( ε > 0) ( N N) ( n N) : n > N a n A < ε.

Transcription

1 CHAPTER 1. LIMITS OF SEQUENCES 1.1. An Ad-hoc Definition. In mathematical analysis, the concept of limits is of paramount importance, in view of the fact that many basic objects of study such as derivatives, integrals, series etc. are defined by pushing some expressions to their limits. There are many forms of limits. Fortunately their (rigorous) definitions all follow the same logical pattern. Once any one of them is well understood, the others are not too difficult to master. To begin with, let us state the standard definition of limits of sequences, without expecting you to understand! Definition (standard). We say that a sequence {a n } n 1 of real numbers converges to A, or A is the limit of this sequence, if the following condition is satisfied: for every given ε > 0, there exists a positive integer N such that a n A < ε hold for all n > N. Using quantifiers, this condition can be rewritten as ( ε > 0) ( N N) ( n N) : n > N a n A < ε. Here N denotes the set of all positive integers. The reader should not worry at this moment if this definition is found to be incomprehensible. Surely the logical structure here is complicated: there are twists and turns in its meaning due the presence of quantifier triple. This may be the main source of difficulties for beginners, who often spend more time struggling with the definition than taking the advantage of it to enhance their understanding or to deal with problems. For this reason we introduce an ad hoc definition of limits designed especially for beginners. It still complies with the general standard of mathematical rigor. It is easier to understand and often easier to work with. In Chapter 3 we shall prove that the ad hoc definition is equivalent to the standard one. By that time you should have no difficulty in understanding and using the standard definition in order to conform your way to the general practice. We begin by considering the following three sequences: 1, 1/2, 1/3, 1/4, 1/5, 1/6,... 1/2, 1/4, 1/8, 1/16, 1/32, , 0.01, 0.001, , ,... Here we only write down the first few terms of each sequence. It is more economic and more precise to express a sequence by writing down a general expression of its nth term 1

2 a n and then putting curly brackets around it to render the sequence into the form {a n }. For example, the nth term of the first sequence above is 1/n and hence this sequence is usually written as {1/n}, or more precisely as {1/n} n 1. Similarly the second sequence can be written as {1/2 n } and the third sequence is {10 n }. All three sequences above have the following common feature: their terms become smaller and smaller and get closer and closer to zero. Technically, becoming smaller and smaller is called decreasing and getting closer and closer to zero called tending to zero. Putting together we simply say decreasing to zero. Definition We say that a sequence {a n } of real numbers is decreasing if a n a n+1 for all n. Thus, the sequence {1/n} is decreasing because we have 1 n > 1 n+1 for all n, which is easy to check. The sequences {1/2 n } and {10 n }, or more generally {a n } for any a with 0 < a < 1, are also decreasing. Definition We say that a sequence {α n } of positive numbers is decreasing to zero if {α n } is decreasing and, for every ε > 0, there exists some n such that α n < ε; (in other words, there is no positive number ε such that α n ε for all n.) We suggest the shorthand α n 0 for the expression α n decreases to zero. Now we come to our ad hoc definition of limits of sequences. Roughly it says that a sequence {a n } converges to some A if the distance between a n and A can be controlled by some positive α n which decreases to zero. Definition ( ad hoc ). We say that a sequence {a n } converges to A and write lim n a n = A or simply lim a n = A, if there is a sequence {α n } of positive numbers decreasing to zero such that a n A < α n for all n. In the above definition, a n A is the distance between a n and A, which is simply a n A when a n A, and is A a n when a n A An Example. Find the limit the sequence { n n } (if it exists). Here n n (which will be rewritten as n 1/n ) stands for the positive number p n such that p n n = n. (We have borrowed a result from the future that guarantees the existence and the uniqueness of the nth root of a positive number.) The problem here is about the behavior of p n n 1/n for large n. Let us take a large concrete n to find out what n 1/n is, say n = (10 10 is certainly a large number explicitly, it is 1 followed by ten zeros): (10 10 ) 1/1010 = (10 10 ) = = 10 (10 9) = = 1. 2

