understood that whatever the starting point is, the elements xn should be de…ned for any n n0 . For 2n , then, we must have example, if the general term of a sequence is xn = n 4 n0 5. A sequence can be given di¤erent ways. 1. List the elements. For example, 1 2 3 ; ; ; ::: . From the elements listed, 2 3 4 the pattern should be clear. n n 2 2. Give a formula to generate the terms. For example, xn = ( 1) . If n! the starting point is not speci…ed, we use the smallest value of n which will work. 3. A sequence can be given recursively. The starting value of the sequence is given. Then, a formula to generate the nth term from one or more 103 104 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES previous terms. For example, we could de…ne a sequence by giving: x1 = 2 1 xn+1 = (xn + 6) 2 Another example is the Fibonacci sequence de…ned by: x1 = 1, x2 = 1 xn = xn 1 + xn 2 for n 3 Like a function, a sequence can be plotted. However, since the domain is a subset of Z, the plot will consist of dots instead of a continuous curve. Since a sequence is de…ned as a function. everything we de…ned for functions (bounds, supremum, in…mum, ...) also applies to sequences. We restate those de…nitions for convenience. De…nition 279 (Bounded Sequence) As sequence (xn ) is said to be bounded above if its range is bounded above. It is bounded below if its range is bounded below. It is bounded if its range is bounded. If the domain of (xn ) is fn 2 Z : n k for some integer kg then the above de…nition simply state that the set fxn : n kg must be bounded above, below or both. De…nition 280 (One-to-one Sequences) A sequence (xn ) is said to be oneto-one if whenever n 6= m then xn 6= xm . De…nition 281 (Monotone Sequences) Let (xn ) be a sequence. 1. (xn ) is said to be increasing if xn xn+1 for every n in the domain of the sequence. If we have xn < xn+1 , we say the sequence is strictly increasing. 2. (xn ) is said to be decreasing if xn the sequence. xn+1 for every n in the domain of 3. A sequence that is either increasing or decreasing is said to be monotone. If it is either strictly increasing or strictly decreasing, we say it is strictly monotone. De…nition 282 (constant) (xn ) is constant if xn+1 = xn 8n. (xn ) is eventually constant if xn+1 = xn 8n > N for some positive integer N . De…nition 283 (Periodic) (xn ) is periodic if 9p 2 Z+ such that 8n, xn+p = xn . Sometimes, when studying at a sequence, it is useful to look only at some of its terms, not all of them. For example, we could take every other term. We could also take some of the terms at random. More speci…cally, given a sequence fan g = fa1 ; a2 ; a3 ; :::g, we can de…ne a new sequence fbk g = fank g, made from 4.2. LIMIT OF A SEQUENCE: DEFINITIONS 105 some of the terms of fan g as follows: Let n1 denote the index of the …rst element of fan g we select and de…ne b1 = an1 .. Let n2 denote the index of the second element of fan g we select and de…ne b2 = an2 . In general, let nk denotes the index of the k th element of fan g we select and de…ne bk = ank . In doing so, note that nk k (as long as n 1 which we can achieve by renumbering). De…nition 284 (subsequence) The sequence fbk g as de…ned above is called a subsequence of fan g. Simply speaking, a subsequence is a sequence which only contains some of the terms of the original sequence. However, it still contains an in…nite amount of terms, in other words, it does not stop. Another way of de…ning a subsequence is given below. De…nition 285 (subsequence) A subsequence of a sequence (xn ) is a sequence of the form x'(n) where ' : N ! N is a strictly increasing function. For example, (x2n ) is a subsequence of (xn ), in this case, ' (n) = 2n. (x2n+1 ) is another subsequence, in this case ' (n) = 2n + 1. Example 286 Given a sequence (xn ) 1. (x2n ) is a subsequence of (xn ), in this case, ' (n) = 2n. 2. (x2n+1 ) is another subsequence, in this case ' (n) = 2n + 1. 4.2 Limit of a Sequence: De…nitions 4.2.1 The Concepts of Eventually and Frequently De…nition 287 (Eventually and frequently) Let (xn ) be a sequence and S R. 1. (xn ) is eventually in S if there exists N > 0 such that n N =) xn 2 S. 2. (xn ) is frequently in S if xn 2 S for in…nitely many values of n. Examples: n (xn ) where xn = ( 1) is frequently in the set f1g and well as frequently in the set f 1g. (xn ) where xn = 1 0 if 1 n 20 is eventually in the set f0g. if n > 20 (xn ) where xn = ln n is eventually is the set [2; 1). The following are simple observations regarding these two notions: If a sequence is in a set, then it is eventually in the set. 106 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES If a sequence is eventually in a set, then it is frequently in the set. (xn ) is eventually in S if and only if it is not frequently in R n S. (xn ) is frequently in S if and only if it is not eventually in R n S. The following is an important theorem regarding subsequences. Theorem 288 Let (xn ) be a sequence and S R. The following are equivalent: 1. (xn ) is frequently in S. 2. (xn ) has a subsequence in S. Proof. We prove both directions. 1. (1 =) 2). Since (xn ) is frequently in S, (xn ) 2 S for in…nitely many n. Let ni be the ith index of (xn ) for which (xn ) 2 S. Then the sequence (yn ) where yi = xni is both a subsequence of (xn ) and is in S. 2. (2 =) 1). Since a subsequence of (xn ) contains an in…nite number of terms of (xn ), the result follows. 4.2.2 De…nitions of Limits In the following, we assume that (xn ) is a sequence of real numbers and L is an extended real number. In this section, We wish to investigate the behavior of xn as n gets arbitrarily large, in other words as n ! 1. This is called …nding the limit of xn as n approaches 1, it is denoted lim xn . We want to know if n!1 the values of xn keep on changing as n ! 1, or if they seem to get closer to some number. Several things can happen. As n ! 1, xn could get arbitrarily large (or small), in other words, xn ! 