Sequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.

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1 Chapter 3 Sequences Both the main elements of calculus (differentiation and integration) require the notion of a limit. Sequences will play a central role when we work with limits. Definition 3.. A Sequence is a function whose domain is the natural numbers. We sometimes denote a sequence as {f(n)} n N, where f(n) is a function on the natural numbers, and we do not specify the range. Example 3.2. Let f(n) = n n n 2 ( ) n. Then f is a function from the natural numbers into 2 2 matrices, and {f(n)} n N is a sequence of 2 2 matrices. Example 3.3. Let f(n) = x n on [0, ]. Then f is a function from the natural numbers to functions on [0, ], and {f(n)} n N is sequence of functions. Example 3.4. Let f(n) = n. Then f is a function from the natural numbers to the real numbers, and {f(n)} n N is a sequence of numbers. Before proceeding to define the limit of a sequence, try to compute the following limits. If you find the limit, ask yourself how you did it and what does the limit represent. Imagine explaining the process to a freshman.. lim 2. lim n + 3n + 2 sin n n 3. lim (ln(n + ) ln n) 4. lim ( + n)/n Answers:. /3; 2. 0; 3. 0; 4.. 7

2 8 3. Sequences 3.. Convergence For the rest of this chapter we will only consider sequences of real numbers, typically denoted {a n } n N. Definition 3.5. A sequence of real numbers, {a n } n N, converges to a real number, a, if and only if, for each ǫ > 0, there exists a natural number N, such that, for all n N, a n a < ǫ. When {a n } n N converges to a we write lim a n = a or a n a as n. In plain language the definition says a n is close to a for large n. We note that implementing the definition requires a candidate for the limit. Later we will see that we can establish the existence of a limit without having to actually know the limit. Example 3.6. Prove lim n = 0. Proof. Let ǫ > 0. By the Archimedean Property we can find a natural number N so that ǫ < N, or N < ǫ. Thus for all n N n N < ǫ, or n 0 < ǫ, for n N. Sometimes a limit does not exist. We need methods to deal with limits not existing. One possibility would be to negate the definition. Lemma 3.7. A sequence {a n } n N does not converge to the number a if there exists a ǫ 0 > 0 such that, for all N N, there exists a n 0 N with a n0 a ǫ 0. Proof. In logic notation the definition of convergence of a sequence is ( ǫ > 0)( N N)( n N)( a n a < ǫ) or ( ǫ > 0)( N N)( n N)(n N a n a < ǫ). Negating the two, we find or ( ǫ 0 > 0)( N N)( n 0 N)( a n a ǫ) ( ǫ 0 > 0)( N N)( n 0 N)(n 0 N and a n a ǫ). Example 3.8. Prove the sequence {( ) n } n N does not have a limit. Proof. The difficulty in applying the lemma is that we have to show there is no limit for each real number a. Suppose a =. Set ǫ 0 =. Then, for all N N, one of a N or a N+ is 2, and for all N N there exists a n 0 N with a n0 ǫ 0. A similar argument follows for a =. Finally suppose a is not equal to or. Let ǫ 0 = min{ a, a + } n N. Then given any N N we have a n a ǫ 0 for all n N.

3 3.2. Some Topology 9 Here is another proof which uses contradiction. Proof. Suppose the limit did exist. Then, for ǫ = /2, there exists a N N such that a n a < /2 for n N. But 2 = + = ( a) + (a + ) a + a + = a + a < =, a contradiction. Theorem 3.9. The limit of a sequence, if it exists, is unique. Proof. Suppose there were two limits, say a and b. Given any ǫ > 0 there exists an N and N 2 such that a n a < ǫ/2 and a n b < ǫ/2 for all n max{n, N 2 }. Then a b = a a n + a n b a n a + a n b < ǫ, for all ǫ > 0. By trichotomy, either a b = 0 for a b > 0. If the latter holds, set ǫ = a b /2. Then a b < ǫ implies a b < a b /2 or < /2, a contradiction. Thus a b = 0 or a = b Some Topology A large part of mathematics consists of recognizing and generalizing patterns. If we look carefully at the definition of convergence for sequences we see that we really do not need the field axioms. All we require is the notion of closeness. We can have that concept without the use of addition or subtraction (or even absolute value). Definition 3.0. A subset O of R is called open in R if, for each point x O, there is a r > 0 such that all points y in R satisfying x y < r also belong to the set O. Example 3.. The set O = {x 0 < x < } is open in R. Example 3.2. The set O = {x 0 < x } is not open in R. We state without proof the following theorems. Theorem 3.3. Open Set Properties. (a) The empty set and R are open in R. (b) The intersection of any two open sets is open in R. (c) The union of any collection of open sets is open in R. Definition 3.4. in R. (a) A subset of R is called closed in R if its complement is open (b) A neighborhood of x R is any set containing an open set containing x. (c) A point x R is called an boundary point of a set S R if every neighborhood of x contains a point in S and a point not in S. (d) A point x R is called an interior point of a set S R if there exists a neighborhood of x contained in S. (e) A point x R is called an exterior point of a set S R if there is a neighborhood of x which is entirely contained in complement of S.