3 This suggests that n 1/n converges to 1. The question is, can we prove this rigorously? Notice that p n 1 for all n. (Otherwise we would have p n < 1, which would lead to n = p n n < 1 n = 1, contradicting n 1.) Since we anticipate that, for large n, p n is only slightly larger than 1, we have to estimate the difference between p n and 1. Now we use a special trick called telescoping to get this estimate. This trick is used so many times in this book that at some point the reader will realize that it should be called a technique. (Calling it a trick or a technique largely depends on how many times you use it.) For notational simplicity, let us replace p n by a so that p n n becomes a n. For n 2, we express a n as a telescoping sum as follows: a n = a n a n 1 + a n 1 a n a 2 a + a = a n 1 (a 1) + a n 2 (a 1) + + a(a 1) + (a 1) + 1. (1.2.1) Since a = p n 1, the n numbers a n 1, a n 2,..., a, 1 are all 1. Thus (1.1) gives a n n(a 1) + 1. (1.2.2) This inequality tells us something about the size of a. Indeed, substituting a back by p n, we have n = p n n n(p n 1), or p n 2. Unfortunately, this is not good enough. We can sharpen (1.2.2) as follows. Changing n in (1.2.2) to k, we can rewrite this inequality as a k k(a 1) + 1. So a k k(a 1). Now we let k in a k k(a 1) runs from n 1 down to 1 and apply the resulting inequalities to the right hand side of (1.2.1). We obtain a n (n 1)(a 1) 2 + (n 2)(a 1) (a 1) 2 + (a 1) + 1 = {(n 1) + (n 2) } (a 1) 2 + a 1 2 n(n 1)(a 1)2. Replacing a by p n, we get n = p n n 1 2 n(n 1)(p n 1) 2. Hence 0 p n 1 α n for n 2, where α n = 2/(n 1) (n 2). Now α n decreases to zero and p n 1 α n. Therefore p n converges to 1. Remark. Rearranging (1.2.1) as a n 1 = (a 1)a n (a 1)a + (a 1) and pulling out the common factor a 1, we get a n 1 = (a 1)(a n 1 + a n a 2 + a + 1), (1.2.3) which may be familiar to you. In case a 1, we can rewrite it as 1 + a + a a n 2 + a n 1 = 1 an 1 a, (1.2.4) 3

4 which is the usual recipe telling us how to find the sum of a geometric series. Another variant of (1.2.1) is obtained by replacing a by a/b in (1.2.4) and then multiplying both sides by b n : a n b n = (a b)(a n 1 + a n 2 b + + ab n 2 + b n 1 ). The proof in the above example requires rather strenuous effort to accomplish. Certainly you may blame the fact lim n n 1/n = 1 for being nontrivial. But you may also say that it is due to our lack of tools at present and the first principle is the only thing allowed to be used here. The purpose of presenting this example here is to illustrate the proper way of showing convergence of sequences according to our definition, and also to point out the need of a good theory to deal with many problems of convergence. Why do we have to trouble ourselves with getting a rigorous definition of convergence? Well, without it, we cannot produce a correct argument. An incorrect argument is one of the most woeful things in mathematics, because it may lead us to erroneous conclusions with disastrous consequences (such as a rejected manuscript, a failed exam, or a sinking ship). Take a look of Gardener in Lewis Carroll s Sylvie and Bruno Concluded [CAR], who sang his verse about finding an error in his argument that would proved he was the Pope. A dreadful fact that extinguished all hope, he lamented, with tears streaming down his cheeks Elementary Properties. In Definition we have used the absolute value a n a of a n a for the distance between a n and a. You may say that a is a just a nice way to write down the awkward expression a 2. A proper definition of the absolute value of a real number a is given by a = { a if a 0; a if a < 0. Here are some elementary facts about absolute values that you should keep in mind: (A1) Triangular inequality: a + b a + b (for all real a and b.) (A2) a b = b a, ab = a b. (A3) a a 0 ε ε a a 0 ε a 0 ε a a 0 + ε. (Here stands for if and only if.) These elementary facts are easy to verify. Indeed, accepting the usage of the square root function, (A2) just says (a b) 2 = (b a) 2 and (ab)2 = a 2 b 2. Also, (A1) can be rewritten as (a + b) 2 a 2 + b 2, which can be checked by squaring both sides. Now we try to avoid the square root function (which is still undefined) in proving the triangular inequality. Notice that every nonzero real number a can be uniquely written as σ a a, where σ a is the sign of a, which is either +1 or 1. Also 4