1. Another possibility is that the values of xn will get closer and closer to some number we will call L. But, it is also possible that the values of xn will keep on changing forever. If we allow L to be an extended real number, then the …rst two cases can be summarized in lim xn = L. We now give a more precise de…nition of limits as well as partial n!1 limits. De…nition 289 (Limit) lim xn = L if for every neighborhood U of L, (xn ) n!1 is eventually in U . In this case, we write one of the following: lim xn = L n!1 lim xn = L xn ! L as n ! 1 xn ! L 4.2. LIMIT OF A SEQUENCE: DEFINITIONS 107 If L is a real number, we say that (xn ) converges to L or that L is a …nite limit for (xn ). Otherwise, we sat that (xn ) diverges. Note that (xn ) may diverge for two reasons. The …rst one is if L = 1, in this case, we say that (xn ) has an in…nite limit. The second is if (xn ) has no limit at all. De…nition 290 (Partial limit) L is a partial limit of (xn ) if for every neighborhood U of L, (xn ) is frequently in U . We now give equivalent conditions for a sequence to have a limit and a partial limit. In each case, we consider whether the limit (partial limit) is …nite or in…nite. Theorem 291 (Finite Limit) Let (xn ) be a sequence, and L 2 R. The following are equivalent: 1. L is a limit of (xn ) in other words lim xn = L. n!1 2. 8 > 0, (xn ) is eventually in (L ; L + ). 3. 8 > 0,9N > 0 such that n N =) xn 2 (L 4. 8 > 0,9N > 0 such that n N =) jxn ; L + ). Lj < . Proof. We prove (1 =) 2 =) 3 =) 4 =) 1). 1. (1 =) 2). Let > 0 be given. Then (L ; L + ) is a neighborhood of L. Thus, if lim xn = L it follows by de…nition that (xn ) is eventually in n!1 (L ; L + ). 2. (2 =) 3). This is the de…nition of eventually. 3. (3 =) 4). xn 2 (L =) jxn Lj < . ; L + ) =) L < xn < L + =) < xn L< 4. (4 =) 1). We need to show that (xn ) is eventually in U where U is any neighborhood of L. If U is any neighborhood of L, then by de…nition, U contains (L ; L + ) for some > 0. We can …nd N > 0 such that n =) N =) jxn < xn =) L =) xn 2 (L =) Lj < L< < xn < L + ;L + ) xn 2 U Thus xn is eventually in U . Since U was arbitrary, the result follows. 108 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Remark 292 Part 4 of theorem 291 is known as the limit. de…nition of the Theorem 293 (Finite Partial Limit) Let (xn ) be a sequence, and L 2 R. The following are equivalent: 1. L is a partial limit of (xn ). 2. 8 > 0, (xn ) is frequently in (L ; L + ). 3. 8 > 0, xn 2 (L ; L + ) for in…nitely many n. 4. 8 > 0,jxn for in…nitely many n. Lj < Proof. See problems Theorem 294 (In…nite Limit) Let (xn ) be a sequence, and L = 1 (the case L = 1 is similar and left to the reader). The following are equivalent: 1. 1 is a limit of (xn ) in other words lim xn = 1 n!1 2. 8M > 0, (xn ) is eventually in (M; 1). 3. 8M > 0, 9N > 0 such that n N =) xn 2 (M; 1). 4. 8M > 0, 9N > 0 such that n N =) xn > M Proof. See problems Theorem 295 (In…nite Partial Limit) Let (xn ) be a sequence, and L = 1 (the case L = 1 is similar and left to the reader). The following are equivalent: 1. 1 is a partial limit of (xn ). 2. 8M > 0, (xn ) is frequently in (M; 1). 3. 8M > 0, xn 2 (M; 1) for in…nitely many n. 4. 8M > 0, xn > M for in…nitely many n. 5. (xn ) is unbounded above. Proof. See problems 4.2. LIMIT OF A SEQUENCE: DEFINITIONS 4.2.3 109 Examples Before we look at examples and elementary theorems, let us make some remarks. jxn Lj simply represents the distance between xn and L. If we think of as a very small quantity, then the above de…nition says that we can make the distance between xn and L as small as we want, simply by taking n large enough . In fact, no matter how small is, we can make the distance smaller. In essence, this is saying that if n is large enough, xn and L become identical. Think of this as a game between two persons. The other person gives you a value for , that it the other person tells you how close he or she wants xn to be from L. You must be able to …nd N such that if n N , xn will be within the prescribed distance. Not only xn , but also all the terms after xn will be within the prescribed distance. Graphically, if we draw the horizontal lines y = L + and y = L , then saying n N ) jxn Lj < means that all the values of xn will fall within the region between the two lines whenever n N . It should be clear to the reader that to prove the limit is L, is arbitrary. Given any , the reader must prove the number N in the de…nition can be found. Usually, N will depend on . We now look at some examples to see how this is done. Example 296 Let (xn ) be such that xn = c, for some constant c. Show that xn ! c. We need to show that for any > 0, (xn ) is eventually in (c ; c + ). Since xn = c for any n, (xn ) is always in fcg (c ; c + ) for any > 0. Hence, (xn ) is eventually in (c ; c + ). Example 297 This example is similar to the previous one, the di¤ erence is that (xn ) is eventually constant, that is there exists N > 0 such that xn = c whenever n N . Again, show that xn ! c. The proof is similar. Here, we have that (xn ) is eventually in fcg (c ; c + ). n Example 298 Let (xn ) be such that xn = ( 1) . Prove that (xn ) does not have a limit (…nite or in…nite) but has two partial limits: 1 and 1. First, we prove that 1 is not a limit of (xn ). We can …nd M > 0 such that (xn ) is not eventually in (M; 1). Any number M > 1 will do. Next, we prove that 1 is not a limit of (xn ).We can …nd M < 0 such that (xn ) is not eventually in ( 1; M ). Any number M < 1 will do. Next, we prove that no real number can be a limit of (xn ). We note that xn only has two values, 1 and 1. For a real number x to be a limit of (xn ), both 1 and 1 would have to be eventually in (x ; x + ), for any 1 > 0. Clearly, if = , this cannot happen, no matter what x is. So, no 2 real number can be a limit of (xn ). 110 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Next, we prove that 1 is a partial limit. We need to prove that (xn ) is frequently in (1 ; 1 + ), for any > 0. By theorem 288, it is enough to show (xn ) has a subsequence in (1 ; 1 + ), for any > 0. The subsequence (x2n ) is in f1g (1 ; 1 + ), for any > 0. The result follows. Finally, we prove that 1 is a partial limit. This is similar. We simply note that the subsequence (x2n 1 ) is in f 1g. Example 299 Let (xn ) be such that xn = n. Prove that xn ! 