Sequences of Real Numbers

Chapter 8 Sequences of Real Numbers In this chapter, we assume the existence of the ordered field of real numbers, though we do not yet discuss or use the completeness of the real numbers. In the next chapter, we will discuss how the real number system is constructed from the rationals. The reader who wishes to not use the real number system until it has been rigorously defined can essentially replace every occurrence of the word real in this chapter with rational. After all, we have already discussed the ordered field properties for Q, and thus all the properties of the real number system used in this chapter are also properties of the rational number system. 8. Limit of a Sequence You probably first studied sequences in second-semester calculus as a prelude to learning about series of numbers, power series of functions, and Taylor s Theorem. The intuitive idea is that the sequence a = {a n } has limit L if the terms of a are as close as we please to L for n sufficiently large. If a has a limit, we say it is convergent. Otherwise we say a is divergent. We want to turn this idea into a precise definition that makes explicit what condition must be satisfied by a convergent sequence. Before we give this precise definition, we pause to look at a few examples and to develop some tools for exploring sequences. 8.. Numerical and graphical exploration In this section, we explore two sequences, a and b, given by n-th term formulas a n = ( )n n 2n +, n 0 b n = 3n 2n +, n 0.

2 CHAPTER 8. SEQUENCES OF REAL NUMBERS One way to explore a sequence is to generate some of its term. At this point, it is a triviality for us to write some code to do this. For sequence a, we might write: MAX_ELEMENTS = 20 #This is the number of terms we ll generate a = [] #start with an empty list for n in range(0,max_elements): a.append(((-.0)**n)*n/(2.0*n+.0)) print(a) Run this code for yourself to generate all 20 terms. The last 4 (rounded) are a[6] = 0.4848485, a[8] = 0.4864865, a[7] = 0.485743, a[9] = 0.487795. The terms with even index appear to be approaching 0.5 and those with odd index appear to be approaching 0.5. Because there does not appear to be a single number that the terms approach, it appears that this sequence does not have a limit. We can look at the same information graphically. To plot this sequence, we simply add two lines to our code: plot(a,".") #the "." tells the program to plot discrete points show() Here s the graph.

0.6 0.4 0.2 0.0 0.2 0.4 0.6 0 5 0 5 20

4 CHAPTER 8. SEQUENCES OF REAL NUMBERS On the other hand, the last four terms of b (rounded) are, b[6] =.4545455, b[7] =.457429, b[8] =.4594595, b[9] =.465385, with graph:

.6.4.2.0 0.8 0.6 0.4 0.2 0.0 0 5 0 5 20

6 CHAPTER 8. SEQUENCES OF REAL NUMBERS For sequence b, the numerical and graphical evidence suggests that the terms will be as close to.5 as we like for n sufficiently large. Exercise 8... Write your own code to generate the first 20 terms of the sequence b and the graph. We move towards a more quantitative approach; let us figure out how large n must be in order for b[n] to be within 00 of 3 2. That is, we want 3n 2n + 3 2 < 00. Because we must find n so that Thus we require 3n 2n + 3 2 = = 3 4n + 2 < 00. 6n 3(2n + ) 2(2n + ) 3 4n + 2, n > 288 4 = 72. To summarize, we have shown that the terms of b will be within 00 of the (suspected) limit 3 2 if n is at least 73. We could do this calculation for errors ε other than 00. If for any ε > 0 we can find an N so that, for all n N, b[n] 3 < ε, we will have shown that the sequence indeed has limit 3 2 Exercise 8..2. For sequence b above, take ε = b[n] 3 2 < ε if n N. Exercise 8..3. Consider the sequence with a n = 2 n. 500 2. and find an N such that. This sequence is decreasing. Write the one-line proof of this fact. 2. It is not hard to guess that this sequence has limit 0. For each of ε = 5, 50, 500, find a value of N such that if n N, 2 n 0 < ε. 8..2 The precise definition of a limit We have thus motivated the version of the definition of limit we will use. Definition 8... The sequence {a n } has limit L if, for every ε > 0, there exists N N such that for all n N, a n L < ε. Example 8... Let us finish the example from the previous section by using the definition to prove that lim n 3n 2n + = 3 2.

8.. LIMIT OF A SEQUENCE 7 Proof. Let ε > 0 be given. We must show that there exists N such that, if n N, 3n 2n + 3 2 < ε. Observe, 3n 2n + 3 3 2 = 4n + 2 < 3 4n. This will be less than ε if n > 3 4ε. We may thus take N to be the smallest natural number larger than 3 4ε. We make an important remark about this proof: The definition of a limit does not require us to find the smallest N that works for a given ε - it only requires us to show that some N exists. Thus although we may get a smaller N by solving the inequality 3 4n + 2 < ε exactly (as we did above in the special case of ε = 00 ), we chose instead to find the smallest n for which the larger expression 3 is less than ε. This idea is 4n used very often in analysis proofs and is thus worth understanding thoroughly. Exercise 8..4. Consider a n = 2n 2 n 2 +4. Note that a n 0 for all ngeq. (a) Find a simple function b n such that 0 a n b n for all n. (b) Now take ε > 0. Find an N such that, if n N, 0 a n < ε. The idea, as discussed above, is to find an N so that, if n N, the larger function b n is less than ε. We consider another example. Example 8..2. We claim lim n n n2 + =. Proof. Let ε > 0 be given. We must show that there exists N such that, if n N, n n2 + < ε.