5 notice that a = σ a a and σ ab = σ a σ b. (As an exercise, the reader should check these.) To show (A1), we may assume that a, b and a + b are nonzero; (otherwise this inequality is trivial.) Letting ε = σ a+b, we have a + b = ε(a + b) = εa + εb = εσ a a + εσ b b a + b. (Since εσ a is either +1 or 1, we have εσ a a a. Similarly, εσ b b b.) This proves (A1). Finally, (A3) can be easily checked by considering two cases: a a 0 and a < a 0. You should convince yourself of the following Observation 1. If {α n } and {β n } are sequences of positive numbers decreasing to zero, then so are the sequences {α n + β n } and {Kα n }, for any K > 0. To verify this observation rigorously, first we note that it follows from α n α n+1 and β n β n+1 that α n + β n α n+1 + β n+1 and Kα n Kα n+1. This shows that the sequences {α n + β n } and {Kα n } are indeed decreasing. Next, let ε > 0 be given. Then there exist positive integers n 1, n 2 and k such that α n1 < ε/2, β n2 < ε/2 and α k < K 1 ε. Take any integer m greater than both n 1 and n 2, we have α m + β m α n1 + β n2 < ε/2 + ε/2 = ε. This shows that {α n +β n } is indeed decreasing to zero. The inequality Kα k < K.K 1 ε = ε tells us that {Kα n } decreases to zero. From this simple observation we can show Fact 1. If {a n } and {b n } are sequences of real numbers converging to a and b respectively, then the sequence {a n + b n } converges to a + b and {a n b n } converges to ab. Proof: By assumption, there are sequences {α n } and {β n } in of positive numbers decreasing to zero, such that a n a < α n and b n b < β n for all n. Hence (a n + b n ) (a + b) = (a n a) + (b n b) a n a + b n b < α n + β n for all n. Since α n + β n decreases to zero, we conclude that a n + b n tends to a + b as n. Next, notice that a n = (a n a) + a a n a + a α n + a α 1 + a. So a n b n ab = a n b n a n b + a n b ab = a n (b n b) + b(a n a) a n b n b + b a n a ( a + α 1 )β n + b α n. Our observation tells us that b α n +( a +α 1 )β n decreases to zero. Hence a n b n converges to ab. (Do you notice the telescoping trick used in this argument?) Q.E.D. 5

6 Fact 2. If lim a n = a and if a n [c, d] for all n, then a [c, d]. In other words, if c a n d for all n and if lim n a n = a, then c a d. Proof: By assumption, we have c a n d and a n a α n for all n, where α n decreases to zero. From a n a α n we have α n a n a and hence a a n +α n d+α n for all n. We claim a d. Indeed, if a > d, we would have α n a d > 0 for all n, contradicting the fact that α n decreases to zero. On the other hand, a n a α n also gives us a n a α n and hence a + α n a n c for all n. A similar argument shows that a c. Q.E.D. Fact 3. If lim a n = lim n b n = A and if a n x n b n for all n, then lim x n = A. (One may describe this fact picturesquely as follows: a and b are two police officers, x is a thief, and A is a police station. The condition a n x n b n means that, at any moment, the officers are grabbing the thief from both sides. The condition lim n an a n = lim n b n = A means that both officers is marching resolutely to the police station. The conclusion lim n x n = A means that the thief approaches to the police station as well, without any choice.) Proof: The assumption lim a n = lim b n = A tells us that there are sequences {α n } and {β n } decreasing to zero such that a n A < α n and b n A < β n for all n. It is enough to check that, for each n, x n A max{ a n A, b n A } ( ) (from which it will follow that x n A a n A + b n A < α n + β n ). Let us consider two cases: A x n and A > x n. In the first case we have A x n b n and hence x n A = x n A b n A = b n A. In the second case we a n x n < A and hence x n A = A x n A a n = a n A. Now ( ) is clear. Q.E.D. According to our definition, we have lim n a n = a if there is a sequence of positive numbers α n decreasing to zero such that a n a < α n for all n. last requirement a bit: instead of asking We may relax the a n a < α n holds for all n, we only require this inequality for n starting from certain stage. This is not a substantial improvement. However it gives us flexibility in some arguments. Observation 2. If the inequality a n a < α n holds for all n > n 0, where α n decreases to zero and n 0 is some positive integer, then lim n a n = a. Proof: Define a new sequence {β n } decreasing to zero as follows. For n > n 0, let β n = α n. For n n 0, let β n = α 1 + max{ a j a : 1 j n 0 } + 1. Clearly β n decreases to zero as n and a n a < β n holds for all n. 6 Q.E.D.