1. We need to show that for any M > 0, there exists N > 0 such that n N =) xn > M . We want xn > M . This will happen when n > M (since xn = n). So, we see that if N is any integer greater than M , we will have n > M =) xn > M . Hence, the result. Example 300 Let (xn ) be such that xn = n2 . Prove that xn ! 1. We need to show that for any M > 0, there exists N > 0 such that n N =) xn > M . 2 2 We want xp n > M . This will happen when n > M (since xn =pn ), that is when n > M . So, we see that if N is any integer greater than M , we will have n > M =) xn > M . Hence, the result. n Example 301 Let (xn ) be such that xn = ( 1) n2 . Prove that (xn ) has no limit and that 1 are partial limits. 1. First, we see why neither 1 nor 1 can be limits of (xn ). (xn ) cannot be eventually in (M; 1) for any number M > 0 because if xn 2 (M; 1), xn+1 2 = (M; 1). For the same reason, (xn ) cannot be eventually in ( 1; M ) for any number M < 0. 2. Then, we see why a real number x cannot be a limit for (xn ). Let x 2 R. n For x to be a limit of this sequence, ( 1) n2 has to be eventually in (x ; x + ) for any > 0. We consider several cases. x = 0. jxn j increases without bounds, so xn will eventually get out of any interval of the form ( ; ). x 6= 0. Let there (why?) = jxj . If xn is in (x 2 ; x + ) then xn+1 cannot be 3. Finally, we see why 1 and 1 are partial limits of (xn ). The subsequence (x2n ) is eventually (M; 1) for any number M > 0 thus (xn ) is frequently in (M; 1) for any number M > 0. This means that 1 is a partial limit of (xn ). Similarly, the subsequence (x2n+1 ) is eventually in ( 1; M ) for any number M < 0. This means that 1 is a partial limit of (xn ). 1 Example 302 Show that the sequence whose general term is xn = converges n to 0. 4.2. LIMIT OF A SEQUENCE: DEFINITIONS Given > 0, we need to …nd N such that n 1 n we usually start with what we want, i.e. 111 N ) 1 n 0 < . To do this, 0 < , and we try to …nd the condition n has to satisfy for it to happen. 1 n 0 < , , 1 < n ,n> So, we see that if N = 1 , then n 1 < n N) 1 1 n 0 < . 1 Example 303 Show that the sequence whose general term is xn = 2 converges n to 0. 1 Given > 0, we need to …nd N such that n N ) 2 0 < . To do this, n 1 we usually start with what we want, i.e. 0 < , and we try to …nd the n2 condition n has to satisfy for it to happen. 1 n2 0 < , , 1 < n2 1 < n2 , n2 > 1 1 , n > p or n < 1 p 1 1 Since n is a natural number, we only retain n > p . So, we see that if N = p , then n N) 1 n 0 < . Example 304 Same example as above, using a slightly di¤ erent approach. 1 1 1 When we had 2 < . We could have noticed that if n > 1 we have 2 < . n n n 1 1 So, if we make < then what we need, 2 < will also happen. We can n n 1 1 1 make < simply by taking n > , it will follow that 2 0 < . This time, n n 1 we …nd that N = works. 112 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Remark 305 It seems that if we do the same problem di¤ erent ways, we …nd a di¤ erent solution. We really don’t. When we …nd N , the only thing we guarantee is that if n N , then jxn Lj < . It might also be true if n is slightly less than N . In other words, we are not …nding the smallest N that will work. If we use approximations like in example 304 , we may not get the best estimate for N. Remark 306 The idea is the following. Forget mathematics for a second (just a second). Imagine you are driving on a dirt road and you come to a bridge. You are not sure if the bridge can withstand the weight of your car. If ahead of you is a vehicle you know to be heavier than yours which just cleared the bridge, then you know your vehicle can clear the bridge as well. Remark 307 The above example illustrates a general technique to …nd N given . In general, the general term of the sequence is too complicated for us to be able to solve for n: If we call xn the general term of the sequence, we do the following: Start with jxn Lj and simplify it as much as possible. We get an expression involving n, call it E1 (n). If we still cannot solve for n in E1 (n) < , we try to …nd an upper bounds of jxn Lj, call it E2 (n) for which we can solve E2 (n) < . The result will follow. Since jxn Lj = E1 (n) < E2 (n), if we can have E2 (n) < , we will also have jxn Lj < . Obviously, E2 (n) has to be small enough so we can make it as small as we want. We illustrate this with the next example. Example 308 Show that the sequence fxn g whose general term is xn = 2n 1 3n + 2 2 converges to . 3 Again, as before, we look at what we need, and see if we can derive the condition n must satisfy. 2n 1 3n + 2 2 3 < () () 6n 3 6n 7 3 (3n + 2) 10 < 3 (3n + 2) Though here we could solve for n, we can also use an upper bound instead. We notice that 10 10 = 3 (3n + 2) 9n + 6 10 < 9n 10 < , it will follow that 9n 10 happens when n > . So, given > 0, n > 9 2n 1 2 proves that lim = . 3n + 2 3 Thus, if we make 2n 1 3n + 2 10 =) 9 2 < . The former 3 2n 1 2 < this 3n + 2 3 4.2. LIMIT OF A SEQUENCE: DEFINITIONS 113 p p Example 309 Prove that lim n + 1 n = 0. Again, as before, we look at what we need, and see if we can derive the condition n must satisfy. p p p p p p n+1 n n+1+ n p < n+1 n 0 < () p n+1+ n We work on the left side of the inequality and …nd an upper bound. p p p p n+1 n n+1+ n 1 p = p p p n+1+ n n+1+ n 1 p < 2 n p p 1 So, if we make p < , we will have what we need, that is n + 1 n < . 2 n p 1 1 The former happens when n > or n > 2 (since n > 0). Now, we see 2 4 p p 1 that given > 0, if n > 2 then n+1 n 0 < thus proving that 4 p p lim n + 1 n = 0. 4.2.4 Exercises 1 . Prove that xn ! 3. n 2 2. Let xn = 3 + . Prove that xn ! 3. n 1 3. Let xn = for n 2 N. If x 2 R : x 6= 0, prove that x is not a partial limit n of (xn ). 8 n n is a multiple of 3 < ( 1) n3 if 0 if n is one more than a multiple of 3 . Prove 4. Let xn = : 4 if n is two more than a multiple of 3 that the partial limits of (xn ) are 1; 1; 0; 4. 