4 20 3. Sequences Example 3.5. The set O = {x 0 x } is closed in R. Example 3.6. Consider the set O = {x 0 x }. The boundary points are x = 0 and x =. Moreover x = /2 is an interior point. Again without proof we have Theorem 3.7. Let O R. The following statements are equivalent (a) O is open in R; (b) every point in O is an interior point of O. (c) O is a neighborhood of each of its points. Theorem 3.8. A set is closed if and only if it contains all of its boundary points. Theorem 3.9. A subset of R is open if and only if it is the union of a countable collection of open intervals. Finally we can recast the definition 3.5. The advantage of doing so it that we do not need the field axioms to define an open set, and we could, if we had more time, generalize the notion of convergence to sets with less structure than the ones we consider here. A sequence {a n } n N eventually has a certain property if it has that property for all n N for some N N. Definition A sequence of real numbers {a n } n N converges to a number a if and only if, for each neighborhood of a, a n is eventually in that neighborhood Algebra of Limits In this section we establish some familiar properties of sequences. Theorem 3.2. Suppose {a n } n N and {b n } n N are convergent sequences, with limits a and b respectively. Then (a) For all α R, lim n = αa. (b) lim n + b n ) = a + b. (c) lim nb n = ab. (d) If b n = 0 for all n N and b = 0, then lim a n b n = a b. Proof. We will prove (c) and (d) and leave the other two as exercises. We are given the following. Given any ǫ > 0 there exists N, N 2, natural numbers, such that a n a < ǫ and b n b < ǫ for all n max{n, N 2 } n N. We calculate a n b n ab = (a n a)b n + a(b n b) b n a n a + a b n b. We know a n a and b n b are small. We have to worry about b n becoming large. Let us suppose for the moment that b n M for all n N and some M > 0. Then a n b n ab M a n a + a b n b.

5 3.3. Algebra of Limits 2 Let ǫ > 0 be given. In one case we choose ǫ = ǫ/2m, and in the other ǫ = ǫ/2a. Then a n b n ab M a n a + a b n b < ǫ 2 + ǫ 2 = ǫ for all n max{n, N 2 }. We show convergent sequences are bounded in the next lemma. The proof of (d) is similar. Indeed, a n a b n b = ba n ab n bb n = ba n ba + ba ab n bb n a n a + a b n b. b n b b n Again we must worry about / b n becoming large. Let us suppose for the moment that / b n M for n N 3, for some N 3 N and for some M. Then, as in Part (c), let ǫ > 0 be given. We choose ǫ = ǫ/(2m) for the bound on a n a and ǫ = b ǫ/(2 a ) for the bound on b n b. For n max{n, N 2, N 3 }, we have a n a b n b < ǫ 2 + ǫ 2. We have two bounds to establish. Lemma Suppose {b n } n N converges to b. b n M for all n N. Then a M > 0 exists so that Proof. Let ǫ = in Definition 3.5. Then a natural number N exists so that b n b < for all n N. The triangle inequality implies, for such n, b n b bn b bn b. That is, b n + b for n N. Set M = max{ b, b 2,..., b N, + b }, and we find b n M for n N. Lemma Suppose {b n } n N converges to b with b n = 0 for all n N and b = 0. Then a N N and a M > 0 exists so that M for all n N. b n Proof. Here we have to worry about b n becoming small. Let ǫ = b /2 in Definition 3.5. A N N exists such that b n b < b /2 for n N. The triangle inequality shows b n = b n b + b b b n b > b 2 for all n N. Set M = 2/ b and the result follows. Theorem (Squeeze Theorem) Let {x n } n N,{y n } n N, and {z n } n N be three sequences of real numbers such that x n y n z n for all n. Suppose further that {x n } n N and {z n } n N converge to x. Then {y n } n N converges to x.