7 Now we use this simple observation to show the following: Fact 5. If lim b n = b, lim a n = a, (a n 0, a 0), then lim b n /a n = b/a. Proof: In view of Fact 1 above, it is enough to show lim a 1 n = a 1. By assumption, there is a sequence {α n } decreasing to zero such that a n a < α n for all n. Now a 1 n a 1 = a n a a n a = a n a a n a < α n a n a. (1.3.1) [Aside: We cannot expect that the right hand side α n / a n a is decreasing to zero. The trouble is caused by a n in the denominator of this expression. In the next step we show that it stays away from zero for large n.] Now a = a a n +a n a a n + a n, or a n a a n a. Since α n decreases to zero and since a /2 is positive, there is a positive integer n 0 such that α n0 < a /2. Thus, for n > n 0, we have a n a < α n α n0 < a /2 and hence a n a a n a > a ( a /2) = a /2. Returning to (1.3.1), we conclude a 1 n a 1 < Kαn for all n > n 0, where K = 2/ a 2. Since {Kα n } is a sequence decreasing to zero, Observation 2 tells us that lim a 1 n = a 1. Q.E.D Series. For many practical problems, it is convenient to express a sequence in the form of a series. By a series of numbers we mean a formal infinite sum a 1 +a 2 +a 3 +, which is often written as n=1 a n (or its variants, such as k=0 a k.) A sum of finitely many numbers is a number, but a formal sum is just an expression (not a number) of grouping some numbers by the addition sign without performing the actual addition. It is a bit hard to swallow, but you have to take it that way. Given a series n=1 a n, we may form the sequence of partial sums s 1, s 2, s 3,... which are given by s 1 = a 1, s 2 = a 1 + a 2, s 3 = a 1 + a 2 + a 3, etc. In general, s n = a 1 + a a n ( = n k=1 a k Conversely, given a sequence {s n } n, if we let a 1 = s 1 and a n = s n s n 1 for n 2, then s n are partial sums of the series n=1 a n. (Notice that s n = a 1 + a a n = s 1 + (s 2 s 1 ) + + (s n s n 1 ), which is a telescoping sum.) This shows that there is a one-one correspondence between sequences and series, which are related by telescoping. Therefore we can transplant all concepts about sequences of numbers to series of numbers. For example, we say that the series n=1 a n converges to s (we also say n=1 a n is summable with s as its sum) if its partial sums s n n k=1 a n converge to s as n. Similarly we say that a series diverges if its corresponding sequence of partial sums diverges. 7 ).

8 For example, let a be a fixed number and consider the geometric series n=0 an. The nth partial sum of this series is s n = 1 + a + a a n. [Note that the zeroth partial sum is s 0 = a 0 = 1.] We can find a closed form of this sum by the following well known trick. Multiplying both sides of s n = 1 + a + a 2 + a a n 1 + a n by a, we obtain as n = a + a 2 + a 3 + a a n + a n+1. Hence the difference between s n and as n is s n as n = 1 a n+1. Thus, in case a 1, we arrive at s n = 1 an+1 1 a. (We have seen this identity in 1.2. Here we use a different argument to derive this.) Suppose a < 1. Then lim n a n = 0 and hence lim n s n = 1 1 a. We conclude Fact. If a < 1, then n=0 an converges to 1 1 a Uniform Convergence. In the present section we study sequences of functions with a common domain, say X. Here X is an arbitrary nonempty set. In order to visualize our theory, normally we let X to be an interval in the real line so that we can draw a picture of a function defined on X as a planar curve. Now we consider two modes of convergence of a sequence {f n } of functions defined on X to a function f on X: uniform convergence and pointwise convergence. Definition We say that f n converges to f uniformly if there is a sequence {ε n } of numbers decreasing to zero such that f n (x) f(x) ε n for all x in X and for all n. (1.5.1) We say that f n (n 1) converges to f pointwisely if, for each x X, the numerical sequence {f n (x)} n 1 converges to f(x). Thus {f n } converges to f pointwisely if, for every x X, there is a sequence of numbers ε x n (the superscript x added to ε n here is to emphasize its dependence on x) decreasing to zero such that f n (x) f(x) < ε x n for all n; see Figure