1. Let xn = 3 + 3 + 2n . Prove that xn ! 2. 5+n 8 1 > < if n = 2k 2n 6. Let xn = for any integer k. Prove that xn ! 1 > : 2 if n = 2k + 1 n +1 0. 5. Let xn = 7. Let xn = 1 n2 + 3n + 1 . Prove that xn ! . 2n2 + n + 4 2 8. Explain and justify the following observations made in the notes: 114 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES (a) If a sequence is in a set, then it is eventually in the set. (b) If a sequence is eventually in a set, then it is frequently in the set. (c) (xn ) is eventually in S if and only if it is not frequently in R n S. (d) (xn ) is frequently in S if and only if it is not eventually in R n S. 9. Prove theorem 288. 10. Prove theorem 293. 11. Prove theorem 294. 12. Prove theorem 295. 4.3. LIMIT OF A SEQUENCE: THEOREMS 4.3 115 Limit of a Sequence: Theorems These theorems fall in two categories. The …rst category deals with ways to combine sequences. Like numbers, sequences can be added, multiplied, divided, ... Theorems from this category deal with the ways sequences can be combined and how the limit of the result can be obtained. If a sequence can be written as the combination of several "simpler" sequences, the idea is that it should be easier ti …nd the limit of the "simpler" sequences. These theorems allow us to write a limit in terms of easier limits. however, we still have limits to evaluate. The second category of theorems deal with speci…c sequences and techniques applied to them. Usually, computing the limit of a sequence involves using theorems from both categories. 4.3.1 Limit Properties We begin with a few technical theorems. They do not play an important role in computing limits, but they play a role in proving certain results about limits. Theorem 310 Let x be a number such that 8 > 0, jxj < , then x = 0. Proof. See problems at the end of the section. Theorem 311 If a sequence converges, then its limit is unique. Proof. We do a proof by contradiction. Assume that an ! L1 and an ! L2 . Given > 0 choose N1 such that n N1 =) jan L1 j < . Similarly, choose 2 N2 such that n N2 =) jan L2 j < . Let N = max (N1 ; N2 ). If n N, 2 then jL1 L2 j = jL1 jan < By theorem 310 , it follows that L1 2 + an + an L1 j + jan 2 L2 j L2 j = L2 = 0, that is L1 = L2 . Theorem 312 If a sequence converges, then it is bounded, that is there exists a number M > 0 such that jan j M for all n. Proof. Choose N such that n N =) jan Lj < 1. By the triangle inequality, we have jan j jLj jan Lj <1 Thus, if n N , jan j < 1 + jLj. Let M1 = max (ja1 j ; ja2 j ; :::; jaN j). Let M = max (M1 ; 1 + jLj). Then, clearly, jan j < M 116 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Theorem 313 If a sequence converges, then 8 > 09N : m; n jan am j < Proof. Given > 0, we can choose N such that n; m N =) jan and jam Lj < jan 2 N =) Lj < 2 . Now, am j = jan L+L = j(an L) j(an < 2 + am j (am L)j + jam 2 L)j Lj by the triangle inequality = The above theorem simply says that if a sequence converges, then the difference between any two terms gets smaller and smaller. It should also be clear to the reader that if an ! L, then so does an+k where k is any natural number. The above theorem can be used to prove that a sequence does not converge by proving that the di¤erence between two of its terms does not get smaller and smaller. Example 314 Find lim cos n We suspect the sequence diverges, as its values will oscillate between -1 and 1 . We can actually prove it using theorem 313. We notice that for any n, jcos 2n cos ((2n + 1) )j = 2. For the sequence to converge, this di¤ erence should approach 0. Hence, the sequence diverges. Theorem 315 Suppose that (an ) converges. Then, any subsequence (ank ) also converges and has the same limit. Proof. Suppose that an ! L. Let bk = ank be a subsequence. We need to prove that given > 0, there exists N such that k N =) jbk Lj < . Let > 0 be given. Choose N such that nk N =) jank Lj < . Now, if k N , then nk N therefore jbk Lj = jank Lj < This theorem is often used to show that a given sequence diverges. To do so, it is enough to …nd two subsequences which do not converge to the same limit. Alternatively, once can …nd a subsequence which diverges. Example 316 Study the convergence of cos n The subsequence cos 2n converges to 1, while the subsequence cos (2n + 1) converges to 1. Thus, cos 2n must diverge. In the next two sections, we look at theorems which give us more tools to compute limits. 4.3. LIMIT OF A SEQUENCE: THEOREMS 4.3.2 117 Limit Laws The theorems below are useful when …nding the limit of a sequence. Finding the limit using the de…nition is a long process which we will try to avoid whenever possible. Since all limits are taken as n ! 1, in the theorems below, we will write lim an for lim an . n!1 Theorem 317 Let (an ) and (bn ) be two sequences such that an ! a and bn ! b with a and b real numbers. Then, the following results hold: 1. lim (an bn ) = (lim an ) (lim bn ) = a b 2. lim (an bn ) = (lim an ) (lim bn ) = ab an bn 3. if lim bn = b 6= 0 then lim = lim an a = lim bn b 4. lim jan j = jlim an j = jaj 5. if an 0 then lim an 6. if an bn then lim an 7. if lim an = a 0 lim bn p p p 0 then lim an = lim an = a Proof. We prove some of these items. The remaining ones will be assigned as problems at the end of the section. 1. We prove (lim an ) 9N : n such that jbn lim (an + bn ) = (lim an ) + (lim bn ). The proof of lim (an bn ) = (lim bn ) is left as an exercise. We need to prove that 8 > 0, N =) jan + bn (a + b)j < . Let > 0 be given, choose N1 n N1 =) jan aj < . Choose N2 such that n N2 =) 2 bj < . Let N = max (N1 ; N2 ). If n N , then 2 jan + bn (a + b)j = j(an jan < 2 + a) + (bn aj + jbn 2 b)j bj by the triangle inequality = 2. We need to prove that 8 > 0, 9N : n N =) jan bn abj < . Since an converges, it is bounded, let M be the bound i.e. jan j < M . Choose N1 such that n N1 =) jan aj < . Choose N2 such that 2 (jbj + 1) 118 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES n N2 =) jbn bj < jan bn 2 (M + 1) . Let N = max (N1 ; N2 ). If n abj = jan bn an b + an b = jan (bn <M < 2 bj + jbj jan 2 (M + 1) + 2 abj b) + b (an jan j jbn + jbj N then a)j aj 2 (jbj + 1) = 3. See problems 4. We need to prove that jan j ! jaj that is 8 > 0, 9N : n jjan j jajj < . Let > 0 be given, choose N such that n jan aj < (since an ! a) . If n N , then we have: jjan j jajj < jan N =) N =) aj by the triangle inequality < 5. We prove it by contradiction. Assume that an ! a < 0. Choose N such 1 a. Then, that n N ) jan aj < 2 an 1 a 2 a< 1 a 2 ) an < 0 ) an < which is a contradiction. 