6 22 3. Sequences Proof. A good place to start would be to write down what we know: Given any ǫ > 0 there exists N i N, i =, 2, such that x n x < ǫ for all n N and z n x < ǫ for all n N 2. The above inequalities imply ǫ < x n x y n x z n x < ǫ for n max{n, N 2 }. That is, y n x < ǫ for such n. Corollary Given any ξ Q c, there exists sequence of rational numbers {r n } n N which converge to ξ. That is, irrational numbers can by approximated with rational numbers. Proof. Clearly ξ < ξ + n. By the density of rationals there exists r n such that ξ < r n < ξ + n. The squeeze theorem implies {r n} n N converges to ξ. While the supremum of a set may Not be in the set, it can however be approximated by elements in the set. Corollary Let E R be non empty and bounded from above. sequence {x n } n N exists in E such that x n α = sup E as n. Then a Proof. The set E has a supremum, α, by the completeness axiom. Since α n is no longer an upper bound there exists w n E with α n < w n α (Theorem 2.). The squeeze theorem shows {w n } n N converges to α Monotone Convergence Theorem We will develop two techniques to show a sequence converges without having to find the limit and apply Definition 3.5. The application of the Monotone Convergence Theorem is one of the techniques. It is equivalent to the completeness axiom. Theorem (Monotone Convergence Theorem) Suppose {a n } n N is a monotone increasing, bounded sequence of real numbers. That is, a n a n+ and a n M some M and for all n N. Then {a n } n N has a limit. Proof. To apply the definition for convergence we need a candidate for the limit. Set E = {a, a 2,...}. The set E is non empty and bounded. By the completeness axiom E has a supremum. Set a = sup E. Let ǫ > 0. Then a ǫ is no longer an upper bound for E. Hence a a N exist such that a ǫ < a N a. Since the sequence is increasing a ǫ < a n a for all n N. That is 0 a a n < ǫ or a a n < ǫ for all n N. Example Consider the sequence The sequence is bounded since a n = n + + n n. a n n + + n n + = n n +.

7 3.4. Monotone Convergence Theorem 23 It is also monotone since a n+ a n = n n n + 2n + + 2n + 2 n + + n n = 2n + 2(n + ) = 2 (n + )(2n + ) 0 By the monotone convergence theorem, the sequence {a n } n N has a limit (in fact, we will see later that lim a n = ln 2). Example Suppose 0 < a <. Prove lim an = 0. Proof. Set a n = a n. The sequence is a monotone decreasing sequence in this case. The monotone convergence theorem still applies (see Exercise 3.2). We need to show the sequence is bounded from below. This is easy since a n = a n = a n 0. We show the sequence is monotone decreasing. That is, a n+ a n. We do this by induction. Since a <, a 2 < a. This is the base step. Suppose a n < a n. That is, a n < a n. Multiplying by a, we find a n+ < a n. That is, a n+ a n. By induction a n+ a n for all n N, and the sequence is decreasing. By the monotone convergence theorem, {a n } n N has a limit. To find the limit, we apply the algebra of limits (we had to know the limit existed to do this!). Set lim a n = L. Then L = lim an+ = lim aan = a lim an = al. That is, L( a) = 0. This implies L = 0. Example Suppose the sequence {x n } n N is defined recursively by x n+ = 2 /x n with x = 2. Prove that {x n } n N has a limit and find the limit. Proof. Notice x n 2 for all n N. Indeed, the inequality is true for n =. Suppose x n 2. Then x n = 2 /x n, and under the assumption on x n, x n 2. By induction, the inequality is true for all n. The sequence is also decreasing. Indeed, for all n x n+ x n = 2 x n = (x n ) 2 < 0. x n x n By the monotone convergence theorem the sequence has a limit. Moreover, by the algebra of limits, the limit satisfies x = 2 /x, or x =. Example 3.3. Find a rational sequence converging to 2. Proof. We know by Corollary 3.25 such a sequence exists. Newton s method for finding zeros of a differentiable function is x n+ = x n f(x n) f (x n ).

