Chapter 5 Limits for sequences Introduction Up until this point, we have mostly used limits without defining exactly what they are1. This approach matches the historical development. Indeed, scientist working in the 18th century, and earlier, such as Archimedes, Oresme, Men- goli, Euler and even Newton, worked with limits without ever giving a proper definition for what this meant. In this chapter we follow the historical devel- opment into the 19th century when we define pre- Fig.1. The definition of the limit was cisely what we mean by the limit of a sequence, actually first widely introduced into and use it to develop a proper theory that puts mainstream mathematics to more ef- our intuition on a firm mathematical ground. ficiently teach French soldiers calcu- 1Well, this is not really true, in Chapter 4.2, we tried lus in the early 1800s. to trick you into taking the definition of the limit out for aspin–seeexercises4.27,4.32,4.36and4.42 Remark 5.1 (Selected problems from previous exams based on this chapter) 1. Give the definition for limk ak =0, and use this definition to show that !1 sin k lim =0. k p !1 k 2. Give the definition for limk ak = , and use this definition to show that !1 1 n! lim = . k 20n 1 !1 209 210 CHAPTER 5. LIMITS FOR SEQUENCES 5.1 The definition for the limit of a sequence Step 1 of 2: What it means for positive sequences to converge to zero We begin by formulating the following special case of the definition of the limit in order to capture the main idea of what is going on in this chapter, and perhaps the main idea of what is going on in these lecture notes. Definition 5.2 The positive sequence (an)n1=1 is said to converge to 0 if: For every ✏>0,thereexistsN>0 so that for all n N it holds that 2 n>N = an <✏. If this holds, we write ) lim an =0 or an 0 as n . n !1 ! !1 If (a ) does not converge to 0, then we write a 0. n n1=1 n 6! The Czech mathematician Bernard Bolzano was the first to properly state this rather abstract definition of what we ought to mean by a limit. However, living in the mathematical backwater of Prague, it took around 50 years for anyone to notice (which also gave others – such as the French – the time to discover it independently). When you read the example below, notice how Definition 5.2 can be thought of as a game of Fig. 2. Bernard Bolzano (1781 – "challenges and responses". 1848). Never afraid of a challenge. Example 5.3 We prove that 1 lim =0. n n !1 2 To do this, we need to show that the conditions of Definition 5.2 are satisfied when n an =1/2 . That is, we must prove that given any ✏>0, we can find a number N,so that 1 n>N = <✏. ) 2n (The hard part is usually to understand that this is what the definition asks of us.) While this proof can be done in one line, let us take our time to figure out what is going on. We begin by taking a specific epsilon, say ✏ =1/5. Since 21 =2, 22 =4, 23 =8, 24 = 16, 5.1. THE DEFINITION FOR THE LIMIT OF A SEQUENCE 211 it seems that if n>2, then we have 1/2n <✏. But what if ✏ is something else, such as ✏ =1/10? Well, in this case, by the same computation, n>3 implies 1/2n <✏. Fig.3. In the language of the definition, for ✏ =1/5,wechooseN =2, and for ✏ =1/10,wechooseN =3. Notice how we can think of being challenged with an ✏ and then having to respond by giving a suitable N. The point is this: to be able to say that a 0, we need to show that for all "✏- n ! challenges" (in the above figures, the challenge is represented by the height of the green line), we need to be able to respond by saying how far to the right do we have to go before every an is smaller than ✏. That is, we have to figure out how large N has to be so that an <✏whenever n>N. But how to do this for every ✏>0? Let us make two observations: It is hopeless to work with specific ✏, like we did above (that was just for fun). • However, to simplify life, we like to pretend that we can. Indeed, we usually assume that ✏ is a fixed, but unknown, number given to us by some person we respect, but are a little afraid of (like, say, my grandmother). This is supposed to remind us