6. We apply the results found in parts 1 and 5 to the sequence an bn . 7. See problems Remark 318 Parts 1, 2 and 3 of the above theorem hold even when a and b are extended real numbers as long as the right hand side in each part is de…ned. You will recall the following rules when working with extended real numbers: 1. 1 + 1 = 1 2. 1 1 = ( 1) ( 1) = 1 1 = ( 1) 1 = 1 ( 1) = 3. If x is any real number, then (a) 1 + x = x + 1 = 1 1 4.3. LIMIT OF A SEQUENCE: THEOREMS 119 (b) 1+x=x 1= 1 x x (c) = =0 1 1 x 1 if x > 0 (d) = 1 if x < 0 0 (e) 1 x=x (f ) ( 1) 1 if x > 0 1 if x < 0 1= x=x 1 if 1 if ( 1) = x>0 x<0 4. However, the following are still indeterminate forms. Their behavior is unpredictable. Finding what they are equal to requires more advanced techniques such as l’Hôpital’s rule. (a) 1 + 1 and 1 (b) 0 1 and 1 0 1 and (c) 1 0 0 1 Remark 319 When using theorems from this category, it is important to remember previous results since these theorems allow us to write a limit in terms of other limits, we hopefully know. The more limits we know, the better o¤ we are. Example 320 If c 6= 0, …nd lim nc . We know from an example in the previous section that lim n1 = 0. Therefore lim c n = c lim = c = 1 n 0 0 Example 321 Find lim n12 . In the previous section, we computed this limit using the de…nition. We can also do it as follows. lim 1 n2 1 n 1 = lim n = 0 0 = = lim 1 n lim 1 n 0 Remark 322 From the above example, we can see that if p is a natural number, lim n1p = 0. 120 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES 2 Example 323 Find lim n2n+3n 2 +1 . This problem involves using a standard technique you should remember. We show all the steps, then we will draw a general conclusion. We begin by factoring the term of highest degree from both the numerator and denominator. lim n2 + 3n 2n2 + 1 = lim n2 1 + n3 n2 2 + n12 = lim 1 + n3 2 + n12 = lim (1) + 3 lim = 1 Now, lim 1 + 3 n 1 n and lim 2 + 1 n2 = lim (2) + lim = 2 1 n2 Since both the limit of the numerator and denominator exist and the limit of the denominator is not 0, we can write lim n2 + 3n 2n2 + 1 = = lim 1 + n3 lim 2 + n12 1 2 Remark 324 The same technique can be applied to every fraction for which the numerator and denominator are polynomials in n. We see that the limit of such a fraction will be the same as the limit of the quotient of the terms of highest degree. 4.3.3 More Theorems on Limits In example 304 , we used an approximation to simplify the problem a little bit. In this particular example, the approximation was not really necessary, it was more to illustrate a point. Sometimes, if the problem is more complicated, it may be necessary to use such an approximation in order to be able to …nd the condition n has to satisfy. In other words, when we try to satisfy jxn Lj < , we usually simplify jxn Lj to some expression involving n. Let E1 (n) denote this expression. This gives us the inequality E1 (n) < which we have to solve for n. If it is too hard, we then try to …nd a second expression we will call E2 (n) such that E1 (n) < E2 (n) < . E2 (n) should be such that solving the inequality E2 (n) < is feasible and easier. In order to achieve this, several tricks are used. We recall some useful results, as well as some theorems below. 4.3. LIMIT OF A SEQUENCE: THEOREMS 121 Theorem 325 (Bernoulli’s inequality) If x n ber, then (1 + x) 1 + nx. 1, and n is a natural num- n Theorem 326 (binomial theorem) (a + b) = an +nan n (n 1) (n 3! 2) an 3 3 b + ::: + nabn 1 n Corollary 327 (1 + x) = 1 + nx + nxn 1 + xn . 1 b+ n (n 1) n a 2 2 2 b + + bn . n (n 1) 2 n (n x + 2 1) (n 3! 2) x3 + ::: + n In particular, when x 0, then (1 + x) is greater than any part of the right n hand side. For example, we obtain Bernoulli’s inequality: (1 + x) 1 + nx. n (n 1) (n 2) 3 n (n 1) n n x2 or (1 + x) x . We could also write (1 + x) 2 3! n And so on. This is useful to get approximations on quantities like 3 . We rewrite it as n 3n = (1 + 2) Theorem 328 (squeeze theorem) If an ! L, cn ! L and an bn cn , then bn ! L Proof. We need to prove that 8 > 09N : n N =) jbn Lj < . Let > 0 be given. Choose N1 such that n N1 =) jan Lj < or < an L < . Similarly, choose N2 such that n N2 =) < cn L < . Let N = max (N1 ; N2 ). If n N then an bn () () () jbn cn () an < an L < bn L< L bn L cn bn L cn L< L Lj < Theorem 329 If 0 < a < 1 then an ! 0 1 1 1. Then, x > 0 and a = . For n Proof. Let x = a 1+x 1 an = n (1 + x) 1 by Bernoulli’s inequality 1 + nx 1 < nx To show that an ! 0, we need to show that jan j < , for any that is 1 an < () < nx 1 () n > x 1, whenever n N, 122 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES So, given > 0, N; = 1 will work. x We look at a few more examples, to see how all these results come into play. 2n2 + 1 using the de…nition n2 + 1 Of course, we can …nd this limit by using the theorems on limits. Here, we do it using the de…nition, as asked. We think the limit is 2. We want to show that 2n2 + 1 for every > 0, there exists N such that n N =) 2 < . First, n2 + 1 we simplify the absolute value. Example 330 Find lim 2n2 + 1 n2 + 1 1 +1 1 = 2 n +1 1 < 2 n 1 < if n > 1 n 2 = n2 1 1 < , which happens when n > , then we will have n 1 2 < . So, N = will work. So, we see that if 2n2 + 1 n2 + 1 1 n2 + 2n 4 We think the limit is 0. We need to prove that for every > 0, there exists N 1 such that n N =) 2 < . We begin by noticing that n + 2n 4 Example 331 Find lim n2 + 2n 4 = n2 + 2 (n 2) 2 > n if n > 2 Therefore, if n > 2, n2 1 + 2n 4 = n2 1 + 2n 4 1 < 2 n 1 < n If n > max 2; 1 , then n2 1 + 2n 4 < . So, N = max 2; 1 will work. Remark 332 The two examples above could have been done using the techniques discussed in the previous subsection, that is without using the de…nition. 