8 CHAPTER 8. SEQUENCES OF REAL NUMBERS Suppose n. Observe, n n2 + = < = n n 2 + n2 + n n 2 + n + n 2 + n 2 n + n 2 + n(n + n 2 + ) < n 2, where the last inequality comes from replacing n 2 + in the denominator with the smaller quantity 0. The last expression is less than ε if n 2 > ε n > ε. We may thus take N to be any natural number larger than ε. Again, we summarize the logic of the argument: We find an N such that, if n N, the larger expression n is less than ε. Thus the smaller expression 2 n n 2 + is also less than ε. Exercise 8..5. Use the definition of limit to prove 8.2 Properties 2 n lim n n + =. In a calculus course, you probably learned to do limit calculations algebraically. n 3n Example 8.2.. Let s find lim 2 n 2n 2 +2. The idea is to multiply by a convenient form of : n 3n 2 lim n 2n 2 + 2 n 3n 2 = lim n 2n 2 + 2 = lim n n 3 2 + 2 n = 0 3 2 + 0 = 3 2. This calculation uses several propositions that we have not yet proved. Most importantly, it uses the fact that the limit of the quotient is the quotient of the limits if both limits exist and the limit of the denominator is not zero. The goal of this section is to prove such results so that we can do limit calculations more easily. n 2 n 2

8.2. PROPERTIES 9 Exercise 8.2.. The example also used the fact that, if A is any constant, A lim n n = 0. Prove this. Note that A could be zero. See if you can write a proof that does not treat the case A = 0 separately. Our first proposition will be used in the proofs of some of the limit laws. It is also useful in its own right, for it gives a necessary condition for convergence of a sequence. Proposition 8.2.. Recall that a sequence {a n } is bounded if there exists M such that a n M for all n. If {a} is convergent, then {a} is bounded. Proof. The idea of the argument is simple; because any finite set is bounded, if we find a bound for the tail of the sequence, we can find a bound for the entire sequence. Thus suppose L is the limit of the sequence. Then for ε =, there exists N such that for all n N, a n L <. Thus by the reverse triangle inequality, for all n N, a n L +. Let M = max{ a 0,..., a N, L + }. For all n, a n M. Exercise 8.2.2. Draw a figure illustrating the idea of this proof. Exercise 8.2.3. Use this proposition to prove that the sequence a n = n 2 is divergent. Proposition 8.2.2. Let {a n } and {b n } be convergent sequences with lim n a n = L and lim n b n = K. Let A be a real constant. Then (a) The sequence {Aa n } is convergent with lim Aa n = AL. n (b) The sequence {a n + b n } is convergent with lim (a n + b n ) = L + K. n (c) The sequence {a n b n } is convergent with (d) If K 0 the sequence lim a nb n = LK. n { } a n bn is convergent with a n lim = L n b n K.

0 CHAPTER 8. SEQUENCES OF REAL NUMBERS Proof. We leave parts (a) and (d) as exercises and prove (b) and (c). Proof of (b). Let ε > 0 be given. Because lim n a n = L, for the positive number ε 2, there exists N such that, if n N, a n L < ε 2. Similarly, there exists N 2 such that, if n N 2, b n K < ε 2. Thus if n N = max{n, N 2 }, by the triangle inequality (a n + b n ) (L + K) = (a n L) + (b n K) a n L + b n K < ε 2 + ε 2 = ε. Thus {a n + b n } converges, with limit L + K. Proof of (c). Let ε > 0 be given. We mush show that there exists N such that for all n N, a n b n LK < ε. (8.2.) We don t know how to relate a n b n directly to LK, but we do know something about the relationship between a n and L. This motivates our next step, i.e., adding and subtracting Lb n inside the absolute value sign. Observe, a n b n LK = a n b n + ( Lb n + Lb n ) LK = (a n L)b n + L(b n K) a n L b n + L b n K. As in the proof of (b), we d like to use the convergence of {a n } to L to argue that the first term is less than ε 2 for sufficiently large n, and similarly for the second term. Indeed, note first that because {b n } is convergent, by Proposition 8.2., there exists M > 0 such that b n M for all n. Now, for the positive number ε 2M there exists N such that, if n N, a n L < ε 2M. Similarly, for the ε positive number 2( L +) there exists N ε 2 such that, if n N 2, b n K < 2( L +). Thus if n N = max{n, N 2 }, a n L b n + L b n K < ε 2M M + L ε 2( L + ) < ε 2 + ε 2 = ε. Because, for n N, a n b n LK < ε, the sequence {a n b n } is convergent with limit LK. Arguments like the above are quite standard in analysis and thus worth understanding thoroughly. The next exercises are thus particularly important. Exercise 8.2.4. Prove (a). See if you can write a proof that does not treat the case A = 0 separately. Exercise 8.2.5. Prove (d).