that we can use ✏ in our formulas – as if it was a concrete number – but that we are not allowed to change ✏ or assuming anything about it (except that it is strictly positive), since that would make grandmother angry. We cannot expect a specific N to work for every ✏. Indeed, the typical situation is • that the smaller ✏ becomes, the larger N has to be. In other words, N is going to be a function of ✏ (sometimes we even write N(✏) to emphasise this). So, we assume that we have our concrete, but unknown, challenge ✏.Ourgoalisthe same as for the concrete values we considered above: to respond with an N.Todothis, we need to understand how large n needs to be for the inequality 1/2n <✏to hold. That is, we need to solve this inequality for n. This is exactly the type of thing we did in Chapter 0: 1 1 <✏ 2n > 2n () ✏ 1 log 2n > log () ✏ 212 CHAPTER 5. LIMITS FOR SEQUENCES n log 2 > log ✏ () − log ✏ n> . () −log 2 That is, given ✏, a suitable choice for N is log(✏)/ log(2). Since this works for every − n ✏>0, we have proved that by Definition 5.2 it holds that limn (1/2 )=0. !1 After reading the above example, you may be excused for thinking that the formula we found for N (as a function of ✏) is rather ugly and uninformative. But nothing could be farther from the truth. Indeed, let us see what happens for ✏ =1/5 and ✏ =1/10. This gives us: log ✏ n N ✏ =1/5= n =2,32... =2 n>2 ) ≥−log 2 ) log ✏ n N ✏ =1/10 = n =3,32... =2 n>3. ) ≥−log 2 ) But this is exactly what we got when we did this "by hand"! Moreover, the formula will tell us exactly which N we are supposed to choose no matter the ✏. Sort of nice, no? Exercise 5.4 Answer the following questions on the above example. (a) Use the formula to find suitable N when ✏ =1/1000 and 1/10000. (b) Use a while-loop to write a program in Python that for any given ✏ produces the first n so that an <✏. Verify that it matches what you found in (a). (c) As we are challenged with smaller and smaller ✏, does N become bigger or smaller? Does this match the formula for ✏ from the above example? Can you explain this graphically? (d) We seem to be focused on small ✏. What is a valid choice for N for large ✏? Say, ✏ = 100? Try to justify your answer using both your Python program, the inequality obtained above, and by thinking graphically. Exercise 5.5 (a) Use the definition of the limit (as in Example 5.3) to prove 5 lim =0. n !1 2n +3 (b) Make a program using a while-loop in Python to verify that your formula for N in part (a) makes sense. Exercise 5.6 Here is an exercise that at first may seem awkward. Prove that 1 lim =0. n !1 n! 5.1. THE DEFINITION FOR THE LIMIT OF A SEQUENCE 213 In the next example, now consider a sequence that has no limit. Example 5.7 We show that the sequence a =1+( 1)n does not converge to zero. n − This formula may look a bit strange, but if we check the first few numbers, we notice that (an)n1=1 =(0, 2, 0, 2, 0, 2, 0,...). Intuitively, it should be quite clear that the sequence does not converge to zero. But how to prove this? Well, suppose that we are challenged with ✏ =1. Then the point is that no matter how large we choose N,we can always find a number n>Nso that an >✏. For instance, this holds if we choose n =2N. Therefore, we have found one value of ✏ for which the definition cannot hold, and so it is impossible that an 0. (Notice, we ! Fig. 4. Here, ✏ =1. How far do we have have not proven that this sequence diverges to move to the right so that the a <✏? – which it does – but only that it does not n converge to 0.) Exercise 5.8 We are given two sequences an (red) and bn (blue) as in Figure 5. (a) For what value of N does it seem that n>N = a + b <✏for the indicated ) n n epsilons? (b) Suppose that both an and bn converge to 0. Also, suppose that we are challenged with some unknown, but fixed, ✏>0.Explain why there must exist some number N so that an + bn <✏whenever n>N.Try to be as concrete as possible so that your explanation counts as a proof. Exercise 5.9 Here is a warm up for the next section: (a) Give a definition for limn an =0when !1 all an are negative.