8 24 3. Sequences We do not need to study Newton s method - we just use it to find a candidate for the sequence. Applying the method to the function f(x) = x 2 2, we find (3.) x n+ = x n x n We arbitrarily choose x 0 = 2. Then x = 3/2, x 2 = 7/2, x 3 = 577/ You can check that the sequence of rational numbers seems to converge (rapidly) to 2. To prove the convergence we apply the monotone convergence theorem. We first show the sequence {x n } n N is bounded. Clearly, 2 x 0 = Suppose 2 xn 2. Then, using (3.), min x + 2 x n+ max x x 2 2 x 2 x 2 2 x The function g(x) = 2 x + 2 x has a minimum of 2 at x = 2. The right side is bounded by max 2 x 2 2 x + 2 x x max 2 x max 2 x 2 By induction 2 x n 2 for all n N. x Next we prove {x n } n N is monotone decreasing. Clearly x x 0 = /2 0. Suppose x n x n 0. Again using (3.), x n+ x n = (x n x n ) 2. x n x n Since x n 2 for all n N, 2 x n x n Hence, x n+ x n 0, and, by induction, x n x n 0 for all n N. The monotone convergence applies, and {x n } n N has a limit. Suppose the limit is L. Then, using the algebra of limits, L = lim x n+ = lim x n + 2 = L x n 2 L Solving for L, we find L = 2. We have one final application of the monotone convergence theorem. As we will see shortly, we need to know when an infinite intersection of intervals is non empty. Example Set I n = (0, /n) = {x 0 < x < /n}. The set {I n } n N is a sequence of intervals. We will be interested in the infinite intersection E := = {x x I n, n N}. In this case note E =. n=

9 3.5. Subsequences and The Bolzano-Weierstrass Theorem 25 Example Set I n = (n, ) = {x n < x}. Again the set {I n } n N is a sequence of intervals. Here E := = {x x I n, n N} =. n= Definition A sequence of intervals {I n } n N is called nested if and only if I I 2 I Here are sufficient conditions for the intersection to be non empty. Theorem (Nested-Cell Theorem) Let {I n } n N be a sequence of nested, closed, bounded intervals. Then = {x x I n, n N} is non empty. n= Proof. By assumption I n = [a n, b n ] with a a 2... and b b For any natural numbers n and k a k a n b n for k < n a k b k b n for k n. The two inequalities imply a k b n for all k N and any n. Thus the monotone increasing sequence {a n } n N is bounded by b n for any n. By the monotone convergence theorem lim a n = a := sup{a n }. The same two inequalities above now imply that a is a lower bound for {b n } n N. Since {b n } n N is monotone decreasing and bounded below, the monotone convergence theorem implies lim b n = b := inf{b n }. Moreover a b. Finally, let x [a, b]. Then a i a x b b i for all i N. That is, x n= I n Subsequences and The Bolzano-Weierstrass Theorem A subsequence of a sequence {a n } n N is a subset of {a n } n N which is ordered. More specifically, Definition Let {a n } n N be a sequence and {n k } k N be any sequence of natural numbers such that n < n 2 < n 3 <.... The sequence {a nk } k N is called a subsequence of {a n } n N. Example The sequence with a n = +( )n 2 does not converge. However, if n k = 2k, {a 2k } k N = {0, 0, 0,...}. Similarly, n k = 2k produces the subsequence {a 2k } k N = {,,,...}. Theorem A sequence converges to a if and only if each of its subsequences converges to a. In fact, if every subsequence converges, they all converge to the same limit. Proof. If all subsequences converge to a, the sequence converges to a since the sequence is a subsequence of itself. Conversely, suppose the sequence converges to a and {a nk } k N is a subsequence. Let ǫ > 0. There exists a natural number N such that a n a < ǫ for all n N. By the construction of the {n k } k N we have k n k for all k N. Thus, for k N,