8.3. CAUCHY SEQUENCES 8.3 Cauchy Sequences We now know what it means for a sequence {a n } to have limit L. Suppose now we are given a sequence that we suspect has no limit. What would we have to prove? We would need to show that for every possible limit L, there exists ε > 0 such that for all N N there exists n N such that a n L ε. Consider, for example, the sequence whose n-th term is a n = ( ) n. This sequence seems not to have a limit because all terms with odd index are equal to and all terms with even index are equal to. Writing a proof using the negation of the definition of a limit is, however, a bit of a pain. We must begin with an arbitrary L. Suppose first that L. Thus L > 0. Take ε = L /2, and N N. For n = 2N, a n =, and so a n L = L > L 2 = ε, proving that L is not the limit. Next consider L =. Take ε = and N N. Then if n = 2N +, a n = and a n L = = 2 > = ε. Thus L = is not the limit. We conclude that {a n } does not have a limit. Although the above is a rather simple example of a divergent sequence, the proof was somewhat awkward. We would like a strong necessary condition for convergence of sequences that gives cleaner proofs that a given sequence is divergent. The condition is the Cauchy condition. Definition 8.3.. The sequence {a n } is a Cauchy sequence if, given ε > 0, there exists N N such that for all m, n N, a m a n < ε. Thus a sequence is Cauchy if, given any error ε, if we go far enough out, any two terms are within ε of one another. The divergent sequence above certainly did not have this property and, indeed, we have found a necessary condition for convergence. Proposition 8.3.. If the sequence {a n } is convergent, then the sequence is Cauchy. Proof. Let ε > 0 be given. We must show that there exists N N such that for all m, n N, a n a m < ε. By hypothesis, {a n } has a limit. Call it L. Thus for the positive number ε 2, there exists M such that for all n M, a n L < ε. We claim we may take N = M. Indeed, for all m, n M, a n a m = (a n L) + (L a m ) Thus {a n } is Cauchy. a n L + L a m triangle inequality < ε 2 + ε 2 = ε.

2 CHAPTER 8. SEQUENCES OF REAL NUMBERS 8.3. Showing that a sequence is Cauchy At this point, you may wonder why we would want to show that a sequence is Cauchy; we know that it is a necessary condition for convergence, but we have not determined whether it is sufficient. Thus proving that a sequence is in fact Cauchy only tells us that it has cleared this first hurdle. In the next chapter we will see that, in fact, within the real number system, the Cauchy condition is in fact sufficient for convergence. It is of great theoretical importance because it gives a characterization of convergent sequences of real numbers. It is, however, also of practical significance, for it will allow us to determine whether a sequence has a limit even if we have no idea what that limit might be. This situation is much more common than one might guess from looking at exercises in calculus and elementary analysis texts. We therefore give an example and some exercises in which we prove directly that a given sequence is Cauchy. Example 8.3.. The sequence given by a n = n 2 is Cauchy. Proof. Let ε > 0 be given. We must show that there exists N N such that, if m, n N, a n a m < ε. Assume without loss of generality that m > n. Then a m a n = m 2 n 2 = n 2 m 2 < n 2. If we take N to be any natural number greater than ε, then for m > n N, a m a n < n 2 N 2 < ε. Exercise 8.3.. Prove directly from the definition that the sequence given by a n = 3( )n is Cauchy. n 8.3.2 Showing that a sequence is divergent We have shown that if {a n } has a limit, then it is Cauchy. The contrapositive of this statement gives an important tool for showing that a sequence is divergent. Even though it is logically equivalent to Proposition 8.3., we state it here for emphasis. Proposition 8.3.2. If {a n } is not Cauchy, then {a n } does not have a limit. In order to use this, we must begin by carefully negating the definition of a Cauchy sequence.