4.3. LIMIT OF A SEQUENCE: THEOREMS 123 4n Example 333 Prove that lim =0 n! The trick we use is worth remembering. We will use the squeeze theorem. We note that 4 4 4 ::: 4 4n = n! 1 2 3 ::: n 43 4 4 4 ::: 4 since 3! n 4 5 ::: n 1 0 3 But lim 43! 4 n 1 = 0 hence the result follows by the squeeze theorem. 4n 2n 4n We suspect this sequence diverges to in…nity. We need to prove that grows 2n without bounds, that is for every M > 0, there exists N such that n N =) 4n > M . First, we notice that 2n Example 334 Find lim n 4n (1 + 3) = 2n 2n n (n 1) 32 2 > by using the binomial theorem 2n 9 (n 1) > 4 We can see that 9 (n 1) 4M > M () n 1 > 4 9 4M + 9 () n > 9 so, N = 4M + 9 will work. 9 n (n 1) 32 n Remark 335 We used the binomial theorem to deduce that (1 + 3) . 2 Often, when using this theorem, students do not know which term to keep, the term in x, or x2 , or x3 ,... The answer is that it depends on what we are trying 4n > M for any M , in other words, to achieve. Here, we wanted to show that 2n n 4 can be made as big as we want simply by taking n large enough. Since we 2n 4n could not solve for n, we replaced by a smaller term. That smaller term 2n should have similar properties, in other words, the smaller term should also get arbitrarily large. If we take the smaller term too small, it will not work. For 124 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES 4n 1 1 > . However, cannot be made as large as 2n 2n 2n n (n 1) 32 n we want, so this does not help us. If instead of using (1 + 3) 2 n (terms of second degree), we had used (1 + 3) 1 + 3n, then we would have obtained example, it is true that 1 + 3n 2n > () () () M () 2nM < 1 + 3n 2nM 3n < 1 n (2M 3) < 1 1 n< 2M 3 So, this does not work. This tells us that 1+3n 2n > M only for a few values of n. This is why we used terms of higher degree. p p Example 336 Find lim n + 2 n p p p p p p n+2 n n+2+ n p lim n + 2 n = lim p n+2+ n 2 = lim p p n+2+ n 2 p = p lim n + 2 + n =0 Remark 337 In many proofs or problems, di¤ erent versions of the triangle inequality are often used. As a reminder, here are the di¤ erent versions of the triangle inequality students should remember. jjaj jbjj ja bj jaj + jbj jjaj jbjj ja + bj jaj + jbj and Finally, we give a theorem which generalizes some of the examples we did above. Theorem 338 All limits are taken as n ! 1. 1 = 0. np p 2. If p > 0 then lim n p = 1. p 3. lim n n = 1. 1. If p > 0 then lim 4. If p > 1 and 2 R then lim n = 0. pn 4.3. LIMIT OF A SEQUENCE: THEOREMS 125 5. If jpj < 1 then lim pn = 0 6. 8p 2 R, lim pn = 0. n! Proof. We prove each part separately. 1. We can use the de…nition. Let > 0 be given. We show there exists 1 N > 0 : n N =) p 0 < . We begin with what we want to achieve. n 1 np < 0 1 < np 1 () () np > () 1 n> p p So, we see that if N is the smallest ineteger larger than follow. 1 p p , the result will 2. See problems p 3. Let xn = n n p 1. Proving the result amounts to proving that lim xn = 0. From xn = n n 1, we can write p n n = xn + 1 n = (xn + 1) n Using the binomial theorem, we see that if n n n (n 1) 2 xn 2 Thus 2 x2n It follows that for all n Since lim 2 n 1 n 1 2 we have 0 q 2 r xn 2 n 1 = 0, it follows using the squeeze theorem that lim xn = 0. 4. Here again, we will use the binomial theorem. Since p > 1, we can write p = 1 + q with q > 0. Therefore, if k is a positive integer such that k > , we have pn = n (1 + q) n (n 1) ::: (n > k! k + 1) qk 126 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Now, if n > 2k, then k < 12 n. It follows that n k + 1 > 12 n + 1 > 21 n. It follows that n (n 1) ::: (n k + 1) nk > k k! 2 k! and therefore 2k k! 1 n 0 pn qk nk Since 2k k! n!1 q k lim 1 nk = = 2k k! 1 lim k k q n!1 n 0 It follows from the squeeze theorem that lim 5. See problems (hint: write p = = 0). 1 q n =0 pn and use part 4 of the theorem with 6. See problems. 4.3.4 Exercises 1. Prove that 5n ! 0. n! 2. Prove that n! ! 0. nn 3. Finish proving theorem 338. 4. Consider (xn ) a sequence of real numbers such that lim xn = x where n!1 x > 0, prove there exists an integer N > 0 such that n N =) xn > 0. 5. Consider (xn ) a sequence of real numbers such that xn 0 for any n. If x is a partial limit of (xn ), prove that x 0. Prove the same result if x = lim xn . Use this to show that if xn yn then lim xn lim yn . 6. Consider (xn ) a sequence of real numbers and let x 2 R. Prove the two conditions below are equivalent. (a) xn ! x as n ! 1. (b) jxn xj ! 0 as n ! 1. 7. Consider (xn ) a sequence of real numbers and let x 2 R. If lim xn = x, prove that lim jxn j = jxj. 4.3. LIMIT OF A SEQUENCE: THEOREMS 127 8. Consider (xn ) a sequence of real numbers and let x 2 R. Suppose that lim xn = x. De…ne yn = xn+p for some integer p. Prove that lim yn = x. 9. Given that an 1. bn for every n and that lim an = 1, prove that lim bn = 10. Suppose that (an ) and (bn ) are sequences of real numbers such that jan bn j < 1 for every n. Prove that if 1 is a partial limit of (an ) then 1 is also a partial limit of (bn ). 11. Two sequences are said to be eventually close if 8 > 0, 9N > 0 : n N =) jan bn j < . (a) Prove that if two sequences (an ) and (bn ) are eventually close and if a number x is the limit of the sequence (an ) then x is also the limit of the sequence (bn ). (b) Prove that if two sequences (an ) and (bn ) are eventually close and if a number x is a partial limit of the sequence (an ) then x is also a partial limit of the sequence (bn ). 12. Determine if the sequences below have limits. In each case, if a limit L exists, given > 0, …nd N such that n N =) jxn Lj < . (a) (b) (c) (d) (e) (f) (g) (h) n 1 n+1 3n2 + 1 n2 + 1 5n3 + n 4 2n3 + 3 n3 n+1 4n2 + 1 n3 + n 1 2 n + 2 + ::: + 2 n2 n n 1 2n 1 ( 1) 1 2n 1 1 2 1 1 3 1 1 4 ::: 1 1 n 5n 2n p (j) Prove that if a > 1 then a2 > a > a > 1 (i) (k) Prove that n n2 n3 n4 ! 0, ! 0, ! 