10 26 3. Sequences n k N. This implies a nk a < ǫ for all k N, and the subsequence {a nk } k N converges to a. A restatement of part of the theorem gives Corollary If {a n } n N converges, all of its subsequences converge to the same limit. This gives us another technique for showing a sequence does not converge. We need only find two convergent subsequences with distinct limits. Example Show the sequence a n = ( ) n does not converge using three different methods. Proof. We have already shown non convergence using two difference methods in Example 3.8. Note that {a 2k } k N and {a 2k } k N converge to and respectively. Thus {a n } n N does not converge. The importance of the following theorem cannot be overstated. Theorem 3.4. (Bolzano-Weierstrass Theorem) Every bounded sequence of real numbers has a convergent subsequence. Proof. Let us denote the sequence {x n } n N. We apply the Nested-Cell Theorem, Theorem We will construct inductively a sequence of intervals such that I I 2 I Since the sequence is bounded, a x n b for all n N and some a and b. Set I = [a, b]. Choose a x n I. We define I = b a. Consider I = I I = a, a + b a + b 2 2, b. One of I or I contains infinitely many of the x n. Call it I 2. Then I 2 = (b a)/2. Choose x n2 I 2 with n 2 > n. Note that I 2 I. This is the base step. Suppose I,..., I k have been chosen so that I I 2 I 3... I k with I k = b a 2 k 2, and x n I, x n2 I 2,..., x nk I k, with n < n 2 <... < n k. By construction I k contains infinitely many of the x n. Bisect it as before, I k = I I with one of the I or I containing infinitely many of the x n. Call the interval I k. Choose x nk I k I k with n k > n k. Note that I k = b a 2 k. By induction this defines I k and x nk for all k N. The {I k } k N satisfy the hypothesis of the Nested-Cell Theorem. Thus their intersection is non empty and there is an x I n. We need to show the subsequence {x nk } converges to x as k. Let ǫ > 0. By Example 3.29 there is a N n=

11 3.6. Cauchy Sequences 27 such that b a 2 k < ǫ for all k N. Since x nk I k and x I k x nk x I k = b x 2 k < ǫ (see HW 2.) for all k N. There is a nice geometric interpretation of the Bolzano-Weierstrass Theorem. Definition Let E R. A point a R is called a cluster point or accumulation point of E if and only if E (a r, a + r) contains infinitely many points for all r > 0. That is, a is a cluster point of E if and only if every neighborhood of a contains a point in E distinct from a, or for all neighborhoods, U of a, (U \{a}) E =. Example If E = (0, ), the cluster points are [0, ]. Note that a cluster point need not be in E. Example If E = Q, the cluster points are all of R, since Q is dense in R. Example The set E = {, 2, 3} has no cluster points. In fact, all points in E are isolated points of E (see Exercise 3.20). Example Suppose {x n } n N is a convergent sequence of real numbers converging to x R. Then x is a cluster point of the set {x, x 2,...}. We may restate the Bolzano-Weierstrass Theorem as follows. Theorem (Bolzano-Weierstrass Theorem for Sets) Every bounded infinite subset of R has a cluster point. Proof. We will call the infinite set E. We first construct a sequence of distinct points in E. Since E is non empty, a x E exists. Consider the set E/{x }. It too is non empty. Thus a x 2 exists in E/{x }. Suppose the distinct points x, x 2,..., x k have been similarly chosen. Then the set E/{x, x 2,..., x k } is non empty, and there exists a x k E/{x, x 2,..., x k }. By induction we have constructed a sequence {x n } n N of distinct points in E. The sequence {x n } n N is bounded since E is bounded. By the Bolzano- Weierstrass theorem, a subsequence of {x n } n N, {x nk } k N exists and converges to some x R. Given any neighborhood of x, the sequence {x n } n N is eventually in that neighborhood. Thus every neighborhood of x contains a point in E, and x is a cluster point of E Cauchy Sequences There is yet another way to show a sequence has a limit without finding the limit (the monotone convergence theorem was the other). Definition A sequence {x n } n N is called Cauchy (pronounced Co She) if and only if for all ǫ > 0 there exists a N N such that x n x m < ǫ for all n, m N.