9 We should emphasize that the difference between these two modes of convergence lies on the sequence {ε n } of positive numbers decreasing to zero satisfying f n (x) f(x) < ε n. If it can be chosen in a way independent of x, then f n converges to f uniformly. Otherwise we only get the pointwise convergence. It is clear that uniform convergence implies pointwise convergence. Soon we shall see an example showing that the converse is false. Geometrically, inequality (1.5.1) says that the graph of f n lies in the ribbon of width 2ε n, which shrinks to the graph of f as n. Now let us consider the special case with X = N, the set of all positive integers. In this case a function f on X gives rise to a (numerical) sequence {f(1), f(2), f(3),...}. Conversely, each numerical sequence {a k } naturally associates with a function f on N: just let f(k) = a k for all k N. Thus functions on N can be considered as (numerical) sequences and vice versa. Now we show that uniform convergence for a sequence of sequences (i.e. functions on N) goes well with taking limits: Proposition Suppose that {f n } is a sequence of functions on N converging uniformly to a function f on N, and suppose that the limits lim k f n (k) = a n and lim k f(k) = a exist for all n. Then {a n } converges to a. Proof: By the uniform convergence of f n, we know that there is a sequence of positive numbers ε n decreasing to zero such that f n (k) f(k) ε n for all k and n. We can rewrite the last inequality as f n (k) f(k) [ ε n, ε n ]. For each fixed n, when k, f n (k) f(k) converges to a n a and hence a n a [ ε n, ε n ], in view of Fact 2 in 1.3. Thus a n a ε n for each n, which shows lim n a n = a. Q.E.D. Example. Define a sequence of functions f n on N by putting { 0, if k n; f n (k) = 1, if k > n. The sequence {f n } can be visualized in the following way: put down the infinite matrix The nth row of this matrix is a sequence of numbers corresponding to the function f n on N. To find out the pointwise convergence of f n, read off the entries from each column, from top to (endless) bottom. For each k N, we find from the kth column that lim n f n (k) = 0. 9

10 Thus f n converges pointwisely to the function f, which is identically zero. On the other hand, a n lim k f n (k) = 1 for each n and a lim k f(k) = 0. Since a n 1 does not converge to a 0, according to the above proposition, here f n converges to f pointwisely, not uniformly. From this example we learn two things: first, uniform convergence implies pointwise convergence, but not vice versa; second, the hypothesis of uniform convergence of f n in Proposition cannot be weakened to pointwise convergence. For two functions f and g defined on X, we can form their sum f +g and their product fg, which are also functions on X, given by (f + g)(x) = f(x) + g(x), (fg)(x) = f(x)g(x) for all x in X, which says, the value of the sum f + g (the product fg resp.) at a point is the sum (the product resp.) of the values of f and g at that point. We leave the easy proof of the following proposition to the reader. Proposition If f n and g n converge uniformly to f and g respectively, then f n + g n and f n g n converge uniformly to f + g and fg respectively. It also makes sense to talk about the uniform convergence of a series of functions, say n=1 f n, where f n are functions defined on a set X. We say that such a series of functions is uniformly convergent to a function S (also defined on X) if its partial sums S n defined by S n (x) = f 1 (x) + f 2 (x) + + f n (x) form a sequence {S n } of functions converging uniformly to S. Thus, the sum S of a uniformly convergent series n=1 f n of functions on X is a function on X such that (f 1 (x) + f 2 (x) + + f n (x)) S(x) α n for all x X, where α n (n 1) are positive numbers forming a sequence decreasing to zero. EXERCISES 1. We say that a sequence {a n } is bounded if there is some M > 0 such that a n M for all n. Show that convergent sequences are bounded. Give an example to show that a bounded sequence is not necessarily convergent. 2. Verify a b a b (by using other proven inequalities in this chapter). Use this inequality to show that lim a n = a implies n lim a n = a. n 10