0, !0 2n 2n 2n 2n 128 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES 13. Prove theorem 310 14. Finish proving theorem 317 15. Let (xn ) and (yn ) be two sequences. Assume further that lim xn = x and y is a partial limit of (yn ). Prove that x + y is a partial limit for (xn + yn ). 16. State and prove similar results for subtraction, multiplication and division. 17. Give an example of two divergent sequences (xn ) and (yn ) such that (xn + yn ) is convergent. 18. Give an example of two sequences (xn ) and (yn ) such that xn ! 0, yn ! 1 and: (a) xn yn ! 0 (b) xn yn ! c where 0 < c < 1 (c) xn yn ! 1 (d) (xn yn ) is bounded but has no limit. 19. Let (xn ) be a sequence of real numbers such that xn ! 0. Prove that x1 + x2 + ::: + xn 7! 0 n 20. Let (xn ) be a sequence of real numbers such that xn ! x for some real number x. Prove that x1 + x2 + ::: + xn 7! x n 21. Show by examples that if jxn j ! jxj then xn does not necessarily converge. If it does converge, it does not necessarily converge to x. 22. Prove that jxn j ! 0 () xn ! 0. 23. Show by examples that if (xn ) and (yn ) are divergent sequences then (xn + yn ) is not necessarily divergent. Do the same for (xn yn ) 24. Show that if (xn ) ! 0 and (yn ) is bounded then (xn yn ) ! 0. 25. Prove that if (xn ) is a sequence in a set S limit of (xn ) must belong to S. 26. Prove that if S that xn ! x. 27. Let S R, then every …nite partial R and x 2 S then there exists a sequence (xn ) in S such R. Prove that the following conditions are equivalent: (a) S is unbounded above. (b) There exists a sequence (xn ) in S such that xn ! 1. 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 4.4 129 Monotone Sequences and Cauchy Sequences 4.4.1 Monotone Sequences The techniques we have studied so far require we know the limit of a sequence in order to prove the sequence converges. However, it is not always possible to …nd the limit of a sequence by using the de…nition, or the limit rules. This happens when the formula de…ning the sequence is too complex to work with. It also happens with sequences de…ned recursively. Furthermore, it is often the case that it is more important to know if a sequence converges than what it converges to. In this section, we look at two ways to prove a sequence converges without knowing its limit. We begin by looking at sequences which are monotone and bounded. These terms were de…ned at the beginning of this chapter. You will recall that in order to show that a sequence is increasing, several methods can be used. 1. Direct approach, simply show that an+1 2. Equivalently, show that an+1 3. Equivalently, show that positive. an an+1 an an for every n. 0 for every n. 1 for every n if both an and an+1 are 4. If an = f (n), one can show a sequence (an ) is increasing by showing that f is increasing that is by showing that f 0 (x) 0. 5. By induction. We now state and prove an important theorem about the convergence of increasing sequences. Theorem 339 An increasing sequence (an ) which is bounded above converges. Furthermore, lim an = sup fan g. n!1 Proof. We need to show that given > 0, there exists N such that n N =) jan Aj < where A is the limit. Let > 0 be given. Since (an ) is bounded above, by theorem 158, fan g has a supremum. Let A = sup fan g. Let > 0 be given. then, A < A. By theorem 151, there exists an element of fan g, call it aN , such that A < aN A. Since (an ) is increasing, we have an aN for every n N . Therefore, if n N , A < an () 0 =) jan A () A an < Aj < < an A 0 130 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES Corollary 340 A decreasing sequence (an ) which is bounded below converges. Furthermore, lim an = inf fan g. n!1 Proof. The proof is similar to the proof of the theorem. Theorem 341 A monotone sequence converges if and only if it is bounded. Proof. The proof follows from results proven in the previous section and the previous theorem and its corollary. Suppose we have a monotone sequence (xn ). If we assume that (xn ) converges, then it follows that it is bounded by theorem 312 that it is bounded. Conversely, if we assume that (xn ) is bounded then it is both bounded above and below. Since (xn ) is monotone, then it is either increasing or decreasing. If it is increasing, it will converge by the previous theorem since it is also bounded above. If it is decreasing, then it will converge by the corollary since it is bounded below. Example 342 Prove that the sequence whose general term is an = n 1 P conk=0 k! verges. We try to establish this result by showing that this sequence is non-decreasing and bounded above. (an ) is increasing: an = 1 + 1 + an+1 1 1 1 + + ::: + . Therefore, 2 3! n! an = 1 >0 (n + 1)! (an ) is bounded above. For this, we use the fact that n! 2n n 1. The proof of this fact is left as an exercise. Therefore, 1+1+ 1 1 1 1 1 1 + + ::: + < 1 + 1 + + 2 + ::: + n 2 3! n! 2 2 2 0 1 1 1 1 <1+ + + 2 2 2 n 1 1 2 <1+ 1 1 2 1 <1+ 1 1 2 <3 1 for every 1 2 + ::: + 1 2 n 1 Since (an ) is bounded above and increasing, it must converge. We will call the limit e0 . 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 131 Example 343 Find lim an where (an ) is de…ned by: n!1 a1 = 2 1 an+1 = (an + 6) 2 If we knew the limit existed, …nding it would be easy. We must …rst establish that it exists. We do this by showing that this sequence is increasing and bounded above. This part is left as an exercise. Once this fact has been established, then we know the sequence must converge. Let L be its limit, we must …nd L. Before proceeding, we will recall the following fact: lim an+1 = lim an . Therefore, if for 1 every n we have an+1 = (an + 6), then we must also have: 2 lim an+1 = lim 1 (an + 6) 2 1 (lim an + 6) 2 1 L = (L + 6) 2 2L = L + 6 L= L=6 n 1 . Show lim an exists. This limit is in fact n the number e, but we won’t show that. Again, to show that (an ) converges, we show that it is increasing and bounded above. Example 344 Let an = 1+ (an ) bounded above. To establish this, we use the following fact: If k is an integer such that 1 < k n, then n (n 1) (n 2) ::: (n nk k + 1) = 1 1 n 1 2 n ::: 1 k 1 n The proof of this is left as an exercise. Using the binomial theorem, we have: 2 1 n (n 1) 1 n (n + + ::: + n 2! n 1 1 1 1 =1+1+ 1 + ::: + 1 2! n k! n 1 1 < 1 + 1 + + ::: + 2! n! < e0 an = 1 + n 1) (n 1 2) ::: (n k! 2 n ::: 1 k + 1) k 1 n 1 n k + ::: + + ::: + where e0 is the limit found in the exercise above. Thus, (an ) is bounded by e0 . 