8.3. CAUCHY SEQUENCES 3 Definition 8.3.2. The sequence {a n } is not Cauchy if there exists ε > 0 such that for all N N, there exist m, n N with a m a n ε. Let us revisit the divergent sequence {a n } considered above, with n-th term formula a n = ( ) n. We claim that this sequence is not Cauchy. Indeed, consider ε = and let N be arbitrary. Consider m = 2N + and n = 2N. Then m, n N and a m a n = ( ) 2N+ ( ) 2N = = 2 ε. The sequence is not Cauchy and thus has no limit. Exercise 8.3.2. Prove directly from the definition that the sequence given by a n = n is not Cauchy. Exercise 8.3.3. Prove that the sequence a n = ( )n n 2n+ chapter is divergent. considered earlier in the 8.3.3 Properties of Cauchy sequences We collect here a number of propositions about Cauchy sequences of real numbers that are analogous to propositions about convergent sequences of real numbers. It is not surprising that such analogues exist because, in the real number system, a sequence is convergent if and only if it is Cauchy (see the next chapter). However, we can prove these results directly from the definition of a Cauchy sequence, and we wish to do so because we will use several of these results when we discuss the construction of the real number system from the rational number system. Proposition 8.3.3. If {a n } is Cauchy, then {a n } is bounded. Proof. Because {a n } is Cauchy, for ε =, there exists N such that for all n, m N, a n a m <. In particular, for all n N, a n a N. By the reverse triangle inequality, for all n N, a n a N +. If we set M = max{ a 0,..., a N, a N + }, then for all n, a n M. Proposition 8.3.4. Let {a n } and {b n } be Cauchy and let A be a constant. (a) {Aa n } is Cauchy. (b) {a n + b n } is Cauchy. (c) {a n b n } is Cauchy. (d) Suppose there exists δ > 0 such that a n δ for all n. Then Cauchy. { } a n is

4 CHAPTER 8. SEQUENCES OF REAL NUMBERS Proof. We prove only part (d), leaving the proofs of the other parts to the reader. Let ε > 0 be given. Because {a n } is Cauchy, for the positive number εδ 2, there exists N such that, for all m, n N, a m a n < εδ 2. Then for m, n N, a m a n = a n a m a m a n = a m a n a m a n { Thus } a n is Cauchy. < δ 2 εδ2 = ε. Exercise 8.3.4. Prove parts (a)-(c) of Proposition 8.3.4 8.4 Problems. Determine (with justification) whether each of the following sequences is convergent or divergent. If the sequence is convergent, find its limit. (a) a n = ( )n n 2n 2 + (b) a n = n2 n + 4n 2 (c) a n = n 2 + 2n + n (d) a n =. (Recall that x is the greatest integer less than or equal 3 to x.) 2. Let a be a sequence of real numbers. We say that a tends to and write lim n a n = if given M, there exists N N such that for all n N, a n > M. (a) Formulate an analogous definition of a sequence tending to. (b) Prove or give a counterexample: If {a n } is not bounded, then either lim n a n = or lim n a n =. (c) Let p be a non-constant polynomial. Define a sequence {p(n)} by evaluating the polynomial at each non-negative integer. What can you say about lim n p(n)? 3. Let r be a real number. Consider the geometric sequence g = {r n }. We wish to explore how the convergence or divergence of the sequence depends on the value of r.

8.4. PROBLEMS 5 (a) For which (if any) values of r is the sequence g bounded? (b) For which (if any) values of r is g convergent? For such r, what is the limit? (c) For which (if any) values of r is lim n r n =? For which (if any) values of r is lim n r n =? (See the previous problem for definitions of these notations.) 4. Let a and b be sequences of real numbers. Define a sequence c by setting c n = a n b n. We proved that if a and b are both convergent, then so it c. In this problem, you will explore the convergence or divergence of c under other hypotheses on the sequences a and b. (a) Suppose lim n a n = 0. Must c be convergent? (Either prove that it must be or give a counter example.) (b) Suppose lim n a n = 0. Could c be convergent even if b is unbounded? Could c be convergent with limit unequal to zero? (c) Could c be convergent even if both a and b are divergent? 5. Part (d) { of } Proposition 8.2.2 implies that if lim n a n = L and L 0, then a n is convergent with limit L. Consider the converse. In other { words, if is convergent, must {a n } be? a n } Programming Project: Ratios of Fibonacci Numbers. Consider the Fibonacci sequence {F n } defined recursively by F 0 = 0, F =, and F n = F n + F n 2 for all n 2. (8.4.). Write code to generate the first 20 terms of the sequence. 2. Let r n = Fn F n for n. Write code to generate r,..., r 9. Plot this sequence {r n } of ratios. 3. Does the graph suggest that {r n } is monotone (non-increasing or nondecreasing)? Eventually monotone? Convergent? With what limit? 4. In the next chapter we will show that {r n } is convergent. Assume this, and find lim n r n. (Suggestion: Obtain an equation relating r n and r n.) 5. In Chapter 3 we discussed linear recurrence relations such as the above. In Exercise??, you were asked to solve the recurrence defining the Fibonacci sequence. The answer is known as Binet s formula and says F n = ( + 5 2 ) n + ( Use this explicit formula to find lim n r n. 5 2 ) n. (8.4.2)