12 28 3. Sequences Theorem (Cauchy Theorem) A sequence {x n } n N converges if and only if it is Cauchy. Proof. Suppose {x n } n N converges. Let ǫ > 0. Then there exists N N such that x n x < ǫ/2 for all n N. Then x m x n = x m x + x x n x m x + x n x < ǫ 2 + ǫ 2 = ǫ for all n, m N. That is, {x n } n N is Cauchy. Suppose {x n } n N is Cauchy. We need a candidate for the limit. We apply the Bolzano-Weierstrass theorem to find the candidate. To do so we need to show the sequence is bounded. We copy the proof of Lemma 3.22 for this. Let ǫ = in the definition of a Cauchy sequence. Then a natural number N exists such that x m x n < for all m, n N. That is, x m = x m x N + x N x m x N + x N < + x N for all m N. Thus x m max{ x, x 2,..., x N, + x N } for all m N. The Bolzano-Weierstrass theorem applies and there exists a subsequence {x nk } k N converging to some x R. We need to show the whole sequence converges to x. Let ǫ > 0. There exists N and N 2 such that x m x n < ǫ 2 m, n N, and x nk x < ǫ 2 k N 2. Set N = max{n, N 2 }, and fix K N. Then n K N, and x n x = x n x nk + x nk x x n x nk + x nk x < ǫ 2 + ǫ 2 = ǫ, for all n N. Example Consider the sequence given recursively by x n = x n 2 + x n 2 with x = and x 2 = 2. One can check by writing out the first few terms that the sequence is not monotone. Indeed, the sequence oscillates. We could try to obtain a candidate for the limit by assuming it exists and using the algebra of limits. If lim x n = L, taking the limit of x n above, we would find L = (L + L)/2, which gives us no useful information. Our only hope is to show the sequence is Cauchy (if we believe it converges). Note that x n x n+ = x n x n + x n 2 = 2 x n x n. Homework Exercise 3.30 shows that the sequence {x n } n N is Cauchy. Moreover, Cauchy s theorem shows that the sequence has a limit. We need only find the limit.

13 3.6. Cauchy Sequences 29 Recall that, if the sequence converges, all subsequences converge to the same limit (Corollary 3.39). We will find the limit of the subsequence {x 2n+ } n N. We claim that x 2n x 2n = 4 n. Indeed, the equality is true for n =. Suppose it is true for n. Then x 2(n+) x 2(n+) = x 2n+2 x 2n+ = 2 (x 2n + x 2n+ ) x 2n+ = 2 (x 2n x 2n+ ) = x 2n 2 2 (x 2n + x 2n ) = 4 (x 2n x 2n ) = 4 n, and the result follows by induction. We have discovered that x 2n = x 2n + 4 n. Thus x = x 3 = 2 ( + 2) = + 2 x 5 = 2 (x 4 + x 3 ) = x x 3 x 7 = 2 (x 6 + x 5 ) = x x 5 = Proceeding inductively, we find Taking the limit, we find x 2n+ = + 2 = + 2 = = n. 4 4 n lim x n = lim x 2n+ = 5 3.

14 30 3. Sequences 3.7. Review Techniques to show limit converges Apply the definition of a limit Algebra of limits Show sequence monotone and bounded Show sequence Cauchy Techniques to show no limit Negate the definition of a limit Show sequence not bounded Show sequence not Cauchy Find a contradiction More than one convergent subsequence with different limits

15 3.7. Review 3 Summary of theorems so far. Try to understand the idea behind the proof of each theorem. The Completeness Axiom Monotone-Convergence Theorem Nested Cell Bolzano-Weierstrass Archimedean Principle Cauchy Thm Denisty of Q