1 n! 1 n! n! 1 n 1 n n 1 2 n ::: 1 n n 132 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES (an ) increasing. The kth term in the expansion of an is 1 k! 1 1 n 1 2 n 1 2 1 1 1 ::: 1 k! n+1 n+1 j j >1 . Hence, an+1 > an . n+1 n The kth term in the expansion of an+1 is j j < , it follows that 1 n+1 n So, (an ) is increasing. Since Since (an ) is increasing and bounded above, it follows that (an ) converges. We de…ne this limit to be the number e. 4.4.2 Cauchy Sequences De…nition 345 (Cauchy Sequence) A sequence (xn ) is said to be a Cauchy sequence if for each > 0 there exists a positive integer N such that m; n N =) jxm xn j < . We begin with some remarks. Remark 346 These series are named after the French mathematician Augustin Louis Cauchy (1789-1857). Remark 347 It is important to note that the inequality jxm xn j < must be valid for all integers m; n that satisfy m; n N . In particular, a sequence (xn ) satisfying jxn+1 xn j < for all n N may not be a Cauchy sequence. Remark 348 A Cauchy sequence is a sequence for which the terms are eventually close to each other. Remark 349 In theorem 313, we proved that if a sequence converged then it had to be a Cauchy sequence. In fact, as the next theorem will show, there is a stronger result for sequences of real numbers. We now look at some examples. 1 Example 350 Consider (xn ) where xn = . Prove that this is a Cauchy n sequence. Let > 0 be given. We want to show that there exists an integer N > 0 such that m; n < N =) jxm xn j < . That it we would like to have jxn Since xm j < () 1 n 1 n 1 < m 1 1 1 < + m n m 1 1 1 + < , the result will follow. This will happen if both < n m n 2 2 1 2 that is when n > and < that is when n > . So, we see that if N is an m 2 2 integer larger than then m; n > N =) jxm xn j < . If we make ::: 1 k k 1 . n+1 1 n . 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 133 n X 1 Example 351 Consider (xn ) where xn = k2 k=1 Let > 0 be given. We want to show that there exists an integer N > 0 such that m; n < N =) jxm xn j < (without loss of generality, let us assume that n > m). That it we would like to have jxn xm j < () n X () k=m+1 n X 1 k2 k=1 xm j = = < n X k=m+1 n X k=m+1 n X k=m+1 = < k=1 1 < k2 Remembering telescoping sums and the fact that jxn m X 1 < k2 1 1 < , we see that k2 k (k 1) 1 k2 1 k2 1 k 1 1 k 1 1 m n 1 1 + m n So, as we saw in the previous example, if N is an integer larger than m; n < N =) jxm xn j < . 2 , then We now look at important properties of Cauchy sequences. Theorem 352 Every Cauchy sequence is bounded. Proof. See problems. Theorem 353 A sequence of real numbers converges if and only if it is a Cauchy sequence. Proof. We have already proven one direction. Let (xn ) be a sequence of real numbers. If (xn ) converges, then we know it is a Cauchy sequence by theorem 313. Assume (xn ) is a Cauchy sequence. We must prove that it converges. By theorem 352, we know that fxn g is bounded. Therefore, by the completeness axiom, for each n, the number an = inf fxn ; xn+1 ; xn+2 ; :::g 134 CHAPTER 4. SEQUENCES AND LIMIT OF SEQUENCES exists. Furthermore, the sequence (an ) is increasing and bounded (why?). Therefore, by theorem 341, (an ) converges. Let a = lim an . We prove that lim xn = a. Let > 0 be given. Since (xn ) is a Cauchy sequence, we can …nd N > 0 such that m; n N =) jxn xm j < : It follows that 2 i h (why). both (xn ) and (an ) are contained in the interval xN ; xN + 2 2 Thus, the limit a is also in this interval. Therefore, for all n N , we have: jxn aj < jxn xN j + jxN aj This proves that lim xn = a, thus (xn ) converges. Remark 354 The key to this theorem is that we are dealing with a sequence of real numbers. The fact that a Cauchy sequence of real number converges is linked to the fact that R is complete. In fact, this is sometimes used as a de…nition of completeness. Some texts say that a set is complete if every Cauchy sequence converges in that set. It is possible to …nd a Cauchy sequence of rational numbers which does not converge in Q. We …nish this section with an important theorem. De…nition 355 A nested sequence of intervals is a sequence fIn g of intervals with the property that In+1 In for all n. Theorem 356 (Nested Intervals) If f[an ; bn ]g is a nested sequence of closed intervals then there exists a point z that belongs to all the intervals. Furthermore, if lim an = lim bn then the point z is unique. Proof. We provide a sketch of the proof and leave the details as homework. First, prove that fan g and fbn g are monotone and bounded. Thus, they must converge. Let a = lim an and b = lim bn . Then, explain why a b. Conclude from this. 4.4.3 Exercises 1. Prove that n! 2n 1 for every n 1 2. Prove that if k is an integer such that 1 < k n (n 1) (n 2) ::: (n nk k + 1) = 1 1 n n, then 1 2 n ::: 1 k 1 n 3. Prove that the sequence given by a1 = 2 1 an+1 = (an + 6) 2 is increasing and bounded above by 6. (hint: use induction for both). 4.4. MONOTONE SEQUENCES AND CAUCHY SEQUENCES 135 4. Show that the sequence de…ned by a1 = 1 an+1 = 3 1 an is increasing and satis…es an < 3 for all n. Then, …nd its limit. 5. Let a and b be two positive numbers such that a > b. Let a1 be their a+b arithmetic mean, that is a1 = . Let b1 be their geometric mean, that 2 p p an + bn is b1 = ab. De…ne an+1 = and bn+1 = an bn . 2 (a) Use mathematical induction to show that an > an+1 > bn+1 > bn . (b) Deduce that both (an ) and (bn ) converge. (c) Show that lim an = lim bn . Gauss called the common value of these limits the arithmetic-geometric mean. 6. Prove theorem 352. 7. Answer the why? parts in the proof of theorem 353. 8. Finish proving theorem 356.