11 3. Show that, if lim n a n = a and if a n 0 for all n, then a 0. Hint: Prove by contradiction: assume a < 0 at the beginning. 4. Show that, if lim n a n = a and if a > 0, then a n is eventually positive in the sense that there is a positive integer N such that a n > 0 for all n N. 5. Show that if lim n a n = a, then lim n 1 2 (a n + a n+1 ) = a. Also show that the converse is false. Hint: If α n is decreasing to zero, then so is α n+1. To show the falsehood of the converse, let a n = ( 1) n. 6. Show that if a n 0 for all n and if lim n a 2 n = a 2 with a > 0, then lim n a n = a. (In other words, if lim n b n = b with b n 0 and b > 0, then lim n bn = b.) 7. Prove Proposition Consider the sequence of functions f n defined on [0, 1] by f n (x) = x n. Show that {f n } is pointwise convergent but not uniformly convergent. 9. (a) Show that the series n=1 1 n(n + 1) (b) Find the sum of the series n=1 Hint: For part (a), write converges to 1. 1 n(n + 1)(n + 2). 1 n(n + 1) = 1 n 1 n + 1 and telescope. 10. Show that if n=1 a n and n=1 b n are convergent, then so is the series n=1 (λ a n+ µb n ), where λ and µ are arbitrary constants. Does the convergence of n=1 a n and n=1 b n imply the convergence of n=1 a nb n? 11. Show that, if {α n } is a sequence of positive numbers decreasing to zero, then so is {β n }, where β n = α 1 + α α n. n Use this fact to show the following implication lim a n = a lim n n a 1 + a a n n = a. Also give an example to show that the reverse implication is not true. Remark: We say that a series n=1 a n is (C,1)-summable (here letter C stands for Cesàro) if the limit lim n (s 1 + s s n )/n exists, where s n stands for the 11

12 n-th partial sum of this series: s n = a a n. The (C,1)-method is well-known for its use in summing Fourier series. 12. Show that if {a n } is decreasing and if lim n (a 1 + a a n )/n = a, then lim n a n = a. (That {a n } is decreasing means that a n a n+1 for all n). Remark: Assertion of this type is called a Tauberian theorem for (C,1)-method. The monotonicity of {a n } here is too strong a condition for any practical value. However, is suggests a weaker condition called slowly monotonic. 13. We say that a sequence {a n } is of bounded variation if there exists some M > 0 such that n k=1 a k a k+1 M for all n. Show that a sequence of bounded variation is bounded but the converse is not true; (see the definition of bounded sequences in Exercise 1 above). Also show that a sequence {a n } is of bounded variation if and only if there are bounded increasing sequences {b n } and {c n } such that a n = b n c n for all n; (that {b n } is increasing means b n b n+1 for all n). 14. Show that p n = 2 1/n (> 1) converges to 1 as n by telescoping: 2 = p n n = (p n n p n 1 n ) + (pn n 1 pn n 2 ) + + (p 2 n p n ) + (p n 1) + 1 (p n 1) + + (p n 1) + 1 n(p n 1). 15. (a) Prove by the method described below that, if 0 < a < 1, then lim n na n = 0. First, write x n = na n and r n = x n /x n 1. Show that lim n r n = a. Take any b with a < b < 1. Then there exists some positive integer N such that r n < b for all n N. Let n > N and write down the telescoping product na n = x n = x n 1 r n = x n 2 r n 1 r n = = x 1 r 2 r 3 r n x 1 r 2 r N b n N 0 as n. (b) State a generalization for which the same argument works. 16. (a) Fix a positive number b and write s k = 1 + b + + b k 1 = (1 b k )/(1 b). Compute the sum n k=1 s k in two different ways to deduce the identity n + (n 1)b + + 2b n 2 + b n 1 = (n (n + 1)b + b n+1 )/(1 b) 2. (b) Letting b = 1/a, multiply both sides of the above identity by a n 1 to get a closed expression for the sum n k=1 kak 1. Then let n to deduce that, for a with a < 1, the series k=0 kak 1 converges to (1 a) 2. 12