16 32 3. Sequences 3.8. Homework Exercise 3.. Give example of two sequences that do not converge but whose (a) sum converges (b) product converges (c) quotient converges Exercise 3.2. Give an example of a convergent sequence {x n } (0, ], but whose limit is not in (0, ]. Exercise 3.3. Use the Definition 3.5 to establish the following limits. (a) (b) (c) lim lim lim n 2 = 0 2n n + 3 = 2 2n + 4 5n + = 2 5 Exercise 3.4. If a convergent sequence is rearranged, prove that the new sequence has the same limit. Exercise 3.5. Let 0 < x n < 4 for all n N. Show that the sequence {x n } n N has a convergent subsequence. In what interval will the limit of this subsequence lie? Exercise 3.6. Suppose a n a > 0 as n and that a n > 0 for all n N. Show that m > 0 exists so that a n m for all n N. Exercise 3.7. (Cesàro Summable) If the sequence {x n } n N converges to x, then the sequence {σ n } also converges to x. Here σ n = n (x + x x n ). Exercise 3.8. Show that the sequence {sin n} n N does not converge. Exercise 3.9. Let a n = n n odd n even. Does the limit exist? Proof required. Exercise 3.0. Suppose {x n } n N converges to x and satisfies 2 < x n < 3 for all n N. Show that the limit satisfies 2 x 3. Exercise 3.. Let {a n } n N be a sequence of real numbers converging to a R. (a) Prove lim a n = a. (b) If lim a n = 0, then lim a n = 0. (c) Show by example that lim a n may exist, while {a n } n N may not converge. Exercise 3.2. Let x n 0 for all n N with x n x as n. Prove x n x as n. Exercise 3.3. Suppose {x n } n N is a convergent sequence of integers. Show that the sequence is eventually constant.

17 3.8. Homework 33 Exercise 3.4. Let {a n } n N be a sequence of positive real numbers with the ratio a n+ /a n converging to L as n. If L <, show that lim a n = 0. Show a n as n if L >, and no conclusion can be made if L =. Apply this to the sequences (a) x n = rn n!, (b) x n = np n!, p > 0, (c) x n = nn n!. Exercise 3.5. Suppose {x n } n N is a bounded sequence (not necessarily convergent), and {y n } n N is a sequence converging to zero. Prove that {x n y n } n N converges to zero. Show by example that the result is not true if either hypothesis on {x n } n N or {y n } n N is dropped. Exercise 3.6. Establish, by using any definition or theorem, the limits of the following sequences. (a) (b) sin n n n N π + 2 n n N Exercise 3.7. Prove Theorem 3.8. (c) 2n + 4 5n + n N (d) n n + n N (e) lim n 3 2n + n 4 + 5n sin(n3 ) Exercise 3.8. Show that every bounded, non empty closed subset of R has a maximum. (Hint: show the supremum of a set is a boundary point of the set). Exercise 3.9. What are the cluster points of the irrational numbers? Exercise Suppose a R is not a cluster point of set E. Prove there exists a r > 0 such that (E \ {a}) (a r, a + r) =. If a E, a point a with this property is called an isolated point of E. Exercise 3.2. Prove the Monotone Convergence Theorem for a decreasing sequence which is bounded below. Exercise Suppose 0 < a. Prove that the sequence {a /n } n N converges to one. Exercise Show the following recursively defined sequences are convergent and find the limit (a) Let x = and x n = 4 (2x n + 3). (b) Let x = 3 and x n = 2 /x n. (c) Let x = 2 and x n = 2 + x n. Exercise Give an example of a sequence whose set of convergent subsequences converge to {, /2, /3, /4...} {0}. Exercise Give an example of a sequence with no convergent subsequences.

18 34 3. Sequences Exercise Give an example of two sequences {x n } n N and {y n } n N that satisfy {x n } {y n } as sets, but with {x n } n N not a subsequence of {y n } n N. Exercise The next two statements are equivalent - they are contrapositives. Prove one of them. (a) Suppose {x n } n N is a bounded sequence. If all its convergent subsequences have the same limit, then the sequence is convergent. (b) Show that a nonconvergent bounded sequence has two convergent subsequences each with distinct limits. Exercise Suppose {x n } n N is unbounded. Show there exists a subsequence {x nk } such that /x nk 0 as k. Exercise Give an example of a Cauchy sequence with {x n } n N (0, ], but whose limit is not in (0, ]. Exercise Let {x n } n N be a sequence and 0 < r <. Suppose that x n+ x n r x n x n for n 2. Show that the sequence is Cauchy. Exercise 3.3. Use Exercise 3.30 to show the sequence x n = /(2 + x n ) for n 2 and x > 0 is Cauchy. Find the limit. Exercise (b) Show that the sequence is not Cauchy. (a) Negate the definition of a Cauchy sequence. x n = n k= (c) Show that x n+ x n 0 as n (compare this to the definition of a Cauchy sequence). k